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A290943 Number of ways to write n as an ordered sum of 3 generalized pentagonal numbers (A001318). 1

%I

%S 1,3,6,7,6,6,7,12,12,12,9,6,12,12,18,13,12,18,12,18,12,13,18,12,24,12,

%T 12,24,21,30,12,18,18,12,24,18,19,18,24,24,18,24,36,24,18,19,18,24,24,

%U 30,18,12,36,30,24,21,18,36,24,36,24,12,36,36,36,18,25,30,24,24,24,30,24,36,30,24

%N Number of ways to write n as an ordered sum of 3 generalized pentagonal numbers (A001318).

%C Conjecture: every number is the sum of at most k - 4 generalized k-gonal numbers (for k >= 8).

%H Robert Israel, <a href="/A290943/b290943.txt">Table of n, a(n) for n = 0..10000</a>

%H Ilya Gutkovskiy, <a href="/A290943/a290943.pdf">Extended graphical example</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalNumber.html">Pentagonal Number</a>

%H <a href="/index/Pol#polygonal_numbers">Index to sequences related to polygonal numbers</a>

%F G.f.: (Sum_{k=-inf..inf} x^(k*(3*k-1)/2))^3.

%F G.f.: (Sum_{k>=0} x^A001318(k))^3.

%e a(6) = 7 because we have [5, 1, 0], [5, 0, 1], [2, 2, 2], [1, 5, 0], [1, 0, 5], [0, 5, 1] and [0, 1, 5].

%p N:= 100;

%p bds:= [fsolve(k*(3*k-1)/2 = N)];

%p G:= add(x^(k*(3*k-1)/2),k=floor(min(bds))..ceil(max(bds)))^3:

%p seq(coeff(G,x,n),n=0..N); # _Robert Israel_, Aug 16 2017

%t nmax = 75; CoefficientList[Series[Sum[x^(k (3 k - 1)/2), {k, -nmax, nmax}]^3, {x, 0, nmax}], x]

%t nmax = 75; CoefficientList[Series[Sum[x^((6 k^2 + 6 k + (-1)^(k + 1) (2 k + 1) + 1)/16), {k, 0, nmax}]^3, {x, 0, nmax}], x]

%t nmax = 75; CoefficientList[Series[EllipticTheta[4, 0, x^3]^3/QPochhammer[x, x^2]^3, {x, 0, nmax}], x]

%Y Cf. A001318, A002175, A008443, A080995, A093518, A093519, A255350, A255934, A256132, A256171, A280718.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Aug 14 2017

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Last modified February 20 16:25 EST 2018. Contains 299380 sequences. (Running on oeis4.)