A027803 a(n) = 35*(n+1)*binomial(n+4, 7)/4.

35, 350, 1890, 7350, 23100, 62370, 150150, 330330, 675675, 1301300, 2382380, 4176900, 7054320, 11531100, 18314100, 28352940, 42902475, 63596610, 92534750, 132382250, 186486300, 259008750, 355077450, 480957750, 644245875

Offset: 3

Author

Thi Ngoc Dinh (via R. K. Guy)

Keywords

• nonn
• easy

Comments

Number of 12-subsequences of [ 1, n ] with just 4 contiguous pairs.

Links

• G. C. Greubel, Table of n, a(n) for n = 3..1000
• Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Crossrefs

Cf. A053347, A062145.

Programs

Magma

A027803:= func< n | 5*(n+1)*(n+4)*Binomial(n+3,6)/4 >;
[A027803(n): n in [3..45]]; // G. C. Greubel, Mar 11 2025

Mathematica

Table[35 (n+1) Binomial[n+4,7]/4, {n,3,30}] (* or *) Table[Binomial[n+1, 4] Binomial[n+4,4], {n,3,30}] (* Michael De Vlieger, Mar 16 2016 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1}, {35,350,1890,7350,23100, 62370,150150,330330,675675}, 30] (* Harvey P. Dale, May 07 2022 *)

SageMath

def A027803(n): return binomial(n+1,4)*binomial(n+4,4)
print([A027803(n) for n in range(3,46)]) # G. C. Greubel, Mar 11 2025

Formula

a(n) = 35*A053347(n-3).

G.f.: 35*x^3*(1+x)/(1-x)^9.

a(n) = C(n+1, 4)*C(n+4, 4). - Zerinvary Lajos, May 10 2005, corrected by R. J. Mathar, Mar 16 2016

From Amiram Eldar, Jan 25 2022: (Start)

Sum_{n>=3} 1/a(n) = 5929/225 - 8*Pi^2/3.

Sum_{n>=3} (-1)^(n+1)/a(n) = 4*Pi^2/3 - 197/15. (End)

E.g.f.: (1/576)*x^3*(3360 + 5040*x + 2352*x^2 + 448*x^3 + 36*x^4 + x^5 )*exp(x). - G. C. Greubel, Mar 11 2025