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%I #14 Aug 17 2018 10:51:29
%S 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,
%T 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,
%U 6,5,6,9,3,5,6,5,5,7,6,6
%N 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).
%C Conjecture: (i) a(n) > 0 for all n > 0.
%C (ii) lim_n a(n)/(log n)^2 = 1/Pi^2.
%C 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.
%C See also A280455 for a similar conjecture of the author involving the partition function.
%H Zhi-Wei Sun, <a href="/A280472/b280472.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/160p.pdf">Problems on combinatorial properties of primes</a>, 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.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/170o.pdf">A result similar to Lagrange's theorem</a>, J. Number Theory 162(2016), 190-211.
%e a(1) = 1 since 1 = 0*(3*0-2) + 0*(3*0+2) + A000009(2).
%e 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).
%e a(1399) = 1 since 1399 = 1*(3*1-2) + 18*(3*18+2) + A000009(32).
%t Oct[n_]:=Oct[n]=IntegerQ[Sqrt[3n+1]]&&Mod[Sqrt[3n+1],3]==1;
%t q[n_]:=q[n]=PartitionsQ[n];
%t 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}]
%Y Cf. A000009, A000567, A045944, A280386, A280455.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Jan 04 2017
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