This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A000217 Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n. (Formerly M2535 N1002) 3595

%I M2535 N1002

%S 0,1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190,210,

%T 231,253,276,300,325,351,378,406,435,465,496,528,561,595,630,666,703,

%U 741,780,820,861,903,946,990,1035,1081,1128,1176,1225,1275,1326,1378,1431

%N Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n.

%C Also referred to as T(n) but C(n+1,2) or binomial(n+1,2) are preferred (the latter is favored slightly over the former).

%C Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - _Omar E. Pol_, Sep 13 2011 and Aug 04 2012

%C Number of edges in complete graph of order n, K_n.

%C Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n+1 of them are illegal because the parentheses are adjacent. Cf. A002415.

%C For n >= 1, a(n) = n*(n+1)/2 is also the genus of a nonsingular curve of degree n+2, like the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001

%C From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n). - _Benoit Cloitre_, Aug 29 2002

%C Number of tiles in the set of double-n dominoes. - Scott A. Brown (scottbrown(AT)neo.rr.com), Sep 24 2002

%C Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - _James A. Raymond_, Apr 08 2003

%C Maximum number of intersections of n+1 lines which may only have 2 lines per intersection point. Maximal number of closed regions when n+1 lines are maximally 2-intersected in given by a(n-1). Using n+1 lines with k>1 parallel lines, the maximum number of 2-intersections is given by a(n) - a(k-1). - _Jon Perry_, Jun 11 2003

%C Number of distinct straight lines that can pass through n points in 3-dimensional space. - _Cino Hilliard_, Aug 12 2003

%C Triangular numbers - odd numbers = shifted triangular numbers; 1,3,6,10,15,21,... - 1,3,5,7,9,11,... = 0,0,1,3,6,10,... - Xavier Acloque, Oct 31 2003 [Corrected by _Derek Orr_, May 05 2015]

%C Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003

%C Maximum number of lines formed by the intersection of n+1 planes. - _Ron R. King_, Mar 29 2004

%C Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - _Mike Zabrocki_, Aug 26 2004

%C Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - _Olivier Gérard_, Aug 28 2012

%C a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. - _Jon Perry_, Dec 16 2004

%C Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.

%C Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005

%C A110560/A110561 = numerator/denominator of the coefficients of the exponential generating function. - _Jonathan Vos Post_, Jul 27 2005

%C Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - _Philippe Deléham_, Aug 02 2005

%C Each pair of neighboring terms adds to a perfect square. - _Zak Seidov_, Mar 21 2006

%C Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - _Geoffrey Critzer_, Jun 23 2006

%C With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n>=1, rho(n)^a(n)=(-1)^(n+1). Just use the triviality a(2*k+1) = 0 mod (2*k+1) and a(2*k) = k mod (2*k).

%C a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^n. - _Sergio Falcon_, Feb 12 2007

%C a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - _Richard Barnes_, Sep 06 2017

%C The number of distinct handshakes in a room with n+1 people. - _Mohammad K. Azarian_, Apr 12 2007 [corrected, _Joerg Arndt_, Jan 18 2016]

%C Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - _James East_, May 03 2007

%C a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007

%C For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - _Amarnath Murthy_, Apr 22 2001 (edited by _Robert A. Beeler_)

%C a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. _Wolfdieter Lang_, Jun 29 2007

%C From _Hieronymus Fischer_, Aug 06 2007: (Start)

%C Numbers m>=0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.

%C Numbers m>=0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).

%C Numbers m>=0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x>=0). (End)

%C If Y and Z are 3-blocks of an n-set X, then, for n>=6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - _Milan Janjic_, Nov 09 2007

%C Equals row sums of triangle A143320, n>0. - _Gary W. Adamson_, Aug 07 2008

%C a(n) is also a perfect number A000396 if n is a Mersenne prime A000668, assuming there are no odd perfect numbers. - _Omar E. Pol_, Sep 05 2008

%C Equals row sums of triangle A152204. - _Gary W. Adamson_, Nov 29 2008

%C The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008

%C -a(n+1) = E(2)*binomial(n+2,2) (n>=0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - _Peter Luschny_, Jan 06 2009

%C Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009

%C The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - _Peter Luschny_, Jul 12 2009

%C a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - _Franklin T. Adams-Watters_, Aug 06 2009

%C a(A006894(n)) = a(A072638(n-1)+1) = A072638(n) = A006894(n+1)-1 for n >= 1. For n=4, a(11) = 66. - _Jaroslav Krizek_, Sep 12 2009

%C Partial sums of A001477. - _Juri-Stepan Gerasimov_, Jan 25 2010. [A-number corrected by _Omar E. Pol_, Jun 05 2012]

%C The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - _Paul Muljadi_, Jan 25 2010

%C From _Charlie Marion_, Dec 03 2010: (Start)

%C More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and

%C a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).

%C Column sums of:

%C 1 3 5 7 9...

%C 1 3 5...

%C 1...

%C ..............

%C --------------

%C 1 3 6 10 15...

%C Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).

