A253711 Second partial sums of 11th powers (A008455).

1, 2050, 181246, 4554746, 57756371, 473755052, 2867080476, 13850340492, 56214660117, 198578979742, 626254969978, 1796939330902, 4759784085863, 11772194010488, 27434359794488, 60688711622904, 128214959758953, 260009617974234, 508294535087734, 961379452201234, 1764741869856955, 3152422588924004, 5492913065904980

Offset: 1

Author

Luciano Ancora, Jan 10 2015

Keywords

• nonn
• easy

Comments

The formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) are the m-th Faulhaber's formulas.

Links

• Luciano Ancora, Recurrence relation for the second partial sums of m-th powers
• Luciano Ancora, Second partial sums of the m-th powers
• Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Crossrefs

Cf. A008455.

Programs

Mathematica

Table[n (n + 1) (n + 2) (70 n^10 + 700 n^9 + 2310 n^8 + 1680 n^7 - 4655 n^6 - 4410 n^5 + 8240 n^4 + 4120 n^3 - 7819 n^2 + 202 n + 1382)/10920, {n, 1, 20}] (* Vincenzo Librandi, Jan 15 2015 *)
RecurrenceTable[{a[n] == 2 a[n - 1] - a[n - 2] + n^11, a[1] == 1, a[2] == 2050}, a, {n, 1, 20}] (* Bruno Berselli, Jan 15 2015 *)

Formula

a(n) = n*(n+1)*(n+2)*(70*n^10 + 700*n^9 + 2310*n^8 + 1680*n^7 - 4655*n^6 - 4410*n^5 + 8240*n^4 + 4120*n^3 - 7819*n^2 + 202*n + 1382)/10920.

a(n) = 2*a(n-1) - a(n-2) + n^11.

G.f.: x*(1 + 2036*x + 152637*x^2 + 2203488*x^3 + 9738114*x^4 + 15724248*x^5 + 9738114*x^6 + 2203488*x^7 + 152637*x^8 + 2036*x^9 + x^10) / (1 - x)^14. - Vincenzo Librandi, Jan 15 2015