A308584 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c*8^d, where a,b,c,d are nonnegative integers with a <= b.

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

Offset: 1

Author

Zhi-Wei Sun, Jun 08 2019

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0. Equivalently, each n = 1,2,3,... can be written as w^2 + x*(x+1) + 5^y*8^z with w,x,y,z nonnegative integers.

We have verified a(n) > 0 for all n = 1..4*10^8.

See also A308566 for a similar conjecture.

a(n) > 0 for all 0 < n < 10^10. - Giovanni Resta, Jun 10 2019

Examples

			a(13) = 1 with 13 = 3*4/2 + 3*4/2 + 5^0*8^0.
a(48) = 1 with 48 = 5*6/2 + 7*8/2 + 5^1*8^0.
a(87) = 1 with 87 = 1*2/2 + 12*13/2 + 5^0*8^1.
a(90) = 1 with 90 = 4*5/2 + 10*11/2 + 5^2*8^0.
a(423) = 1 with 423 = 9*10/2 + 22*23/2 + 5^3*8^0.
a(517) = 1 with 517 = 17*18/2 + 24*25/2 + 5^0*8^2.
a(985) = 1 with 985 = 19*20/2 + 34*35/2 + 5^2*8^1.
a(2694) = 1 with 2694 = 7*8/2 + 68*69/2 + 5^1*8^2.
a(42507) = 1 with 42507 = 178*179/2 + 223*224/2 + 5^2*8^2.
a(544729) = 1 with 544729 = 551*552/2 + 857*858/2 + 5^5*8^1.
a(913870) = 1 with 913870 = 559*560/2 + 700*701/2 + 5^3*8^4.
a(1843782) = 1 with 1843782 = 808*809/2 + 1668*1669/2 + 5^6*8^1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Crossrefs

Cf. A000217, A000351, A001018, A303656, A303637, A308411, A308547, A308566.

Programs

Mathematica

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