A005449 Second pentagonal numbers: a(n) = n*(3*n + 1)/2.

0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480

Offset

0,2

Comments

Number of edges in the join of the complete graph and the cycle graph, both of order n, K_n * C_n. - Roberto E. Martinez II, Jan 07 2002

Also number of cards to build an n-tier house of cards. - Martin Wohlgemuth, Aug 11 2002

The modular form Delta(q) = q*Product_{n>=1} (1-q^n)^24 = q*(1 + Sum_{n>=1} (-1)^n*(q^(n*(3*n-1)/2)+q^(n*(3*n+1)/2))^24 = q*(1 + Sum_{n>=1} A033999(n)*(q^A000326(n)+q^a(n))^24. - Jonathan Vos Post, Mar 15 2006

Row sums of triangle A134403.

Bisection of A001318. - Omar E. Pol, Aug 22 2011

Sequence found by reading the line from 0 in the direction 0, 7, ... and the line from 2 in the direction 2, 15, ... in the square spiral whose vertices are the generalized pentagonal numbers, A001318. - Omar E. Pol, Sep 08 2011

A general formula for the n-th second k-gonal number is given by T(n, k) = n*((k-2)*n+k-4)/2, n>=0, k>=5. - Omar E. Pol, Aug 04 2012

Partial sums give A006002. - Denis Borris, Jan 07 2013

A002260 is the following array A read by antidiagonals:

0, 1, 2, 3, 4, 5, ...

0, 1, 2, 3, 4, 5, ...

0, 1, 2, 3, 4, 5, ...

0, 1, 2, 3, 4, 5, ...

0, 1, 2, 3, 4, 5, ...

0, 1, 2, 3, 4, 5, ...

and a(n) is the hook sum: Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013

From Klaus Purath, May 13 2021: (Start)

This sequence and A000326 provide all integers m such that 24*m + 1 is a square. The union of the two sequences is A001318.

If A is a sequence satisfying the recurrence t(n) = 3*t(n-1) - 2*t(n-2) with the initial values either A(0) = 1, A(1) = n + 2 or A(0) = -1, A(1) = n-1, then a(n) = (A(i)^2 - A(i-1)*A(i+1))/2^i + n^2 for i>0. (End)

a(n+1) is the number of Dyck paths of size (3,3n+2), i.e., the number of NE lattice paths from (0,0) to (3,3n+2) which stay above the line connecting these points. - Harry Richman, Jul 13 2021

Binomial transform of [0, 2, 3, 0, 0, 0, ...], being a(n) = 2*binomial(n,1) + 3*binomial(n,2). a(3) = 15 = [0, 2, 3, 0] dot [1, 3, 3, 1] = [0 + 6 + 9 + 0]. - Gary W. Adamson, Dec 17 2022

a(n) is the sum of longest side length of all nondegenerate integer-sided triangles with shortest side length n and middle side length (n + 1), n > 0. - Torlach Rush, Feb 04 2024

References

Henri Cohen, A Course in Computational Algebraic Number Theory, vol. 138 of Graduate Texts in Mathematics, Springer-Verlag, 2000.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..2000

A. O. L. Atkin and F. Morain, Elliptic Curves and Primality Proving, Math. Comp., Vol. 61, No. 203 (1993), pp. 29-68.

Charles H. Conley and Valentin Ovsienko, Quiddities of polygon dissections and the Conway-Coxeter frieze equation, arXiv:2107.01234 [math.CO], 2021.

Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, Acta Academiae Scientiarum Imperialis Petropolitanae, Vol. 1780: I, pp. 56-75.

Leonhard Euler, Observatio de summis divisorum, Novi Commentarii academiae scientiarum Petropolitanae, Vol. 5, pp. 59-74.

Leonhard Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 8.

Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.

Alfred Hoehn, Illustration of initial terms of A000326, A005449, A045943, A115067.

D. Suprijanto and I. W. Suwarno, Observation on Sums of Powers of Integers Divisible by 3k-1, Applied Mathematical Sciences, Vol. 8, No. 45 (2014), pp. 2211-2217.

Martin Wohlgemuth, Pentagon, Kartenhaus und Summenzerlegung.

Index entries for two-way infinite sequences

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

Formula

a(n) = A110449(n, 1) for n>0.

