A162899 Partial sums of [A052938(n)^2].

1, 10, 14, 30, 39, 64, 80, 116, 141, 190, 226, 290, 339, 420, 484, 584, 665, 786, 886, 1030, 1151, 1320, 1464, 1660, 1829, 2054, 2250, 2506, 2731, 3020, 3276, 3600, 3889, 4250, 4574, 4974, 5335, 5776, 6176, 6660, 7101, 7630, 8114, 8690, 9219, 9844, 10420

Offset: 0

Author

Rick L. Shepherd, Jul 16 2009

Keywords

• nonn
• easy

Comments

Another plausible solution, besides A115391 and A116955, to A115603: Each additional term of the partial sums here is the square of a number that alternately differs +2, -1, +2, -1, ..., from the previous number that is squared: a(3) = 30 = 1^2 + 3^2 + 2^2 + 4^2, where 1, 3, 2, 4 display this pattern.

Links

• Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Crossrefs

Cf. A115603, A115391, A116955.

Programs

Mathematica

LinearRecurrence[{2,1,-4,1,2,-1},{1,10,14,30,39,64},50] (* Harvey P. Dale, Sep 26 2020 *)

PARI

a(n) = sum(k=0, n, (if(k%2==0, k+2, k+5)/2)^2)

Formula

a(n) = sum(k=0..n, A052938(n)^2).

a(n) = (60-36*(-1)^n+(109-9*(-1)^n)*n+24*n^2+2*n^3)/24. G.f.: (4*x^4-4*x^3-7*x^2+8*x+1) / ((x-1)^4*(x+1)^2). - Colin Barker, Jul 18 2013