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A280472
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Number of ways to write n as the sum of an octagonal number (A000567), a second octagonal number (A045944), and a strict partition number (A000009).
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3
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1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 4, 3, 6, 4, 4, 4, 4, 6, 3, 4, 4, 6, 6, 4, 3, 4, 5, 3, 4, 5, 5, 3, 7, 7, 4, 4, 5, 7, 6, 5, 7, 4, 6, 5, 2, 6, 4, 4, 3, 7, 4, 4, 6, 9, 7, 4, 8, 4, 6, 4, 6, 7, 5, 6, 5, 6, 9, 3, 5, 6, 5, 5, 7, 6, 6
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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.
On 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.
See also A280455 for a similar conjecture of the author involving the partition function.
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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*0-2) + 0*(3*0+2) + A000009(2).
a(50) = 2 since 50 = 4*(3*4-2) + 1*(3*1+2) + A000009(7) = 4*(3*4-2) + 0*(3*0+2) + A000009(10).
a(1399) = 1 since 1399 = 1*(3*1-2) + 18*(3*18+2) + A000009(32).
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MATHEMATICA
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Oct[n_]:=Oct[n]=IntegerQ[Sqrt[3n+1]]&&Mod[Sqrt[3n+1], 3]==1;
q[n_]:=q[n]=PartitionsQ[n];
Do[r=0; m=2; Label[bb]; If[q[m]>n, Goto[cc]]; Do[If[Oct[n-q[m]-x(3x-2)], r=r+1], {x, 0, (Sqrt[3(n-q[m])+1]+1)/3}]; m=m+If[m<3, 2, 1]; Goto[bb]; Label[cc]; Print[n, " ", r]; 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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