G.f.: x*(2+x)/(1-x)^3. E.g.f.: exp(x)*(2*x + 3*x^2/2). a(n) = n*(3*n + 1)/2. a(-n) = A000326(n). - Michael Somos, Jul 18 2003

a(n) = A001844(n) - A000217(n+1) = A101164(n+2,2) for n>0. - Reinhard Zumkeller, Dec 03 2004

a(n) = Sum_{j=1..n} n+j. - Zerinvary Lajos, Sep 12 2006

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

a(n) = 2*C(3*n,4)/C(3*n,2), n>=1. - Zerinvary Lajos, Jan 02 2007

a(n) = A000217(n) + A000290(n). - Zak Seidov, Apr 06 2008

a(n) = a(n-1) + 3*n - 1 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010

a(n) = A129267(n+5,n). - Philippe Deléham, Dec 21 2011

a(n) = 2*A000217(n) + A000217(n-1). - Philippe Deléham, Mar 25 2013

a(n) = A130518(3*n+1). - Philippe Deléham, Mar 26 2013

a(n) = (12/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Vladimir Kruchinin, Jun 04 2013

a(n) = floor(n/(1-exp(-2/(3*n)))) for n>0. - Richard R. Forberg, Jun 22 2013

a(n) = Sum_{i=1..n} 2*i - 1 + (i mod 2). - Wesley Ivan Hurt, Oct 11 2013

a(n) = (A000292(6*n+k+1)-A000292(k))/(6*n+1) - A000217(3*n+k+1), for any k >= 0. - Manfred Arens, Apr 26 2015

Sum_{n>=1} 1/a(n) = 6 - Pi/sqrt(3) - 3*log(3) = 0.89036376976145307522... . - Vaclav Kotesovec, Apr 27 2016

a(n) = A000217(2*n) - A000217(n). - Bruno Berselli, Sep 21 2016

Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/sqrt(3) + 4*log(2) - 6. - Amiram Eldar, Jan 18 2021

From Klaus Purath, May 13 2021: (Start)

Partial sums of A016789 for n > 0.

a(n) = 3*n^2 - A000326(n).

a(n) = A000326(n) + n.

a(n) = A002378(n) + A000217(n-1) for n >= 1. [Corrected by Rémi Guillaume, Aug 14 2024] (End)

From Klaus Purath, Jul 14 2021: (Start)

b^2 = 24*a(n) + 1 we get by b^2 = (a(n+1) - a(n-1))^2 = (a(2*n)/n)^2.

a(2*n) = n*(a(n+1) - a(n-1)), n > 0.

a(2*n+1) = n*(a(n+1) - a(n)). (End)

A generalization of Lajos' formula, dated Sep 12 2006, follows. Let SP(k,n) = the n-th second k-gonal number. Then SP(2k+1,n) = Sum_{j=1..n} (k-1)*n+j+k-2. - Charlie Marion, Jul 13 2024

Example

From Omar E. Pol, Aug 22 2011: (Start)
Illustration of initial terms:
.                                               O
.                                             O O
.                                 O         O O O
.                               O O       O O O O
.                     O       O O O     O O O O O
.                   O O     O O O O     O O O O O
.           O     O O O     O O O O     O O O O O
.         O O     O O O     O O O O     O O O O O
.    O    O O     O O O     O O O O     O O O O O
.    O    O O     O O O     O O O O     O O O O O
.
.    2     7        15         26           40
.
(End)

Maple

A005449:=n->n*(3*n + 1)/2; seq(A005449(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Mathematica

Table[n (3 n + 1)/2, {n, 0, 100}] (* Zak Seidov, Jan 31 2012 *)

Prog

(PARI) {a(n) = n * (3*n + 1) / 2} /* Michael Somos, Jul 18 2003 */
(Magma) [n*(3*n + 1) / 2: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Crossrefs

Cf. A016789 (first differences), A006002 (partial sums).

Cf. A000217, A000320, A000326, A001318, A033568, A049451, A101165, A101166.

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, this sequence, A045943, A115067, A140090, A140091, A059845, A140672-A140675, A151542.

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=3).

Cf. numbers of the form n*((2*k+1)*n+1)/2 listed in A022289 (this sequence is the case k=1).

Sequence in context: A194140 A029888 A194112 * A293401 A323038 A333637

Adjacent sequences: A005446 A005447 A005448 * A005450 A005451 A005452

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved