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A280455
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Number of ways to write n as x*(3x-1)/2 + y*(3y+1)/2 + p(z), where x,y,z are nonnegative integers with z > 0, and p(.) is the partition function given by A000041.
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4
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1, 2, 3, 3, 3, 3, 3, 5, 4, 5, 2, 4, 4, 5, 5, 4, 6, 5, 5, 3, 4, 6, 7, 3, 5, 3, 8, 3, 8, 7, 6, 5, 4, 7, 3, 4, 6, 8, 4, 5, 4, 12, 5, 8, 5, 6, 4, 5, 8, 5, 4, 7, 7, 6, 5, 7, 8, 5, 9, 6, 6, 5, 10, 8, 6, 3, 7, 8, 7, 4, 6, 7, 9, 3, 5, 4, 8, 7, 9, 13
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OFFSET
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1,2
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 0.
(ii) lim_n a(n)/(log n)^2 = 1/Pi^2.
This is similar to the author's conjecture in A280386. At the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified part (i) of the above conjecture for n up to 10^9.
We also have some other similar conjectures. For example, we conjecture that any positive integer can be expressed as the sum of two triangular numbers and a partition number.
As the main term of log p(n) is Pi*sqrt(2n/3), the partition function p(n) eventually grows faster than any polynomial.
See also A280472 for a similar conjecture.
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LINKS
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Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
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EXAMPLE
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a(1) = 1 since 1 = 0*(3*0-1)/2 + 0*(3*0+1)/2 + p(1).
a(2) = 2 since 2 = 1*(3*1-1)/2 + 0*(3*0+1)/2 + p(1) = 0*(3*0-1)/2 + 0*(3*0+1)/2 + p(2).
a(2771) = 1 since 2771 = 35*(3*35-1)/2 + 25*(3*25+1)/2 + p(1).
a(9426) = 1 since 9426 = 4*(3*4-1)/2 + 79*(3*79+1)/2 + p(3).
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
p[n_]:=p[n]=PartitionsP[n];
Pen[n_]:=Pen[n]=SQ[24n+1]&&Mod[Sqrt[24n+1], 6]==1;
Do[r=0; m=1; Label[bb]; If[p[m]>n, Goto[cc]]; Do[If[Pen[n-p[m]-x(3x-1)/2], r=r+1], {x, 0, (Sqrt[24(n-p[m])+1]+1)/6}]; m=m+1; Goto[bb]; Label[cc]; Print[n, " ", r]; Label[aa]; Continue, {n, 1, 80}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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