A308644 Number of ways to write n as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+1)/2, where a and b are nonnegative integers, and c and d are integers.

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

Offset: 1

Author

Zhi-Wei Sun, Jun 13 2019

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n > 0. Also, any positive integer n can be written as (3^a*5^b)^2 + c*(3c+1)/2 + d*(7d+3)/2, where a and b are nonnegative integers, and c and d are integers.

Conjecture 2: Let r be 1 or 3. Then, any positive integer n can be written as (3^a*4^b)^2 + c*(3c+1)/2 + d*(7d+r)/2, where a and b are nonnegative integers, and c and d are integers.

We have verified Conjectures 1-2 for all n = 1..10^6.

See also A308640, A308641 and A308656 for similar conjectures.

Examples

			a(152) = 1 with 152 = (3^0*5^0)^2 + (-4)*(3*(-4)+1)/2 + 6*(7*6+1)/2.
a(129894) = 1 with 129894 = (3^0*5^1)^2 + 154*(3*154+1)/2 + 164*(7*164+1)/2.
a(200963) = 1 with 200963 = (3^1*5^0)^2 + 364*(3*364+1)/2 + 24*(7*24+1)/2.
a(371278) = 1 with 371278 = (3^3*5^1)^2 + (-382)*(3*(-382)+1)/2 + (-196)*(7*(-196)+1)/2.
a(534699) = 1 with 534699 = (3^2*5^2)^2 + 543*(3*543+1)/2 + (-109)*(3*(-109)+1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Universal sums of three quadratic polynomials, Sci. China Math., in press.

Crossrefs

Cf. A000244, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308656.

Programs

Mathematica

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-9^a*25^b-x(7x+1)/2],r=r+1],{a,0,Log[9,n]},{b,0,Log[25,n/9^a]},{x,-Floor[(Sqrt[56(n-9^a*25^b)+1]+1)/14],(Sqrt[56(n-9^a*25^b)+1]-1)/14}];tab=Append[tab,r],{n,1,100}];Print[tab]