A153977 One-fourth of partial sums of A153976.

2, 9, 27, 65, 135, 252, 434, 702, 1080, 1595, 2277, 3159, 4277, 5670, 7380, 9452, 11934, 14877, 18335, 22365, 27027, 32384, 38502, 45450, 53300, 62127, 72009, 83027, 95265, 108810, 123752, 140184, 158202, 177905, 199395, 222777, 248159

Offset

1,1

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(0)=2, a(1)=9, a(2)=27, a(3)=65, a(4)=135; for n>4, a(n) = 5*a(n-1)-10*a(n-2)+ 10*a(n-3)-5*a(n-4)+a(n-5). - Harvey P. Dale, Aug 02 2011

a(n) = (A000217(n-1)^2 + A000217(n+1)^2 - 1)/4. - Richard R. Forberg, Dec 25 2013

Recurrence: (n-1)*(n^2 - n + 6)*a(n) = (n+1)*(n^2 + n + 6)*a(n-1). - Vaclav Kotesovec, Dec 26 2013

a(n) = A000217(A000217(n)) + A000217(n). - Bruno Berselli, May 28 2015

a(n) = (A000217(n)^2 + 3*A000217(n))/2 where A000217(n) is the n-th triangular number. - Frederic Isenmann, Feb 04 2017

Sum_{n>=1} 1/a(n) = 14/9 - 4*tanh(sqrt(23)*Pi/2)*Pi/(3*sqrt(23)). - Amiram Eldar, Aug 23 2022

Maple

A153977:=n->(1/4)*sum(i^3 + (i+2)^3, i=0..n): seq(A153977(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Mathematica

a[n_]:=n^3; lst={}; s=0; Do[s+=(a[n]+a[n+2]); AppendTo[lst, s/4], {n, 0, 6!}]; lst
Accumulate[Array[#^3+(#+2)^3&, 40, 0]]/4 (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {2, 9, 27, 65, 135}, 40] (* Harvey P. Dale, Aug 02 2011 *)

Prog

(PARI) a(n)=(n^4 + 2*n^3 + 7*n^2 + 6*n)/8 \\ Charles R Greathouse IV, Feb 06 2017

Crossrefs

Cf. A000217, A000537, A000578, A003215, A005898, A006007, A027602, A153976.

Sequence in context: A182325 A008910 A239059 * A051746 A277240 A256233

Adjacent sequences: A153974 A153975 A153976 * A153978 A153979 A153980

Keyword

nonn,easy

Author

Vladimir Joseph Stephan Orlovsky, Jan 03 2009

Status

approved