A116955 -id:A116955 - cp's OEIS frontend

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

Rick L. Shepherd, Jul 16 2009

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.

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

Cf. A115603, A115391, A116955.

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

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

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

A164765 Partial sums of [A080782^2].

Original entry on oeis.org

1, 10, 14, 30, 66, 91, 140, 221, 285, 385, 529, 650, 819, 1044, 1240, 1496, 1820, 2109, 2470, 2911, 3311, 3795, 4371, 4900, 5525, 6254, 6930, 7714, 8614, 9455, 10416, 11505, 12529, 13685, 14981, 16206, 17575, 19096, 20540, 22140, 23904, 25585

Offset: 1

Carl R. White, Aug 25 2009

Yet another plausible solution to A115603.
The first differences of A115603 are all squares (assuming a prior term of 0), meaning that any sequence beginning 1,3,2,4 is sufficient to account for them; This solution chooses the permutation of integers A080782 = {1,3,2,4,6,5,7,9,8,...}
Ultimately that means this sequence is equal to A000330 for every two out of three consecutive terms, and is greater by 2n+1 where different.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,2,-4,2,-1,2,-1).

Original puzzle: A115603; Used in this solution: A080782, A000330; Other solutions: A115391, A116955, A162899

Accumulate[Array[#+Mod[#+1,3]&,70,0]^2] (* Harvey P. Dale, Mar 29 2013 *)

Vec(x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2) + O(x^40)) \\ Colin Barker, Aug 03 2020

a(n) = ( n(n+1) + 6 - 8*sin^2(Pi*(n+1)/3) )*(2n+1)/6.
a(n) = Sum_{k=0..n} A080782(k)^2.
From Colin Barker, Aug 03 2020: (Start)
G.f.: x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2).
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5) - a(n-6) + 2*a(n-7) - a(n-8) for n>8.
(End)

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A115603 My teacher gave this as a "riddle".

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

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A164765 Partial sums of [A080782^2].

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