A002561 a(n) = n^5 + 1.

0, 1, 2, 33, 244, 1025, 3126, 7777, 16808, 32769, 59050, 100001, 161052, 248833, 371294, 537825, 759376, 1048577, 1419858, 1889569, 2476100, 3200001, 4084102, 5153633, 6436344, 7962625, 9765626, 11881377

Offset

-1,3

Links

Vincenzo Librandi, Table of n, a(n) for n = -1..1000

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

Formula

a(n) = A000584(n) + 1, for n >= 0.

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), n > 4. - Harvey P. Dale, Aug 13 2013

G.f.: ( 1 - 4*x + 36*x^2 + 56*x^3 + 31*x^4 ) / (x-1)^6. - R. J. Mathar, Jul 28 2014

E.g.f.: (1 + x + 15*x^2 + 25*x^3 + 10*x^4 + x^5)*exp(x). - G. C. Greubel, Oct 24 2018

Sum_{n>=1} 1/a(n) = 1/2 + Sum_{n>=1} (-1)^(n+1) * (zeta(5*n) - 1) = 0.5359628431... - Amiram Eldar, Nov 06 2020

Mathematica

Table[n^5+1, {n, -1, 40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 2, 33, 244, 1025}, 40] (* Harvey P. Dale, Aug 13 2013 *)

Prog

(Magma) [n^5+1: n in [-1..30]]; // Vincenzo Librandi, Jun 07 2013
(PARI) a(n)=n^5+1 \\ Charles R Greathouse IV, Oct 07 2015
(GAP) List([1..30], n->n^5+1); # Muniru A Asiru, Oct 23 2018
(Sage) [n^5 + 1 for n in (-1..30)] # G. C. Greubel, Nov 20 2018

Crossrefs

Cf. A000584.

Sequence in context: A362538 A006558 A228542 * A181547 A030448 A093992

Adjacent sequences: A002558 A002559 A002560 * A002562 A002563 A002564

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved