A254470 Sixth partial sums of fourth powers (A000583).

1, 22, 198, 1134, 4884, 17226, 52338, 141570, 348777, 795652, 1701700, 3444948, 6651216, 12321804, 22011804, 38073948, 63985977, 104782986, 167620090, 262495090, 403165620, 608300550, 902911230, 1320114510, 1903286385, 2708672616, 3808530792, 5294887048

Offset: 1

Author

Luciano Ancora, Feb 15 2015

Keywords

• nonn
• easy

Examples

			First differences:   1, 15,  65, 175,  369,   671, ... (A005917)
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The fourth powers:   1, 16,  81, 256,  625,  1296, ... (A000583)
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First partial sums:  1, 17,  98, 354,  979,  2275, ... (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ... (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ... (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ... (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ... (A254681)
Sixth partial sums:  1, 22, 198,1134, 4884, 17226, ... (this sequence)
		

Links

• Luciano Ancora, Table of n, a(n) for n = 1..1000
• Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
• Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers.
• Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Crossrefs

Cf. A000538, A000583, A005917, A101089, A101090, A101091, A254469, A254471, A254472, A254681.

Programs

Magma

[n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(1+12*n+ 2*n^2)/302400: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Mathematica

Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (1 + 12 n + 2 n^2)/302400,{n, 25}] (* or *) CoefficientList[Series[(- 1 - 11 x - 11 x^2 - x^3)/(- 1 + x)^11, {x, 0, 24}], x]
Nest[Accumulate,Range[30]^4,6] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,22,198,1134,4884,17226,52338,141570,348777,795652,1701700},30] (* Harvey P. Dale, Apr 23 2016 *)

PARI

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400) \\ Derek Orr, Feb 19 2015

Formula

G.f.: (-x - 11*x^2 - 11*x^3 - x^4)/(- 1 + x)^11.

a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400.

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) + n^4.

Sum_{n>=1} 1/a(n) = 3320303/2601 + 1400*Pi^2/17 + (8960/17)*sqrt(2/17)*Pi*cot(sqrt(17/2)*Pi). - Amiram Eldar, Jan 26 2022