A134291 Tenth column (and diagonal) of Narayana triangle A001263.

1, 55, 1210, 15730, 143143, 1002001, 5725720, 27810640, 118195220, 449141836, 1551580888, 4936848280, 14620666060, 40648664980, 106847919376, 267119798440, 638337753625, 1464421905375, 3237143159250, 6917263803450

Offset: 0

Author

Wolfdieter Lang, Nov 13 2007

Keywords

• nonn
• easy

Comments

See a comment under A134288 on the coincidence of column and diagonal sequences.

Kekulé numbers K(O(1,9,n)) for certain benzenoids (see the Cyvin-Gutman reference, p. 105, eq. (i)).

References

• S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 25.

Crossrefs

Cf. A001263, A002378, A134288.

Cf. A134290 (ninth column of Narayana triangle).

Programs

GAP

List([0..30], n-> Binomial(n+10,10)*Binomial(n+9,8)/9); # G. C. Greubel, Aug 28 2019

Magma

[Binomial(n+10,10)*Binomial(n+9,8)/9: n in [0..30]]; // G. C. Greubel, Aug 28 2019

Maple

a := n -> ((n+1)*((n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8)*(n+9))^2*(n+10))/ 1316818944000:
seq(a(n), n=0..19); # Peter Luschny, Sep 01 2016

Mathematica

Table[Binomial[n + 10, 10]*Binomial[n + 10, 9]/(n + 10), {n, 0, 30}] (* Wesley Ivan Hurt, Apr 25 2017 *)

PARI

vector(30, n, binomial(n+9,10)*binomial(n+8,8)/9) \\ G. C. Greubel, Aug 28 2019

Sage

[binomial(n+10,10)*binomial(n+9,8)/9 for n in (0..30)] # G. C. Greubel, Aug 28 2019

Formula

a(n) = A001263(n+10,10) = binomial(n+10,10)*binomial(n+10,9)/(n+10).

O.g.f.: P(9,x)/(1-x)^19 with the numerator polynomial P(9,x) = Sum_{k=1..9} A001263(9,k)*x^(k-1), the ninth row polynomial of the Narayana triangle: P(9,x) = 1 + 36*x + 336*x^2 + 1176*x^3 + 1764*x^4 + 1176*x^5 + 336*x^6 + 36*x^7 + x^8.

a(n) = Product_{i=1..9} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016

From Amiram Eldar, Oct 19 2020: (Start)

Sum_{n>=0} 1/a(n) = 2987553139/196 - 1544400*Pi^2.

Sum_{n>=0} (-1)^n/a(n) = 1179648*log(2)/7 - 114472793/980. (End)