%C (End)

%C A004201(a(n)) = A000290(n); A004202(a(n)) = A002378(n). - _Reinhard Zumkeller_, Feb 12 2011

%C 1/a(n+1), n>=0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the _Stephen Crowley_ formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/A196839. - _Wolfdieter Lang_, Oct 26 2011

%C From _Charlie Marion_, Feb 23 2012: (Start)

%C a(n) + a(A002315(k)*n + A001108(k+1)) = (A001653(k+1)*n + A001109(k+1))^2. For k=0 we obtain a(n) + a(n+1) = (n+1)^2 (identity added by _N. J. A. Sloane_ on Feb 19 2004).

%C a(n) + a(A002315(k)*n - A055997(k+1)) = (A001653(k+1)*n - A001109(k))^2.

%C (End)

%C Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - _J. M. Bergot_, May 04 2012

%C The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - _J. M. Bergot_, May 17 2012

%C (a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - _J. M. Bergot_, May 18 2012

%C a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - _J. M. Bergot_, May 22 2012

%C In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - _Charlie Marion_, Sep 11 2012

%C a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - _J. M. Bergot_, May 22 2012

%C In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - _Charlie Marion_, Sep 11 2012

%C a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - _J. M. Bergot_, May 22 2012

%C Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - _J. M. Bergot_, May 23 2012

%C a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - _Charlie Marion_, Oct 02 2012

%C In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - _Douglas Latimer_, Dec 17 2012

%C From _James East_, Jan 08 2013: (Start)

%C For n>=1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].

%C For n>=3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].

%C (End)

%C For n>=3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - _James East_, Jan 15 2013

%C Conjecture: For n>0, there is always a prime between A000217(n) and A000217(n+1). Sequence A065383 has the first 1000 of these primes. - _Ivan N. Ianakiev_, Mar 11 2013

%C The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - _Charlie Marion_, Mar 28 2013

%C The series Sum_{k>=1} 1/a(k) = 2, given in a formula below by _Jon Perry_, Jul 13 2003, has partial sums 2*n/(n+1) (telescopic sum) = A022998(n)/A026741(n+1). - _Wolfdieter Lang_, Apr 09 2013

%C For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - _Lekraj Beedassy_, May 29 2013

%C Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - _Peter Bala_, Jan 05 2015

%C Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n>1. - _J. M. Bergot_, Jul 24 2013

%C a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - _J. M. Bergot_, Aug 13 2013

%C Dimension of orthogonal group O(n+1). - _Eric M. Schmidt_, Sep 08 2013

%C Number of positive roots in the root system of type A_n (for n>0). - _Tom Edgar_, Nov 05 2013

%C A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - _Gary Detlefs_, Jan 02 2014

%C Also the alternating row sums of A095831. Also the alternating row sums of A055461, for n >= 1. - _Omar E. Pol_, Jan 26 2014

%C For n>=3, a(n-2) is the number of permutations of 1,2...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - _Vladimir Shevelev_, Feb 14 2014

%C a(n) is the dimension of the vector space of symmetric n X n matrices. - _Derek Orr_, Mar 29 2014

%C Non-vanishing subdiagonal of A132440^2/2. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - _Tom Copeland_, Apr 05 2014

%C The number of Sidon subsets of {1,...,n+1} of size 2. - _Carl Najafi_, Apr 27 2014

%C Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - _Tom Copeland_, Apr 27 2014

%C Number of weak compositions of n into three parts. - _Robert A. Beeler_, May 20 2014

%C A228474(a(n))=n; A248952(a(n))=0; A248953(a(n))=a(n); A248961(a(n))=A000330(n). - _Reinhard Zumkeller_, Oct 20 2014

%C Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n>0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k>2, let b(0) = 0, b(1) = 1 and, for n>1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n>0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - _Charlie Marion_, Nov 03 2014

%C Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - _Charlie Marion_, Feb 21 2015

%C Consider the partition of the natural numbers into parts from the set S=(1,2,3...n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - _David Neil McGrath_, May 05 2015

%C a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc),(bbcc),(bccc),(bbbc). - _David Neil McGrath_, May 21 2015

%C Sum_{k=0..n}k*a(k+1) = a(A000096(n+1)). - _Charlie Marion_, Jul 15 2015

%C Let O(n) be the oblong number n(n+1) = A002378(n) and S(n) the square number n^2 = A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). - _Charlie Marion_, Jul 16 2015

%C Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - _Wolfgang Tintemann_, Aug 02 2015

%C Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t}1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t}1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - _Zhi-Wei Sun_, Sep 09 2015

%C The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - _R. J. Mathar_, Nov 29 2015

%C For n>=1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - _Milan Janjic_, Jan 07 2016

%C In this sequence only 3 is prime. - _Fabian Kopp_, Jan 09 2016

%C Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n>0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - _Charlie Marion_, Jan 14 2016

%C Numbers n such that 8n + 1 is a perfect square. - _Juri-Stepan Gerasimov_, Apr 09 2016

%C Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - _Miquel Cerda_, Jun 26 2016

%C For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - _Melvin Peralta_, Aug 29 2016

%C In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - _Stanislav Sykora_, Oct 23 2016

%C Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - _Charlie Marion_, Nov 27 2016

%C a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - _Vladimir Letsko_, Dec 19 2016

%C For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - _Patrick J. McNab_, Dec 25 2016

%C Does not satisfy Benford's law (cf. Ross, 2012). - _N. J. A. Sloane_, Feb 12 2017

%C Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - _Aviel Livay_, Feb 13 2017

%C Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - _David Nacin_, Feb 22 2017

%C Also the Wiener index of the complete graph K_{n+1}. - _Eric W. Weisstein_, Sep 07 2017

%C Number of intersections between the Bernstein polynomials of degree n. - _Eric Desbiaux_, Apr 01 2018

%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

%D C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See Chapter 1.

