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A290943 Number of ways to write n as an ordered sum of 3 generalized pentagonal numbers (A001318). 1
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, 12, 24, 21, 30, 12, 18, 18, 12, 24, 18, 19, 18, 24, 24, 18, 24, 36, 24, 18, 19, 18, 24, 24, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Conjecture: every number is the sum of at most k - 4 generalized k-gonal numbers (for k >= 8).
In 1830, Legendre showed that for each integer m>4 every integer N >= 28*(m-2)^3 can be written as the sum of five m-gonal numbers. In 1994 R. K. Guy proved that each natural number is the sum of three generalized pentagonal numbers. In a 2016 paper Zhi-Wei Sun proved that each natural number is the sum of four octagonal numbers. - Zhi-Wei Sun, Oct 03, 2020
LINKS
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Eric Weisstein's World of Mathematics, Pentagonal Number
FORMULA
G.f.: (Sum_{k=-infinity..infinity} x^(k*(3*k-1)/2))^3.
G.f.: (Sum_{k>=0} x^A001318(k))^3.
EXAMPLE
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].
MAPLE
N:= 100;
bds:= [fsolve(k*(3*k-1)/2 = N)];
G:= add(x^(k*(3*k-1)/2), k=floor(min(bds))..ceil(max(bds)))^3:
seq(coeff(G, x, n), n=0..N); # Robert Israel, Aug 16 2017
MATHEMATICA
nmax = 75; CoefficientList[Series[Sum[x^(k (3 k - 1)/2), {k, -nmax, nmax}]^3, {x, 0, nmax}], x]
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]
nmax = 75; CoefficientList[Series[EllipticTheta[4, 0, x^3]^3/QPochhammer[x, x^2]^3, {x, 0, nmax}], x]
CROSSREFS
Sequence in context: A256936 A021276 A329516 * A067753 A129023 A188883
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Aug 14 2017
STATUS
approved

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Last modified April 25 05:56 EDT 2024. Contains 371964 sequences. (Running on oeis4.)