A101855 a(n) = n*(n+1)*(n+2)*(n+4)*(n+23)/120.

6, 30, 91, 216, 441, 812, 1386, 2232, 3432, 5082, 7293, 10192, 13923, 18648, 24548, 31824, 40698, 51414, 64239, 79464, 97405, 118404, 142830, 171080, 203580, 240786, 283185, 331296, 385671, 446896, 515592, 592416, 678062, 773262, 878787, 995448

Offset: 1

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004

Keywords

• easy
• nonn

Comments

6th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 3: 1,4,1. 6th row of the array shown in A101853. Partial sums of A101854.

Links

• C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Programs

Mathematica

Table[n(n+1)(n+2)(n+4)(n+23)/120,{n,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{6,30,91,216,441,812},40](* Harvey P. Dale, Feb 07 2013 *)

Formula

G.f.: x*(6-6*x+x^2) / (x-1)^6. - R. J. Mathar, Dec 06 2011

a(1)=6, a(2)=30, a(3)=91, a(4)=216, a(5)=441, a(6)=812, 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). - Harvey P. Dale, Feb 07 2013