A234042 a(n) = binomial(n+4,4)*gcd(n,5)/5.

1, 1, 3, 7, 14, 126, 42, 66, 99, 143, 1001, 273, 364, 476, 612, 3876, 969, 1197, 1463, 1771, 10626, 2530, 2990, 3510, 4095, 23751, 5481, 6293, 7192, 8184, 46376, 10472, 11781, 13209, 14763, 82251, 18278, 20254, 22386, 24682, 135751, 29799, 32637, 35673, 38916

Offset

0,3

Comments

The sixth column of the triangle A107711.

Links

Indranil Ghosh, Table of n, a(n) for n = 0..10000

Formula

a(n) = A107711(n+5,5) = binomial(n+5,5)*gcd(n,5)/(n+5), with n >= 0.

O.g.f.: ((1+x^20) + x*(1+x^18) + 3*x^2*(1+x^16) + 7*x^3*(1+x^14) + 14*x^4*(1+x^12) + 121*x^5*(1+x^10)+37*x^6*(1+x^8) + 51*x^7*(1+x^6) + 64*x^8*(1+x^4) + 73*x^9*(1+x^2) + 381*x^10)/(1-x^5)^5. From the 5-section using n = 5*k + j, for j = 0, 1, 2, 3, 4.

Sum_{n>=0} 1/a(n) = 20/3 - 16*sqrt(10-22/sqrt(5))*Pi/5. - Amiram Eldar, Sep 20 2022

Mathematica

a[n_] := Binomial[n + 4, 4] * GCD[n, 5]/5; Table[a[n], {n, 0, 40}] (* Amiram Eldar, Sep 20 2022 *)

Prog

(PARI) a(n) = binomial(n+4, 4)*gcd(n, 5)/5 \\ Charles R Greathouse IV, Feb 16 2017

Crossrefs

Cf. A107711, A208950 (fifth column of A107711), A109009 (gcd(n,5)).

Sequence in context: A146155 A106363 A128658 * A001203 A154883 A302029

Adjacent sequences: A234039 A234040 A234041 * A234043 A234044 A234045

Keyword

nonn,easy

Author

Wolfdieter Lang, Feb 24 2014

Status

approved