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Expansion of (Sum_{k>=2} x^(k^2))^3.
(history; published version)
#4 by N. J. A. Sloane at Mon Jan 16 14:19:14 EST 2017
STATUS

proposed

approved

#3 by Ilya Gutkovskiy at Mon Jan 16 05:56:54 EST 2017
STATUS

editing

proposed

#2 by Ilya Gutkovskiy at Mon Jan 16 05:14:56 EST 2017
NAME

allocated for Ilya Gutkovskiy

Expansion of (Sum_{k>=2} x^(k^2))^3.

DATA

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 3, 0, 3, 0, 0, 1, 0, 6, 0, 0, 0, 3, 3, 0, 3, 0, 6, 0, 0, 3, 0, 3, 3, 6, 0, 0, 1, 6, 6, 0, 0, 0, 6, 0, 6, 6, 0, 3, 0, 6, 6, 0, 0, 6, 3, 3, 3, 6, 6, 0, 3, 0, 6, 1, 3, 12, 6, 0, 0, 6, 3, 6, 6, 0, 3, 0, 3, 15, 6, 0, 0, 6, 12, 0, 3, 3, 6, 6, 0, 12, 3, 0, 6, 6

OFFSET

0,18

COMMENTS

Number of ways to write n as an ordered sum of 3 squares > 1.

LINKS

<a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>

FORMULA

G.f.: (Sum_{k>=2} x^(k^2))^3.

G.f.: (1/8)*(-1 - 2*x + theta_3(0,x))^3, where theta_3 is the 3rd Jacobi theta function.

EXAMPLE

G.f. = x^12 + 3*x^17 + 3*x^22 + 3*x^24 + x^27 + 6*x^29 + 3*x^33 + 3*x^34 + 3*x^36 + ...

a(17) = 3 because we have [9, 4, 4], [4, 9, 4] and [4, 4, 9].

MATHEMATICA

nmax = 105; CoefficientList[Series[Sum[x^k^2, {k, 2, nmax}]^3, {x, 0, nmax}], x]

CoefficientList[Series[(-1 - 2 x + EllipticTheta[3, 0, x])^3/8, {x, 0, 105}], x]

KEYWORD

allocated

nonn

AUTHOR

Ilya Gutkovskiy, Jan 16 2017

STATUS

approved

editing

#1 by Ilya Gutkovskiy at Mon Jan 16 05:14:56 EST 2017
NAME

allocated for Ilya Gutkovskiy

KEYWORD

allocated

STATUS

approved