A303814 Generalized 24-gonal (or icositetragonal) numbers: m*(11*m - 10) with m = 0, +1, -1, +2, -2, +3, -3, ...

0, 1, 21, 24, 64, 69, 129, 136, 216, 225, 325, 336, 456, 469, 609, 624, 784, 801, 981, 1000, 1200, 1221, 1441, 1464, 1704, 1729, 1989, 2016, 2296, 2325, 2625, 2656, 2976, 3009, 3349, 3384, 3744, 3781, 4161, 4200, 4600, 4641, 5061, 5104, 5544, 5589, 6049, 6096, 6576, 6625

Offset: 0

Author

Omar E. Pol, Jun 06 2018

Keywords

• nonn
• easy

Comments

a(25) = 1729 is the Hardy-Ramanujan number.

Numbers k such that 11*k + 25 is a square. - Bruno Berselli, Jun 08 2018

Partial sums of A317320. - Omar E. Pol, Jul 28 2018

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A051876, A317320.

Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), this sequence (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

Programs

Mathematica

With[{pp = 24, nn = 55}, {0}~Join~Riffle[Array[PolygonalNumber[pp, #] &, Ceiling[nn/2]], Array[PolygonalNumber[pp, -#] &, Ceiling[nn/2]]]] (* Michael De Vlieger, Jun 06 2018 *)
Table[(22 n (n + 1) + 9 (2 n + 1) (-1)^n - 9)/8, {n, 0, 50}] (* Bruno Berselli, Jun 08 2018 *)
CoefficientList[ Series[-x (x^2 + 20x + 1)/((x - 1)^3 (x + 1)^2), {x, 0, 50}], x] (* or *)
LinearRecurrence[{1, 2, -2, -1, 1}, {0, 1, 21, 24, 64}, 50] (* Robert G. Wilson v, Jul 28 2018 *)

PARI

concat(0, Vec(x*(1 + 20*x + x^2)/((1 + x)^2*(1 - x)^3) + O(x^40))) \\ Colin Barker, Jun 12 2018

Formula

From Bruno Berselli, Jun 08 2018: (Start)

G.f.: x*(1 + 20*x + x^2)/((1 + x)^2*(1 - x)^3).

a(n) = a(-n-1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).

a(n) = (22*n*(n + 1) + 9*(2*n + 1)*(-1)^n - 9)/8. Therefore:

a(n) = n*(11*n + 20)/4, if n is even, or (n + 1)*(11*n - 9)/4 otherwise.

(2*n - 1)*a(n) + (2*n + 1)*a(n-1) - n*(11*n^2 - 10) = 0. (End)

Sum_{n>=1} 1/a(n) = (11 + 10*Pi*cot(Pi/11))/100. - Amiram Eldar, Mar 01 2022