A308621 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c*5^(3d) with a,b,c,d nonnegative integers.

1, 2, 2, 3, 3, 2, 3, 5, 2, 4, 4, 3, 3, 5, 3, 3, 6, 4, 4, 5, 2, 6, 5, 4, 4, 5, 2, 4, 6, 3, 4, 8, 5, 3, 5, 4, 5, 8, 5, 5, 4, 3, 5, 7, 4, 5, 8, 4, 2, 8, 2, 6, 7, 4, 3, 4, 6, 5, 8, 4, 4, 6, 5, 5, 5, 5, 6, 8, 4, 6, 7, 4, 6, 10, 4, 4, 7, 5, 2, 10, 4, 7, 7, 4, 8, 4, 4, 7, 8, 2, 4, 9, 5, 5, 9, 5, 5, 7, 5, 5

Offset: 1

Author

Zhi-Wei Sun, Jun 11 2019

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0. Equivalently, any positive integer n can be written as a^2 + b*(b+1) + 2^c*5^(3d) with a,b,c,d nonnegative integers.

This was motivated by A308584, and we verified a(n) > 0 for all n = 1..2*10^8. Then, on the author's request, Giovanni Resta verified the above conjecture for n up to 10^10. G. Resta also noted that 7272774228 cannot be written as a*(a+1)/2 + b*(b+1)/2 + 2^c*250^d with a,b,c,d nonnegative integers.

See also A308566, A308584 and A308623 for similar conjectures.

Examples

			a(1) = 1 with 1 = 0*1/2 + 0*1/2 + 2^0*5^(3*0).
a(78210) = 1 with 78210 = 85*86/2 + 385*386/2 + 2*5^3.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000079, A000217, A000351, A308566, A308584, A308623.

Programs

Mathematica

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[TQ[n-5^(3k)*2^m-x(x+1)/2],r=r+1],{k,0,Log[5,n]/3},{m,0,Log[2,n/5^(3k)]},{x,0,(Sqrt[4(n-5^(3k)*2^m)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]