A016781 a(n) = (3*n+1)^5.

1, 1024, 16807, 100000, 371293, 1048576, 2476099, 5153632, 9765625, 17210368, 28629151, 45435424, 69343957, 102400000, 147008443, 205962976, 282475249, 380204032, 503284375, 656356768, 844596301, 1073741824, 1350125107, 1680700000, 2073071593, 2535525376

Offset

0,2

Comments

In general the e.g.f. of {(1 + 3*m)^n}_{m>=0} is E(n,x) = exp(x)*Sum_{m=0..n} A282629(n, m)*x^m, and the o.g.f. is G(n, x) = (Sum_{m=0..n} A225117(n, n-m}*x^m)/(1-x)^(n+1). - Wolfdieter Lang, Apr 02 2017

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, May 13 2012

From Wolfdieter Lang, Apr 02 2017: (Start)

O.g.f.: (1+1018*x+10678*x^2+14498*x^3+2933*x^4+32*x^5)/(1-x)^6.

E.g.f: exp(x)*(1+1023*x+7380*x^2+8775*x^3+2835*x^4+243*x^5). (End)

a(n) = A000584(A016777(n)). - Michel Marcus, Apr 06 2017

Sum_{n>=0} 1/a(n) = 2*Pi^5/(3^6*sqrt(3)) + 121*zeta(5)/3^5. - Amiram Eldar, Mar 29 2022

Mathematica

(3Range[0, 20]+1)^5 (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 1024, 16807, 100000, 371293, 1048576}, 30] (* Harvey P. Dale, May 13 2012 *)

Prog

(Magma) [(3*n+1)^5: n in [0..30]]; // Vincenzo Librandi, Sep 21 2011
(Maxima) A016781(n):=(3*n+1)^5$
makelist(A016781(n), n, 0, 20); /* Martin Ettl, Nov 12 2012 */

Crossrefs

Cf. A016777, A016778, A016779, A016780, A225117, A282629.

Sequence in context: A138334 A268124 A205611 * A247933 A344306 A205349

Adjacent sequences: A016778 A016779 A016780 * A016782 A016783 A016784

Keyword

nonn,easy

Author

N. J. A. Sloane.

Status

approved