cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A298375 Partial sums of A230584.

Original entry on oeis.org

2, 5, 11, 18, 29, 43, 61, 84, 111, 145, 183, 230, 281, 343, 409, 488, 571, 669, 771, 890, 1013, 1155, 1301, 1468, 1639, 1833, 2031, 2254, 2481, 2735, 2993, 3280, 3571, 3893, 4219, 4578, 4941, 5339, 5741, 6180, 6623, 7105, 7591, 8118, 8649, 9223, 9801, 10424
Offset: 1

Views

Author

Gerald Hillier, Jan 18 2018

Keywords

Examples

			For n = 5 then a(5) = 2+3+6+7+11 = 29.
		

Crossrefs

Cf. A230584.

Programs

  • Mathematica
    CoefficientList[ Series[(2 + x - x^2 - x^3 + 2x^5 - x^6)/((x -1)^4 (x + 1)^2), {x, 0, 50}], x] (* Robert G. Wilson v, Jan 18 2018 *)
  • PARI
    Vec(x*(2 + x - x^2 - x^3 + 2*x^5 - x^6) / ((1 - x)^4*(1 + x)^2) + O(x^50)) \\ Colin Barker, Jan 18 2018

Formula

Let g = n + ((n + 1) mod 2), then for n > 1,
a(n) = (g^3 + 6*g^2 + 11*g + 18) / 12 - If(n mod 2 = 1, 0, ((n + 2) / 2)^2 + 2).
From Colin Barker, Jan 18 2018: (Start)
G.f.: x*(2 + x - x^2 - x^3 + 2*x^5 - x^6) / ((1 - x)^4*(1 + x)^2).
a(n) = (n^3 + 6*n^2 + 14*n) / 12 for n>1 and even.
a(n) = (n^3 + 6*n^2 + 11*n + 18) / 12 for n>1 and odd.
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
(End)