%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.

%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 109ff.

%D Marc Chamberland, Single Digits: In Praise of Small Numbers, Chapter 3, The Number Three, p. 72, Princeton University Press, 2015.

%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.

%D L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.

%D Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.

%D James Gleick, The Information: A History, A Theory, A Flood, Pantheon, 2011. [On page 82 mentions a table of the first 19999 triangular numbers published by E. de Joncort in 1762.]

%D Labos E.: On the number of RGB-colors we can distinguish. Partition Spectra. Lecture at 7th Hungarian Conference on Biometry and Biomathematics. Budapest. Jul 06 2005.

%D A. Messiah, Quantum Mechanics, Vol.1, North Holland, Amsterdam, 1965, p. 457.

%D J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%D T. Trotter, Some Identities for the Triangular Numbers, Journal of Recreational Mathematics, Spring 1973, 6(2).

%D D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-93 Penguin Books 1987.

%H N. J. A. Sloane, <a href="/A000217/b000217.txt">Table of n, a(n) for n = 0..30000</a>

%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972.

%H Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>, section 39.7, pp. 776-778

%H Luciano Ancora, <a href="https://upload.wikimedia.org/wikipedia/commons/9/9c/FigurateN.pdf">The Square Pyramidal Number and other figurate numbers</a>, ch. 5.

%H S. Barbero, U. Cerruti, N. Murru, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barbero2/barbero7.html">A Generalization of the Binomial Interpolated Operator and its Action on Linear Recurrent Sequences </a>, J. Int. Seq. 13 (2010) # 10.9.7, proposition 18.

%H Jean-Luc Baril, Sergey Kirgizov, Vincent Vajnovszki, <a href="https://arxiv.org/abs/1803.06706">Descent distribution on Catalan words avoiding a pattern of length at most three</a>, arXiv:1803.06706 [math.CO], 2018.

%H Paul Barry, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Barry/barry84.html">A Catalan Transform and Related Transformations on Integer Sequences</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

%H T. Beldon and T. Gardiner, <a href="http://www.jstor.org/stable/3621134">Triangular numbers and perfect squares</a>, The Mathematical Gazette 86 (2002), 423-431.

%H Michael Boardman, <a href="http://www.jstor.org/stable/3219201">The Egg-Drop Numbers</a>, Mathematics Magazine, 77 (2004), 368-372. [From _Parthasarathy Nambi_, Sep 30 2009]

%H Anicius Manlius Severinus Boethius, <a href="https://archive.org/stream/aniciimanliitor01friegoog#page/n110/mode/2up">De institutione arithmetica libri duo</a>, Book 2, sections 7-9.

%H H. Bottomley, <a href="/A002378/a002378.gif">Illustration of initial terms of A000217, A002378</a>

%H Scott A. Brown, <a href="http://home.neo.rr.com/scottbrown">Brown's Math Page, etc.</a> [Broken link?]

%H P. J. Cameron, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/groups.html">Sequences realized by oligomorphic permutation groups</a>, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.

%H Peter M. Chema, <a href="/A000217/a000217_3.pdf">Illustration of first 25 terms as corners of a double square spiral.</a>

%H Karl Dienger, <a href="/A000217/a000217.pdf">Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung</a>, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]

%H Tomislav Doslic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL8/Doslic/doslic15.html">Maximum Product Over Partitions Into Distinct Parts</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.8.

%H Askar Dzhumadildaev and Damir Yeliussizov, <a href="http://cs.uwaterloo.ca/journals/JIS/VOL16/Yeliussizov/dzhuma6.html">Power Sums of Binomial Coefficients</a>, Journal of Integer Sequences, Vol. 16 (2013), #13.1.1.

%H J. East, <a href="http://dx.doi.org/10.1142/S0218196710005509">Presentations for singular subsemigroups of the partial transformation semigroup</a>, Internat. J. Algebra Comput., 20 (2010), no. 1, 1-25.

%H J. East, <a href="http://dx.doi.org/10.1142/S021819671100611X">On the singular part of the partition monoid</a>, Internat. J. Algebra Comput. 21 (2011), no. 1-2, 147-178.

%H L. Euler, <a href="http://eulerarchive.maa.org/index.html">The Euler archive - E806 D, Miscellanea, Section 87</a>, Opera Postuma Mathematica et Physica, 2 vols., St. Petersburg Academy of Science, 1862.

