OFFSET
1,2
COMMENTS
In general, the r-th successive summation of the fourth powers from 1 to n = (2*n+r)*(12*n^2+12*n*r+r^2-5*r)*(r+n)!/((r+4)!*(n-1)!). Here r = 3. - Gary Detlefs, Mar 01 2013
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = (n*(1+n)*(2+n)*(3+n)*(3+2*n)*(-1+2*n*(3+n)))/840.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^8. [Colin Barker, Apr 04 2012]
a(n)= (2*n+3)*(12*n^2+36*n-6)*(n+3)!/(5040*(n-1)!), n>0 - Gary Detlefs, Mar 01 2013
MATHEMATICA
Nest[Accumulate, Range[50]^4, 3] (* Paolo Xausa, Jun 17 2024 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
EXTENSIONS
Edited by Ralf Stephan, Dec 16 2004
STATUS
approved