A101090 Third partial sums of fourth powers (A000583).

1, 19, 135, 605, 2054, 5778, 14178, 31350, 63855, 121693, 219505, 378027, 625820, 1001300, 1555092, 2352732, 3477741, 5035095, 7155115, 9997801, 13757634, 18668870, 25011350, 33116850, 43375995, 56245761, 72257589, 92026135, 116258680, 145765224, 181469288

Offset: 1

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004

Keywords

• easy
• nonn
• changed

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

Crossrefs

Cf. A101091, A101089.

Programs

Mathematica

Nest[Accumulate, Range[50]^4, 3] (* Paolo Xausa, Jun 17 2024 *)

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)!). - Gary Detlefs, Mar 01 2013

Extensions

Edited by Ralf Stephan, Dec 16 2004