login
This site is supported by donations to The OEIS Foundation.

 

Logo

Many excellent designs for a new banner were submitted. We will use the best of them in rotation.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A000217 Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n.
(Formerly M2535 N1002)
2489
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

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

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

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.

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

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

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

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

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 T(n-1). Using n+1 lines with k>1 parallel lines, the maximum number of 2-intersections is given by T(n)-T(k-1). - Jon Perry, Jun 11 2003

Number of distinct straight lines that can pass through n points in 3-dimensional space. - Cino Hilliard, Aug 12 2003

Triangular numbers - odd numbers = triangular numbers: 0,1,3,6,10,15,21... - 0,1,3,5,7,9,11... = 0,0,0,1,3,6,10... - Xavier Acloque, Oct 31 2003

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

Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004

Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004

Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012

a(n) == 1 mod (n+2) if n is odd and == n/2+2 mod (n+2) if n is even. - Jon Perry, Dec 16 2004

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.

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

A110560/A110561 = numerator/denominator of the coefficients of the exponential generating function. - Jonathan Vos Post, Jul 27 2005

Binomial transform is {0, 1, 5, 18, 56, 160, 432, ... }, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005

a(n) = A111808(n,2) for n>1. - Reinhard Zumkeller, Aug 17 2005

Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006

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

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

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).

a(n) = A126890(n,0). - Reinhard Zumkeller, Dec 30 2006

a(n) is the number of terms in the expansion of (a_1+a_2+a_3)^n. - Sergio Falcon, Feb 12 2007

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

The number of distinct handshakes in a room with n people (n>=2). - Mohammad K. Azarian, Apr 12 2007

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

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

a(n), n>=1, is the number of ways in which n-1 can be written as a sum of three positive integers if representations differing in the order of the terms are considered to be different. In other words a(n), n>=1, is the number of positive integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001

a(n+1), n>=0, 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

Numbers m>=0 such that round(sqrt(2m+1))-round(sqrt(2m))=1. - Hieronymus Fischer, Aug 06 2007

Numbers m>=0 such that ceiling(2*sqrt(2m+1))-1=1+floor(2*sqrt(2m)). - Hieronymus Fischer, Aug 06 2007

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). - Hieronymus Fischer, Aug 06 2007

Sequence allows us to find X values of the equation: 8*X^3 + X^2 = Y^2. To find Y values: b(n)=n(n+1)(2n+1)/2. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007

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

Equals row sums of triangle A143320, n>0. - Gary W. Adamson, Aug 07 2008

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

a(n) = A022264(n) - A049450(n). - Reinhard Zumkeller, Oct 09 2008

Equals row sums of triangle A152204. - Gary W. Adamson, Nov 29 2008

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

-a(n+1) = E(2)*C(n+2,2) (n>=0) where E(n) are the Euler numbers in the enumeration A122045 and C(n,k) are the binomial coefficients A007318. 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

4a(x)+4a(y)+1=(x+y+1)^2+(x-y)^2. - Vladimir Shevelev, Jan 21 2009

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

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

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

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

Partial sums of A001477. - Juri-Stepan Gerasimov, Jan 25 2010. [A-number corrected by Omar E. Pol, Jun 05 2012]

The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010

From Charlie Marion, Oct 15 2010: (Start)

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

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

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.

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

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

(End)

From Charlie Marion, Dec 03 2010: (Start)

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

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

Column sums of:

1 3 5  7  9...

    1  3  5...

          1...

..............

--------------

1 3 6 10 15...

Sum(n=1..infinity, 1/a(n)^2)=4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).

(End)

A004201(a(n)) = A000290(n); A004202(a(n)) = A002378(n). - Reinhard Zumkeller, Feb 12 2011

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

From Charlie Marion, Feb 23 2012: (Start)

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).

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

(End)

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

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

(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

a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012

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

a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012

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

a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012

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

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

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

From James East, Jan 08 2013 (Start):

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].

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].