%H E. T. Frankel, <a href="/A000217/a000217_1.pdf"> A calculus of figurate numbers and finite differences</a>, American Mathematical Monthly, 57 (1950), 14-25. [Annotated scanned copy]

%H Adam Grabowski, <a href="http://dx.doi.org/10.2478/forma-2013-0012">Polygonal Numbers</a>, Formalized Mathematics, Vol. 21, No. 2, Pages 103-113, 2013; DOI: 10.2478/forma-2013-0012; <a href="http://fm.mizar.org/fm21-2/numpoly1.pdf">alternate copy</a>

%H S. S. Gupta, <a href="http://www.shyamsundergupta.com/triangle.htm">Fascinating Triangular Numbers</a>

%H C. Hamberg, <a href="http://staff.imsa.edu/math/journal/volume1/articles/Triangular.pdf">Triangular Numbers Are Everywhere</a>

%H Guo-Niu Han, <a href="http://www-irma.u-strasbg.fr/~guoniu/papers/p77puzzle.pdf">Enumeration of Standard Puzzles</a>

%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a> [Cached copy]

%H A. M. Hinz, S. Klavžar, U. Milutinović, C. Petr, <a href="http://dx.doi.org/10.1007/978-3-0348-0237-6">The Tower of Hanoi - Myths and Maths</a>, Birkhäuser 2013. See page 35. <a href="http://tohbook.info">Book's website</a>

%H J. M. Howie, <a href="http://dx.doi.org/10.1017/S0308210500010647">Idempotent generators in finite full transformation semigroups</a>, Proc. Roy. Soc. Edinburgh Sect. A, 81 (1978), no. 3-4, 317-323.

%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=253">Encyclopedia of Combinatorial Structures 253</a>

%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>

%H Xiangdong Ji, <a href="http://www.physics.umd.edu/courses/Phys741/xji/chap8_12.pdf">Chapter 8: Structure of Finite Nuclei</a>, Lecture notes for Phys 741 at Univ. of Maryland, pp. 139-140 [From _Tom Copeland_, Apr 07 2014].

%H R. Jovanovic, <a href="http://milan.milanovic.org/math/english/triangular/triangular.html">Triangular numbers</a>

%H R. Jovanovic, <a href="http://milan.milanovic.org/math/Math.php?akcija=SviTroug">First 2500 Triangular numbers</a>

%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2003), 65-75.

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary Equations</a>, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.

%H Clark Kimberling and John E. Brown, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL7/Kimberling/kimber67.html">Partial Complements and Transposable Dispersions</a>, J. Integer Seqs., Vol. 7, 2004.

%H J. Koller, <a href="http://www.mathematische-basteleien.de/triangularnumber.htm">Triangular Numbers</a>

%H A. J. F. Leatherland, <a href="http://yoyo.cc.monash.edu.au/~bunyip/primes/triangleUlam.htm">Triangle Numbers on Ulam Spiral</a>

%H Sergey V. Muravyov, Liudmila I. Khudonogova, Ekaterina Y. Emelyanova, <a href="https://doi.org/10.1016/j.measurement.2017.08.045">Interval data fusion with preference aggregation</a>, Measurement (2017), see page 5.

%H A. Nowicki, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Nowicki/nowicki3.html">The numbers a^2+b^2-dc^2</a>, J. Int. Seq. 18 (2015) # 15.2.3

%H Ed Pegg, Jr., <a href="http://www.mathpuzzle.com/MAA/07-Sequence%20Pictures/mathgames_12_08_03.html">Sequence Pictures</a>, Math Games column, Dec 08 2003.

%H Ed Pegg, Jr., <a href="/A000043/a000043_2.pdf">Sequence Pictures</a>, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]

%H Ivars Peterson, <a href="http://www.maa.org/mathtourist/mathtourist_11_7_07.html">Triangular Numbers and Magic Squares</a>.

%H Alexsandar Petojevic, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL5/Petojevic/petojevic5.html">The Function vM_m(s; a; z) and Some Well-Known Sequences</a>, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7

%H Simon Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992.

%H Simon Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">1031 Generating Functions and Conjectures</a>, Université du Québec à Montréal, 1992.

%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polnum01.jpg">Illustration of initial terms of A000217, A000290, A000326, A000384, A000566, A000567</a>

%H David G. Radcliffe, <a href="https://arxiv.org/abs/1606.05398">A product rule for triangular numbers</a>, arXiv:1606.05398 [math.NT], 2016.

%H F. Richman, <a href="http://math.fau.edu/Richman/mla/triangle.htm">Triangle numbers</a>

%H J. Riordan, <a href="/A000217/a000217_2.pdf">Review of Frankel (1950)</a> [Annotated scanned copy]

%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.

%H Kenneth A. Ross, <a href="http://www.jstor.org/stable/10.4169/math.mag.85.1.036">First Digits of Squares and Cubes</a>, Math. Mag. 85 (2012) 36-42. doi:10.4169/math.mag.85.1.36.

