A308662 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

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

Offset: 1

Author

Zhi-Wei Sun, Jun 15 2019

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n > 0.

Conjecture 2: Let r be 1 or 2. Then, any positive integer n can be written as (2^a*5^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.

We have verified Conjectures 1 and 2 for all n = 1..10^8.

Examples

			a(3) = 1 with 3 = (2^0*5^0)^2 + (-1)*(3*(-1)+1) + 0*(3*0+2).
a(7) = 1 with 7 = (2^1*5^0)^2 + (-1)*(3*(-1)+1) + (-1)*(3*(-1)+2).
a(10) = 1 with 10 = (2^0*5^0)^2 + 1*(3*1+1) + 1*(3*1+2).
a(52) = 1 with 52 = (2^0*5^0)^2 + 3*(3*3+1) + (-3)*(3*(-3)+2).
a(98) = 1 with 98 = (2^0*5^1)^2 + 4*(3*4+1) + (-3)*(3*(-3)+2).
a(14596) = 1 with 14596 = (2^3*5^0)^2 + (-36)*(3*(-36)+1) + (-60)*(3*(-60)+2).
a(22163) = 1 with 22163 = (2^3*5^0)^2 + 66*(3*66+1) + (-55)*(3*(-55)+2).
a(150689) = 1 with 150689 = (2^6*5^1)^2 + 117*(3*117+1) + (-49)*(3*(-49)+2).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), No. 7, 1367-1396.

Crossrefs

Cf. A000079, A000351, A001082, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661.

Programs

Mathematica

OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]];
tab={};Do[r=0;Do[If[OctQ[n-4^a*25^b-x(3x+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{x,-Floor[(Sqrt[12(n-4^a*25^b)+1]+1)/6],(Sqrt[12(n-4^a*25^b)+1]-1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]