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A153977
One-fourth of partial sums of A153976.
2
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
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
KEYWORD
nonn,easy
AUTHOR
STATUS
approved