(End)

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

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

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

The series sum(1/a(k),k=1..infinity) = 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

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 9T+1. - Lekraj Beedassy, May 29 2013

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

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

Dimension of orthogonal group O(n+1). - Eric M. Schmidt, Sep 08 2013

Number of positive roots in the root system of type A_n (for n>0). - Tom Edgar, Nov 05 2013

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

Also the alternating row sums of A095831. Also the alternating row sums of A055461, for n >= 1. - Omar E. Pol, Jan 26 2014

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

a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014

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

REFERENCES

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.

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

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

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

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

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

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

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.

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.]

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.

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

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

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

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

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

D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-3 Penguin Books 1987.

LINKS

N. J. A. Sloane, Table of n, a(n) for n = 0..30000

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

T. Beldon and T. Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette 86 (2002), 423-431.

Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372. [From Parthasarathy Nambi, Sep 30 2009]

Anicius Manlius Severinus Boethius, De institutione arithmetica libri duo, Book 2, sections 7-9.

H. Bottomley, Illustration of initial terms of A000217, A002378

Scott A. Brown, Brown's Math Page, etc. [Broken link?]

P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.

Tomislav Doslic, Maximum Product Over Partitions Into Distinct Parts, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.8.

Askar Dzhumadildaev and Damir Yeliussizov, Power Sums of Binomial Coefficients, Journal of Integer Sequences, Vol. 16 (2013), #13.1.1.

J. East, Presentations for singular subsemigroups of the partial transformation semigroup, Internat. J. Algebra Comput., 20 (2010), no. 1, 1-25.

J. East, On the singular part of the partition monoid, Internat. J. Algebra Comput. 21 (2011), no. 1-2, 147-178.

Adam Grabowski, Polygonal Numbers, Formalized Mathematics, Vol. 21, No. 2, Pages 103-113, 2013; DOI: 10.2478/forma-2013-0012; alternate copy

S. S. Gupta, Fascinating Triangular Numbers

C. Hamberg, Triangular Numbers Are Everywhere

Guo-Niu Han, Enumeration of Standard Puzzles

Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]

A. M. Hinz, S. Klavžar, U. Milutinović, C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 35. Book's website

J. M. Howie, Idempotent generators in finite full transformation semigroups, Proc. Roy. Soc. Edinburgh Sect. A, 81 (1978), no. 3-4, 317-323.

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 253

Milan Janjic, Two Enumerative Functions

Xiangdong Ji, Chapter 8: Structure of Finite Nuclei, Lecture notes for Phys 741 at Univ. of Maryland, pp. 139-140 [From Tom Copeland, Apr 07 2014].

R. Jovanovic, Triangular numbers

R. Jovanovic, First 2500 Triangular numbers

H. K. Kim, "On Regular polytope numbers".

Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.

Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.

J. Koller, Triangular Numbers

A. J. F. Leatherland, Triangle Numbers on Ulam Spiral

Ivars Peterson, Triangular Numbers and Magic Squares.

Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Omar E. Pol, Illustration of initial terms of A000217, A000290, A000326, A000384, A000566, A000567

F. Richman, Triangle numbers

Frank Ruskey and Jennifer Woodcock, The Rand and block distances of pairs of set partitions, in Combinatorial Algorithms, 287-299, Lecture Notes in Comput. Sci., 7056, Springer, Heidelberg, 2011.

sci.math, Jun 98 Square numbers that are triangular

James A. Sellers, Partitions Excluding Specific Polygonal Numbers As Parts, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.4.

N. J. A. Sloane, Illustration of initial terms of A000217, A000290, A000326

T. Trotter, Some Identities for the Triangular Numbers, J. Rec. Math. vol. 6, no. 2 Spring 1973.

G. Villemin's Almanach of Numbers, Nombres Triangulaires

Eric Weisstein's World of Mathematics, Triangular Number

Eric Weisstein's World of Mathematics, Absolute Value

Eric Weisstein's World of Mathematics, Composition

Eric Weisstein's World of Mathematics, Distance

Eric Weisstein's World of Mathematics, Golomb Ruler

Eric Weisstein's World of Mathematics, Polygonal Number

Eric Weisstein's World of Mathematics, Line Line Picking

Eric Weisstein's World of Mathematics, Trinomial Coefficient

Eric Weisstein's World of Mathematics, Wiener Index

Wikipedia, Floyd's triangle

Index entries for "core" sequences

Index entries for related partition-counting sequences

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

G.f.: x/(1-x)^3.

