

A046092


4 times triangular numbers: a(n) = 2*n*(n+1).


148



0, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324
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OFFSET

0,2


COMMENTS

Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 + 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754).  Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in (n+1) X (n+1) square grid with all horizontal and vertical segments filled in.  Asher Auel (asher.auel(AT)reed.edu), Jan 12 2000
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): sum(i>a(n)+1,1/i^2) < sum(i>n,1/i^3) < sum(i>a(n),1/i^2).  Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular ngon when n is even.  SenPeng Eu, Apr 05 2001
Number of ways to change two nonidentical letters in the word aabbccdd..., where there are n type of letters.  Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n1)dimensional sides of an (n+1)dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.).  Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a kgon where k is odd. We label a triangle with n=1, a pentagon with n=2, .., a kgon with n = floor(k/2). Imagine every player standing on every vertex of the kgon.
Initially there are 2 frisbees on two neighboring players. Every time they throw the frisbee to their neighbor with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified it by simulating the processes with a computer program. For example a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*{1 + sqrt(2k + 1)}.  Lekraj Beedassy, Jun 04 2006
A diagonal of A059056.  Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2nset X into 2blocks then a(n1) is equal to the number of 2subsets of X containing none of X_i, (i=1,...n).  Milan Janjic, Jul 16 2007
Sequence allows us to find X values of the equation: 2*X^3 + X^2 = Y^2. To find Y values: b(n)=2n(n+1)(2n+1).  Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007
Number of (n+1)permutations of 3 objects u,v,w, with repetition allowed, containing n1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2permutations either in front, or in the middle, or at the end.  Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4,... and the same line from 0, in the direction 0, 12,..., in the square spiral whose vertices are the triangular numbers A000217.  Omar E. Pol, May 03 2008
Twice oblong numbers.  Omar E. Pol, May 03 2008
a(n) is also the least weight of selfconjugate partitions having n different even parts.  Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the SwissKnife polynomials P(n,x) A153641 (P(n,1)(1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = (P(2,1)(1)^k P(2,2k+1))/2. (End)
The sum of squares of n+1 consecutive numbers between a(n)n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2+11^2+12^2=13^2+14^2.  Tanya Khovanova, Jul 20 2009
Number of units of a(n) belongs to a periodic sequence: 0, 4, 2, 4, 0.  Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009
Number of roots in the root system of type D_{n+1} (for n>2).  Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1)  2 = A001844(n)  1.  Jaroslav Krizek, Dec 27 2013
a(n) = A245300(n,n).  Reinhard Zumkeller, Jul 17 2014
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A147973(n+3), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference.  Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n).  Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1unitwide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n)  h(n) for n>=1.  Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n.  Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)cocktail party graph.  Eric W. Weisstein, Jun 29 2017


REFERENCES

T. M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, 1976, page 3.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: AddisonWesley, 1994.
HsienKuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wpcontent/files/2016/12/aathhrr1.pdf. Also Exact and Asymptotic Solutions of a DivideandConquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
H. J. Brothers, Pascal's Prism: Supplementary Material.
Milan Janjic, Two Enumerative Functions
Ron Knott, Pythagorean Triples and Online Calculators
Tanya Khovanova, A Miracle Equation [From Tanya Khovanova, Jul 20 2009]
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 24922501. [From Augustine O. Munagi, Dec 18 2008]
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
D. Suprijanto and Rusliansyah, Observation on Sums of Powers of Integers Divisible by Four, Applied Mathematical Sciences, Vol. 8, 2014, no. 45, 2219  2226.
Eric Weisstein's World of Mathematics, Aztec Diamond
Eric Weisstein's World of Mathematics, Cocktail Party Graph
Eric Weisstein's World of Mathematics, Connected Dominating Set
Eric Weisstein's World of Mathematics, Gear Graph
Eric Weisstein's World of Mathematics, Hamiltonian Path
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

a(n) = A100345(n+1, n1) for n>0.
a(n) = 2*A002378(n) = 4*A000217(n).  Lekraj Beedassy, May 25 2004
a(n) = C(2n, 2)  n = 4*C(n, 2).  Zerinvary Lajos, Feb 15 2005
a(n)=A028896A002378; a(n)=A124080A028896; a(n)=A049598A033996.  Zerinvary Lajos, Mar 06 2007
Array read by rows: row n gives A033586(n), A085250(n+1).  Omar E. Pol, May 03 2008
a(n) = a(n1)+4*n; o.g.f.:4*x/(1x)^3; e.g.f.: exp(x)*(2*x^2+4*x).  Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n)=1/int((x*n+x1)*(step((1+x*n)/n)1)*n*step((x*n+x1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function
sum(1/a(n), n >= 1) = sum(1/((2*n)*(n+1)), n >= 1) = 1/2. (End)
a(0)=0, a(1)=4, a(2)=12, a(n)=3*a(n1)3*a(n2)+a(n3).  Harvey P. Dale, Jul 25 2011
For n>0 a(n)=1/(Integral_{x=0..Pi/2} (sin(x))^(2*n1)*(cos(x))^3).  Francesco Daddi, Aug 02 2011
a(n) = A001844(n)  1.  Omar E. Pol, Oct 03 2011
(a(n)  A000217(k))^2 = A000217(2nk)*A000217(2n+1+k)  (A002378(n)  A000217(k)), for all k. See also A001105.  Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k1)*a(n+1) = a((n+1)*(t(k)t(k1)1)), where k>=2, n>=1, t(k)=A000217(k). (End)
2*a(n)+1 = A016754(n) = A005408(n)^2, the odd squares.  M. F. Hasler, Oct 02 2014
Sum_{n>=1} (1)^(n+1)/a(n) = log(2)  1/2 = A187832.  Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2).  Enrique Navarrete, Aug 30 2017


EXAMPLE

a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25),...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2.  Augustine O. Munagi, Dec 18 2008


MATHEMATICA

Table[2 n (n + 1), {n, 0, 50}] (* Stefan Steinerberger, Apr 03 2006 *)
LinearRecurrence[{3, 3, 1}, {0, 4, 12}, 50] (* Harvey P. Dale, Jul 25 2011 *)
4*Binomial[Range[50], 2] (* Harvey P. Dale, Jul 25 2011 *)


PROG

(PARI) a(n)=binomial(n+1, 2)<<2 \\ Charles R Greathouse IV, Jun 10 2011
(MAGMA) [2*n*(n+1): n in [0..50]]; // Vincenzo Librandi, Oct 04 2011
(Maxima) A046092(n):=2*n*(n+1)$
makelist(A046092(n), n, 0, 30); /* Martin Ettl, Nov 08 2012 */
(Haskell)
a046092 = (* 2) . a002378  Reinhard Zumkeller, Dec 15 2013


CROSSREFS

Cf. A045943, A028895, A002943, A054000, A000330, A007290, A002378, A033996, A124080, A028896, A049598, A005563, A000217, A033586, A085250.
Main diagonal of array in A001477.
Equals A033996/2 Cf. A001844.  Augustine O. Munagi, Dec 18 2008
Cf. A078371, A141530 (see Librandi's comment in A078371).
Cf. A097080, A001845.
Cf. similar sequences listed in A299645.
Sequence in context: A008006 A081937 A088557 * A301059 A008241 A008216
Adjacent sequences: A046089 A046090 A046091 * A046093 A046094 A046095


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane, Eric W. Weisstein


STATUS

approved