%H Frank Ruskey and Jennifer Woodcock, <a href="http://dx.doi.org/10.1007/978-3-642-25011-8_23">The Rand and block distances of pairs of set partitions</a>, in Combinatorial Algorithms, 287-299, Lecture Notes in Comput. Sci., 7056, Springer, Heidelberg, 2011.

%H Sci.math Newsgroup, <a href="http://www.math.niu.edu/~rusin/known-math/98/sq_tri">Square numbers which are triangular</a> [Broken link: <a href="/A000217/a000217_1.txt">Cached copy</a>]

%H James A. Sellers, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL7/Sellers/sellers58.html">Partitions Excluding Specific Polygonal Numbers As Parts</a>, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.4.

%H N. J. A. Sloane, <a href="/A000217/a000217.gif">Illustration of initial terms of A000217, A000290, A000326</a>

%H H. Stamm-Wilbrandt, <a href="https://www.ibm.com/developerworks/community/blogs/HermannSW/entry/sum_of_pascal_s_triangle_reciprocals10">Sum of Pascal's triangle reciprocals</a>

%H T. Trotter, <a href="http://www.trottermath.net/numthry/trident.html">Some Identities for the Triangular Numbers</a>, J. Rec. Math. vol. 6, no. 2 Spring 1973. [Warning: As of March 2018 this site appears to have been hacked. Proceed with great caution. The original content should be retrieved from the Wayback machine and added here. - _N. J. A. Sloane_, Mar 29 2018]

%H G. Villemin's Almanach of Numbers, <a href="http://villemin.gerard.free.fr/Wwwgvmm/Geometri/NbTrianB.htm">Nombres Triangulaires</a>

%H Manuel Vogel, <a href="https://doi.org/10.1007/978-3-319-76264-7_6">Motion of a Single Particle in a Real Penning Trap</a>, Particle Confinement in Penning Traps, Springer Series on Atomic, Optical, and Plasma Physics, Vol. 100. Springer, Cham. 2018, 61-88.

%H Michel Waldschmidt, <a href="http://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/ContinuedFractionsOujda2015.pdf">Continued fractions</a>, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AbsoluteValue.html">Absolute Value</a>, <a href="http://mathworld.wolfram.com/BinomialCoefficient.html">Binomial Coefficient</a>, <a href="http://mathworld.wolfram.com/Composition.html">Composition</a>, <a href="http://mathworld.wolfram.com/Distance.html">Distance</a>, <a href="http://mathworld.wolfram.com/GolombRuler.html">Golomb Ruler</a>, <a href="http://mathworld.wolfram.com/LineLinePicking.html">Line Line Picking</a>, <a href="http://mathworld.wolfram.com/PolygonalNumber.html">Polygonal Number</a>, <a href="http://mathworld.wolfram.com/TriangularNumber.html">Triangular Number</a>, <a href="http://mathworld.wolfram.com/TrinomialCoefficient.html">Trinomial Coefficient</a>, and <a href="http://mathworld.wolfram.com/WienerIndex.html">Wiener Index</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Floyd%27s_triangle">Floyd's triangle</a>

%H <a href="/index/Cor#core">Index entries for "core" sequences</a>

%H <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Pol#polygonal_numbers">Index to sequences related to polygonal numbers</a>

%H <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>

%F G.f.: x/(1-x)^3. - _Simon Plouffe_ in his 1992 dissertation

%F E.g.f.: exp(x)*(x+x^2/2).

%F a(n) = a(-1-n).

%F a(n) + a(n-1)*a(n+1) = a(n)^2. - _Terrel Trotter, Jr._, Apr 08 2002

%F a(n) = (-1)^n*Sum_{k=1..n} (-1)^k*k^2. - _Benoit Cloitre_, Aug 29 2002

%F a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. - _Jon Perry_, Jul 13 2003

%F For n>0, a(n) = A001109(n) - Sum_{k=0...n-1} (2k+1)*A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). - _Charlie Marion_, Jul 18 2003

%F With interpolated zeros, this is n(n+2)*(1+(-1)^n)/16. - _Benoit Cloitre_, Aug 19 2003

%F a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). - _Benoit Cloitre_, Aug 19 2003

%F a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003

%F a(n) = a(n-1) + (1 + sqrt[1 + 8*a(n-1)])/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). - _Carl R. White_, Nov 04 2003

%F a(n) = Sum_{k=1..n} phi(k)*floor(n/k) = Sum_{k=1..n} A000010(k)*A010766(n, k) (R. Dedekind). - _Vladeta Jovovic_, Feb 05 2004

%F a(n) + a(n+1) = (n+1)^2. - _N. J. A. Sloane_, Feb 19 2004

%F a(n) = a(n-2) + 2n - 1. - _Paul Barry_, Jul 17 2004

%F a(n) = sqrt[Sum_{i=1..n}{j=1..n} (i*j)] = sqrt(A000537(n)). - _Alexander Adamchuk_, Oct 24 2004

%F a(n) = sqrt(sqrt(Sum_{i=1..n}{j=1..n} (i*j)^3)) = (Sum_{i=1..n}{j=1..n}{k=1..n} (i*j*k)^3)^(1/6). - _Alexander Adamchuk_, Oct 26 2004

