A280472 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).
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
Offset: 1
Keywords
Examples
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).
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- 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.
- Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Programs
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Mathematica
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}]
Comments