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

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

a(n) = a(n-1)+n. - Zak Seidov, Mar 06 2005

a(n) + a(n-1)*a(n+1) = a(n)^2. - Terry Trotter (ttrotter(AT)telesal.net), Apr 08 2002

a(n) = (-1)^n*sum(k=1, n, (-1)^k*k^2). - Benoit Cloitre, Aug 29 2002

a(n) = ((n+2)/n)*a(n-1)Sum(n=1..infinity, 1/a(n)) = 2. - Jon Perry, Jul 13 2003

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

With interpolated zeros, this is n(n+2)/8*(1+(-1)^n)/2=sum{k=0..n, sum{j=0..k, floor(k^2/4)}}. - Benoit Cloitre, Aug 19 2003

a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j)=C(i+j+1, i). - Benoit Cloitre, Aug 19 2003

a(n) = ((n^3-(n-1)^3)-(n^1-(n-1)^1))/(2^3-2^1) = (n^3-(n-1)^3-1)/6. - Xavier Acloque, Oct 24 2003

a(n) = a(n-1) + (1 + sqrt[1 + 8*a(n-1)])/2. E.g., a(4) = a(3) + (1 + sqrt[1 + 8*a(3)])/2 = 6 + (1 + sqrt[49])/2 = 6+8/2 = 10. 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

a(n)+a(n+1) = (n+1)^2. - N. J. A. Sloane, Feb 19 2004

a(n) = a(n-2)+2n-1. - Paul Barry, Jul 17 2004

a(n) = Sqrt[Sum[Sum[(i*j), {i, 1, n}], {j, 1, n}]] = sqrt(A000537(n)). - Alexander Adamchuk, Oct 24 2004

a(n) = Sqrt[Sqrt[Sum[Sum[(i*j)^3, {i, 1, n}], {j, 1, n}]]]. a(n) = Sum[Sum[Sum[(i*j*k)^3, {i, 1, n}], {j, 1, n}], {k, 1, n}]^(1/6). - Alexander Adamchuk, Oct 26 2004

a(0) = 0, a(1) = 1, a(n) = 2*a(n-1)-a(n-2)+1. - Miklos Kristof, Mar 09 2005

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

a(n) = A108299(n+3,4) = -A108299(n+4,5). - Reinhard Zumkeller, Jun 01 2005

a(n) = floor((2n+1)^2/8). - Paul Barry, May 29 2006

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

(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

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 the third comment by J. M. Bergot dated May 22 2012]. - Charlie Marion, Feb 04 2011

a(n) = A023896(n) + A067392(n). - Lekraj Beedassy, Mar 02 2007

Sum_{k, 0<=k<=n}a(k)*A039599(n,k)=A002457(n-1), for n>=1. - Philippe Deléham, Jun 10 2007

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

If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then a(n)=-f(n,n-1,1), for n>=1. - Milan Janjic, Dec 20 2008

a(n) = A000124(n-1) + n-1 for n >= 2. a(n) = A000124(n) - 1. - Jaroslav Krizek, Jun 16 2009

An exponential generating function for the inverse of this sequence is given by sum((pochhammer(1, m)*pochhammer(1, m))*x^m/(pochhammer(3, m)*factorial(m)), m = 0 .. infinity)=((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 x=0. A000217[n+1]=limit(Diff(((2-2*x)*log(1-x)+2*x)/x^2, x$n),x=0)^-1=limit((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),x=0)^-1. - Stephen Crowley, Jun 28 2009

a(n) = A034856(n+1) - A005408(n) = A005843(n) + A000124(n) - A005408(n). - Jaroslav Krizek, Sep 05 2009

