A164765 Partial sums of [A080782^2].

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

1,2

Comments

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.

Links

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).

Formula

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)

Mathematica

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

Prog

(PARI) 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

Crossrefs

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

Sequence in context: A162899 A115391 A241162 * A116955 A192291 A144967

Adjacent sequences: A164762 A164763 A164764 * A164766 A164767 A164768

Keyword

easy,nonn

Author

Carl R. White, Aug 25 2009

Status

approved