%F a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. - _Miklos Kristof_, Mar 09 2005

%F a(n) = a(n-1) + n. - _Zak Seidov_, Mar 06 2005

%F a(n) = A108299(n+3,4) = -A108299(n+4,5). - _Reinhard Zumkeller_, Jun 01 2005

%F a(n) = A111808(n,2) for n>1. - _Reinhard Zumkeller_, Aug 17 2005

%F a(n)*a(n+1) = A006011(n+1) = (n+1)^2*(n^2+2)/4 = 3*A002415(n+1) = 1/2*a(n^2+2*n). a(n-1)*a(n) = 1/2*a(n^2-1). - _Alexander Adamchuk_, Apr 13 2006 [Corrected and edited by _Charlie Marion_, Nov 26 2010]

%F a(n) = floor((2n+1)^2/8). - _Paul Barry_, May 29 2006

%F For positive n, we have a(8*a(n))/a(n) = 4*(2n+1)^2 = (4n+2)^2, i.e., a(A033996(n))/a(n) = 4*A016754(n) = (A016825(n))^2 = A016826(n). - _Lekraj Beedassy_, Jul 29 2006

%F a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997) p 130]. - _R. J. Mathar_, Nov 22 2006

%F a(n) = A126890(n,0). - _Reinhard Zumkeller_, Dec 30 2006

%F a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by _J. M. Bergot_ dated May 22 2012]. - _Charlie Marion_, Feb 04 2011

%F (sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007

%F a(n) = A023896(n) + A067392(n). - _Lekraj Beedassy_, Mar 02 2007

%F Sum_{k=0..n} a(k)*A039599(n,k) = A002457(n-1), for n>=1. - _Philippe Deléham_, Jun 10 2007

%F 8*a(n)^3 + a(n)^2 = Y(n)^2, where Y(n) = n*(n+1)*(2n+1)/2 = 3*A000330(n). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007 [Edited by _Derek Orr_, May 05 2015]

%F A general formula for polygonal numbers is P(k,n) = (k-2)(n-1)n/2 + n = n + (k-2)*A000217(n-1), for n >= 1, k >= 3. - _Omar E. Pol_, Apr 28 2008 and Mar 31 2013

%F a(3*n) = A081266(n), a(4*n) = A033585(n), a(5*n) = A144312(n), a(6*n) = A144314(n). - _Reinhard Zumkeller_, Sep 17 2008

%F a(n) = A022264(n) - A049450(n). - _Reinhard Zumkeller_, Oct 09 2008

%F If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n>=1. - _Milan Janjic_, Dec 20 2008

%F 4*a(x) + 4*a(y) + 1 = (x+y+1)^2 + (x-y)^2. - _Vladimir Shevelev_, Jan 21 2009

%F a(n) = A000124(n-1) + n-1 for n >= 2. a(n) = A000124(n) - 1. - _Jaroslav Krizek_, Jun 16 2009

%F An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((pochhammer(1, m)*pochhammer(1, m))*x^m/(pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0. A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. - _Stephen Crowley_, Jun 28 2009

%F a(n) = A034856(n+1) - A005408(n) = A005843(n) + A000124(n) - A005408(n). - _Jaroslav Krizek_, Sep 05 2009

%F With offset 1, a(n) = floor(n^3/(n+1))/2. - _Gary Detlefs_, Feb 14 2010

%F a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). - _Bruno Berselli_, May 23 2010

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0,a(1)=1. - _Mark Dols_, Aug 20 2010

%F From _Charlie Marion_, Oct 15 2010: (Start)

%F a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and

%F a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.

%F In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.

%F a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.

%F In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.

%F (End)

%F a(n) = sqrt A000537(n). - _Zak Seidov_, Dec 07 2010

%F For n>0 a(n)=1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). - _Francesco Daddi_, Aug 02 2011

%F a(n) = A110654(n)*A008619(n). - _Reinhard Zumkeller_, Aug 24 2011

%F a(2k-1) = A000384(k), a(2k) = A014105(k), k>0. - _Omar E. Pol_, Sep 13 2011

%F a(n) = A026741(n)*A026741(n+1). - _Charles R Greathouse IV_, Apr 01 2012

%F a(n) + a(a(n)) + 1 = a(a(n)+1). - _J. M. Bergot_, Apr 27 2012

%F a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind, A048994. - _Mircea Merca_, May 03 2012

%F a(n)*a(n+1) = a(Sum_{m=1..n} A005408(m))/2, for n>=1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. - _Ivan N. Ianakiev_, May 27 2012

%F a(n) = A002378(n)/2 = (A001318(n) + A085787(n))/2. - _Omar E. Pol_, Jan 11 2013

%F G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ... , where A(x) = (1 + 3x + 3x^2 + x^3). - _Gary W. Adamson_, Jun 26 2012

%F G.f.: G(0) where G(k)= 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - _Sergei N. Gladkovskii_, Nov 23 2012