With offset 1, a(n)=1/2*floor(n^3/(n+1)). - Gary Detlefs, Feb 14 2010

a(n) = 4*a(floor(n/2))+(-1)^(n+1)*floor((n+1)/2). - Bruno Berselli, May 23 2010

a(n) = 3*a(n-1)-3*a(n-2)+a*(n-3); a(0)=0,a(1)=1. - Mark Dols (markdols99(AT)yahoo.com), Aug 20 2010

a(n) = sqrt((sum(i^3,{i,1,n}))). - Zak Seidov, Dec 07 2010

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

a(n) = A110654(n) * A008619(n). - Reinhard Zumkeller, Aug 24 2011

a(2k-1) = A000384(k), a(2k) = A014105(k), k>0. - Omar E. Pol, Sep 13 2011

a(n) = A026741(n) * A026741(n+1). - Charles R Greathouse IV, Apr 01 2012

a(n) + a(a(n)) + 1 = a(a(n)+1). - J. M. Bergot, Apr 27 2012

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

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

a(n) = A002378(n)/2 = (A001318(n) + A085787(n))/2. - Omar E. Pol, Jan 11 2013

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

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

a(n) = A002088(n) + A063985(n). - Reinhard Zumkeller, Jan 21 2013

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

a(n) + a(n+1) + a(n+2) + a(n+3) + n = a(2n+4). - Ivan N. Ianakiev, Mar 16 2013

a(n) + a(n+1) + ... + a(n+8) + 6n = a(3n+15). - Charlie Marion, Mar 18 2013

a(n) + a(n+1) + ... + a(n+20) + 2n^2 + 57n = a(5n+55). - Charlie Marion, Mar 18 2013

3*a(n) + a(n-1) = a(2n), for n>0. - Ivan N. Ianakiev, Apr 05 2013

a(n+1) = det(C(i+2,j+1), 1 <= i,j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013

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

a(n) = floor(n/2) + ceil(n^2/2) = n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013

a(n) = floor((n+1)/(e^(2/(n+1)) - 1)). - Richard R. Forberg, Jun 22 2013

Sum(n>=1, a(n)/n! ) = 3*exp(1)/2. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - Richard R. Forberg, Jul 15 2013

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

a(a(n)) = a(a(n)-1) + a(n), n>=1. - Vladimir Shevelev, Jan 21 2014

EXAMPLE

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 + ...

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

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.

MAPLE

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

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)

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

A000217:=-1/(z-1)**3; [Simon Plouffe in his 1992 dissertation.]

A000217:=seq(binomial(k+1, 2), k=0..100); [Wesley Ivan Hurt, Jun 23 2013]

MATHEMATICA

Table[(m^2 - m)/2, {m, 54}] (* Zerinvary Lajos, Mar 24 2007 *)

Array[ #*(# - 1)/2 &, 54] (* Zerinvary Lajos, Jul 10 2009 *)

FoldList[#1 + #2 &, 0, Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)

Accumulate[Range[0, 70]] (* Harvey P. Dale, Sep 09 2012 *)

PROG

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

(PARI) is_A000217(n)=n*2==(1+n=sqrtint(2*n))*n  \\ M. F. Hasler, May 24 2012

(PARI) is(n)=ispolygonal(n, 3) \\ Charles R Greathouse IV, Feb 28 2014

(Haskell)

a000217 n = a000217_list !! n

a000217_list = scanl1 (+) [0..] -- Reinhard Zumkeller, Sep 23 2011

CROSSREFS

Cf. A007318, A002024, A000124, A002378, A000292, A000330, A000096, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A051942, A101859, A001477, A046092, A001082, A036666, A062717, A028347, A087475, A006011, A002415, A010054, A210569, A143320, A000396, A000668, A008953, A008954.

A diagonal of A008291. a(n) = A110555(n+2, 2).

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

a(3*n)=A081266, a(4*n)=A033585, a(5*n)=A144312, a(6*n)=A144314. - Reinhard Zumkeller, Sep 17 2008

n-gonal numbers: A000290, A000326, A000384, A000566, A000567, A001106, A001107, A051682, A051624, A051865-A051876.

Column 2 of A195152.

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.

Boustrophedon transforms: A000718, A000746.

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

Sequence in context: A105339 A089594 A161680 * A105340 A176659 A109811

Adjacent sequences:  A000214 A000215 A000216 * A000218 A000219 A000220

KEYWORD

nonn,core,easy,nice,changed

AUTHOR

N. J. A. Sloane

STATUS

approved

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

Content is available under The OEIS End-User License Agreement .

Last modified April 20 18:30 EDT 2014. Contains 240820 sequences.