A008354 -id:A008354 - cp's OEIS frontend

A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

1, 15, 84, 300, 825, 1911, 3920, 7344, 12825, 21175, 33396, 50700, 74529, 106575, 148800, 203456, 273105, 360639, 469300, 602700, 764841, 960135, 1193424, 1470000, 1795625, 2176551, 2619540, 3131884, 3721425, 4396575, 5166336, 6040320, 7028769, 8142575

Offset: 0

Emeric Deutsch, Jun 17 2005

Kekulé numbers for certain benzenoids.
This is the case P(3,n) of the family of sequences defined in A132458. - Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007
Using the triangular numbers 0, 1, 3, ..., create a sequence of advancing sums of k-tuples with k=n*(n+1)/2 of the odd numbers: 0, 1, 15, 84, 300, 825, 1911, 3920, ... . This begins 0, then 1, then 3+5+7=15, then 9+11+13+15+17+19=84, then 21+23+...+39=300 and so on. - J. M. Bergot, Dec 08 2014
Partial sums of A008354. - J. M. Bergot, Dec 19 2014
Coefficients in the terminating series identity 1 - 15*n/(n + 4) + 84*n*(n - 1)/((n + 4)*(n + 5)) - 300*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 2,3,4,.... Cf. A000330. - Peter Bala, Feb 12 2019

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 231, # 33).

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000537, A126274.

Cf. A000217, A000330, A008354 (first diffs.), A132458.

List([0..40],k->(k+1)^2*(k+2)^2*(2*k+3)/12); # Muniru A Asiru, Feb 18 2019

A108674 :=n->(n+1)^2*(n+2)^2*(2*n+3)/12;
seq(A108674(n),n=0..35);

Table[(n^2+4*n^3+5*n^4+2*n^5)/12, {n, 40}] (* Enrique Pérez Herrero, Feb 27 2013 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,15,84,300,825,1911},40] (* Harvey P. Dale, Apr 04 2024 *)

a(n)=(n+1)^2*(n+2)^2*(2*n+3)/12 \\ Charles R Greathouse IV, Feb 27 2013

G.f.: (1+z)*(1+8*z+z^2)/(1-z)^6.
a(n) = Sum_{j=1..n+1} j^2 Sum_{i=1..n+1} i. - Alexander Adamchuk, Jun 25 2006
a(n) = A000330(n+1) * A000217(n+1). - Daniel Suteu, Nov 26 2020
E.g.f.: exp(x)*(12 + 168*x + 330*x^2 + 184*x^3 + 35*x^4 + 2*x^5)/12. - Stefano Spezia, Mar 02 2022
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) 192*log(2) - 132.
Sum_{n>=0} (-1)^n/a(n) = 2*Pi^2 - 48*Pi + 132. (End)

A269237 a(n) = (n + 1)^2(5n^2 + 10*n + 2)/2.

Original entry on oeis.org

1, 34, 189, 616, 1525, 3186, 5929, 10144, 16281, 24850, 36421, 51624, 71149, 95746, 126225, 163456, 208369, 261954, 325261, 399400, 485541, 584914, 698809, 828576, 975625, 1141426, 1327509, 1535464, 1766941, 2023650, 2307361, 2619904, 2963169, 3339106, 3749725, 4197096

Offset: 0

Ilya Gutkovskiy, Apr 09 2016

Partial sums of centered dodecahedral numbers (A005904).

OEIS Wiki, Centered Platonic numbers
Eric Weisstein's World of Mathematics, Platonic Solid
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A005904, A006566, A008354, A014820, A037270, A132366.

[(n + 1)^2*(5*n^2 + 10*n + 2)/2 : n in [0..50]]; // Wesley Ivan Hurt, Oct 15 2017

A269237:=n->(n + 1)^2*(5*n^2 + 10*n + 2)/2: seq(A269237(n), n=0..50); # Wesley Ivan Hurt, Oct 15 2017

Table[(n + 1)^2 ((5 n^2 + 10 n + 2)/2), {n, 0, 35}]
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 34, 189, 616, 1525}, 36]

x='x+O('x^99); Vec((1+29*x+29*x^2+x^3)/(1-x)^5) \\ Altug Alkan, Apr 10 2016

G.f.: (1 + 29*x + 29*x^2 + x^3)/(1 - x)^5.
E.g.f.: exp(x)*(2 + 66*x + 122*x^2 + 50*x^3 + 5*x^4)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
Sum_{n>=0} 1/a(n) = (5 - Pi^2 - sqrt(15)*Pi*cot(sqrt(3/5)*Pi))/9 = 1.0377796966... . - Vaclav Kotesovec, Apr 10 2016

A329650 Numbers that are sums of consecutive centered icosahedral numbers (A005902).

Original entry on oeis.org

1, 13, 14, 55, 68, 69, 147, 202, 215, 216, 309, 456, 511, 524, 525, 561, 870, 923, 1017, 1072, 1085, 1086, 1415, 1484, 1793, 1940, 1995, 2008, 2009, 2057, 2338, 2869, 2899, 3208, 3355, 3410, 3423, 3424, 3472, 3871, 4395, 4926, 4956, 5083, 5265, 5412, 5467, 5480, 5481, 6341, 6525, 6740

Offset: 1

Ilya Gutkovskiy, Nov 18 2019

nonn

Cf. A005902, A008354, A329597, A329634, A329635, A329636.

cp's OEIS Frontend

A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

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A269237 a(n) = (n + 1)^2(5n^2 + 10*n + 2)/2.

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A329650 Numbers that are sums of consecutive centered icosahedral numbers (A005902).

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cp's OEIS Frontend

A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

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GAP

Maple

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Formula

A269237 a(n) = (n + 1)^2*(5*n^2 + 10*n + 2)/2.

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Author

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Magma

Maple

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PARI

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A329650 Numbers that are sums of consecutive centered icosahedral numbers (A005902).

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Author

Keywords

Crossrefs

A269237 a(n) = (n + 1)^2(5n^2 + 10*n + 2)/2.