%F a(n) = A002088(n) + A063985(n). - _Reinhard Zumkeller_, Jan 21 2013

%F G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Mar 14 2013

%F a(n) + a(n+1) + a(n+2) + a(n+3) + n = a(2n+4). - _Ivan N. Ianakiev_, Mar 16 2013

%F a(n) + a(n+1) + ... + a(n+8) + 6n = a(3n+15). - _Charlie Marion_, Mar 18 2013

%F a(n) + a(n+1) + ... + a(n+20) + 2n^2 + 57n = a(5n+55). - _Charlie Marion_, Mar 18 2013

%F 3*a(n) + a(n-1) = a(2n), for n>0. - _Ivan N. Ianakiev_, Apr 05 2013

%F In general, a(k*n) = (2k-1)*a(n) + a((k-1)*n-1). - _Charlie Marion_, Apr 20 2015

%F Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). - _Robert Israel_, Apr 20 2015

%F a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). - _Mircea Merca_, Apr 06 2013

%F G.f.: x + 3*x^2*G(0)/2, where G(k)= 1 + 1/(1 - x/(x + (k+2)/(k+4)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 01 2013

%F a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). - _Wesley Ivan Hurt_, Jun 15 2013

%F a(n) = floor((n+1)/(exp(2/(n+1))-1)). - _Richard R. Forberg_, Jun 22 2013

%F Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - _Richard R. Forberg_, Jul 15 2013

%F G.f.: W(0)*x/(2-2*x), where W(k) = 1 + 1/( 1 - x*(k+2)/( x*(k+2) + (k+1)/W(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 19 2013

%F a(a(n)) = a(a(n)-1) + a(n), n>=1. - _Vladimir Shevelev_, Jan 21 2014

%F a(n) = A245300(n,0). - _Reinhard Zumkeller_, Jul 17 2014

%F a(n) = (-1)^n*Sum_{i=1..n} (-1)^i*i^2. - _Derek Orr_, Jul 30 2014

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . - _Richard R. Forberg_, Aug 11 2014

%F 2/(Sum_{n>=m} 1/a(n)) = m, for m>0. - _Richard R. Forberg_, Aug 12 2014

%F a(a(n)-1) + a(a(n+2)-1) + 1 = A000124(n+1)^2. - _Charlie Marion_, Nov 04 2014

%F a(n) = 2*A000292(n)-A000330(n). - _Luciano Ancora_, Mar 14 2015

%F a(n) = A007494(n-1) + A099392(n) for n > 0. - _Bui Quang Tuan_, Mar 27 2015

%F A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2k) - 2k. - _Charlie Marion_, Dec 10 2015

%F a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from _N. J. A. Sloane_'s a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). - _Ben Paul Thurston_, Dec 28 2015

%F Dirichlet g.f.: (zeta(s-2) + zeta(s-1))/2. - _Ilya Gutkovskiy_, Jun 26 2016

%F a(n)^2 - a(n-1)^2 = n^3. - _Miquel Cerda_, Jun 29 2016

%F a(n) = A080851(0,n-1). - _R. J. Mathar_, Jul 28 2016

%F a(n) = A000290(n-1) - A034856(n-4). - _Peter M. Chema_, Sep 25 2016

%F a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4n + 10). - _Charlie Marion_, Nov 23 2016

%F 2*a(n)^2 + a(n) = a(n^2+n). - _Charlie Marion_, Nov 29 2016

%F G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3x + 6x^2 + 7x^3 + 6x^4 + 3x^5 + x^6). - _Gary W. Adamson_, Dec 03 2016

%F a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. - _Michael Yukish_, Mar 20 2017

%F a(n) = Sum_{k=1 to n} ((2k-1)!!(2n-2k-1)!!)/((2k-2)!!(2n-2k)!!). - _Michael Yukish_, Mar 20 2017

%F a(n) = A110660(n) + A008794(n+1). - _Olivier Pirson_, Nov 05 2017

%F a(n) = a(n-k) + k*n - a(k-1), generalization of the formulas dated Mar 06 2005 and Jul 17 2004. - _Charlie Marion_, Jan 29 2018

%F For any k >= 0, (2*k + 1)^2*a(n) + k*(k + 1)/2 = a((2*k+1)*n+k). - _Stanislav Sykora_, Mar 14 2018

%F a(n) = 2*a(n-k) - a(n-2k) + k^2. - _Charlie Marion_, Jun 01 2018

%F a(n^2+n-1) + a(n) = 2*a(n)^2. Complements formula dated Nov 29 2016. - _Charlie Marion_, Jun 22 2018

%F 8*a(n)+1 = A016754(n). - _Charlie Marion_, Oct 10 2018

%e G.f.: x + 3*x^2 + 6*x^3 + 10*x^4 + 15*x^5 + 21*x^6 + 28*x^7 + 36*x^8 + 45*x^9 + ...

%e When n=3, a(3) = 4*3/2 = 6.

%e Example(a(4)=10): ABCD where A, B, C and D are different links in a chain or different amino acids in a peptide possible fragments: A, B, C, D, AB, ABC, ABCD, BC, BCD, CD = 10.

%e a(2): hollyhock leaves on the Tokugawa Mon, a(4): points in Pythagorean tetractys, a(5): object balls in eight-ball billiards. - _Bradley Klee_, Aug 24 2015

%p A000217 := proc(n) n*(n+1)/2; end;

%p istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return true else return false; end if; end proc; # _N. J. A. Sloane_, May 25 2008

%p ZL := [S, {S=Prod(B, B, B), B=Set(Z, 1 <= card)}, unlabeled]: seq(combstruct[count](ZL, size=n), n=2..55); # _Zerinvary Lajos_, Mar 24 2007

%p isA000217 := proc(n)

%p issqr(1+8*n) ;

%p end proc: # _R. J. Mathar_, Nov 29 2015

%t Table[(m^2 - m)/2, {m, 54}] (* _Zerinvary Lajos_, Mar 24 2007 *)

%t Array[ #*(# - 1)/2 &, 54] (* _Zerinvary Lajos_, Jul 10 2009 *)

%t FoldList[#1 + #2 &, 0, Range@ 50] (* _Robert G. Wilson v_, Feb 02 2011 *)

%t Accumulate[Range[0,70]] (* _Harvey P. Dale_, Sep 09 2012 *)

%t CoefficientList[Series[x / (1 - x)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, Jul 30 2014 *)

%t (* For Mathematica 10.4+ *) Table[PolygonalNumber[n], {n, 0, 53}] (* _Arkadiusz Wesolowski_, Aug 27 2016 *)

%t LinearRecurrence[{3, -3, 1}, {0, 1, 3}, 54] (* _Robert G. Wilson v_, Dec 04 2016 *)

%t (* The following Mathematica program, courtesy of Steven J. Miller, is useful for testing if a sequence is Benford. To test a different sequence only one line needs to be changed. This strongly suggests that the triangular numbers are not Benford, since the second and third columns of the output disagree. - _N. J. A. Sloane_, Feb 12 2017 *)

%t fd[x_] := Floor[10^Mod[Log[10, x], 1]]

%t benfordtest[num_] := Module[{},

%t For[d = 1, d <= 9, d++, digit[d] = 0];

%t For[n = 1, n <= num, n++,

%t {

%t d = fd[n(n+1)/2];

%t If[d != 0, digit[d] = digit[d] + 1];

%t }];

%t For[d = 1, d <= 9, d++, digit[d] = 1.0 digit[d]/num];

%t For[d = 1, d <= 9, d++,

%t Print[d, " ", 100.0 digit[d], " ", 100.0 Log[10, (d + 1)/d]]];

%t ];

%t benfordtest[20000]

%o (PARI) A000217(n) = n * (n + 1) / 2;

%o (PARI) is_A000217(n)=n*2==(1+n=sqrtint(2*n))*n \\ _M. F. Hasler_, May 24 2012

%o (PARI) is(n)=ispolygonal(n,3) \\ _Charles R Greathouse IV_, Feb 28 2014

%o a000217 n = a000217_list !! n

%o a000217_list = scanl1 (+) [0..] -- _Reinhard Zumkeller_, Sep 23 2011

%o (MAGMA) [n*(n+1)/2: n in [0..60]]; // _Bruno Berselli_, Jul 11 2014

%o (MAGMA) [n: n in [0..1500] | IsSquare(8*n+1)]; // _Juri-Stepan Gerasimov_, Apr 09 2016

%o (Sage) [n*(n+1)/2 for n in (0..60)] # _Bruno Berselli_, Jul 11 2014

%o (Scheme) (define (A000217 n) (/ (* n (+ n 1)) 2)) ;; _Antti Karttunen_, Jul 08 2017

%o (J) a000217=: *-:@>: NB. _Stephen Makdisi_, May 02 2018

%Y Cf. A000096, A000124, A000292, A000330, A000396, A000668, A001082, A001477, A001788, A002024, A002378, A002415, A006011, A007318, A008953, A008954, A010054 (characteristic function), A028347, A036666, A046092, A051942, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A062717, A087475, A101859, A143320, A210569, A245031, A245300.

%Y a(n) = A110449(n, 0).

%Y a(n) = A110555(n+2, 2).

%Y A diagonal of A008291.

%Y Column 2 of A195152.

%Y Numbers of the form n*t(n+k,h)-(n+k)*t(n,h), where t(i,h) = i*(i+2*h+1)/2 for any h (for A000217 is k=1): A005563, A067728, A140091, A140681, A212331.

%Y Boustrophedon transforms: A000718, A000746.

%Y Iterations: A007501 (start=2), A013589 (start=4), A050542 (start=5), A050548 (start=7), A050536 (start=8), A050909 (start=9).

%Y Cf. A002817 (doubly triangular numbers).

%Y Cf. A104712 (first column, starting with a(1)).

%Y Some generalized k-gonal numbers are A001318 (k=5), this sequence (k=6), A085787 (k=7), etc.

%K nonn,core,easy,nice,changed

%O 0,3

%A _N. J. A. Sloane_

%E Edited by _Derek Orr_, May 05 2015

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified November 16 19:07 EST 2018. Contains 317275 sequences. (Running on oeis4.)