A308641 Number of ways to write n as (2^a*5^b)^2 + c*(3c+1)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers with c*(3c+1)/2 <= d*(3d+1)/2.

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

Offset: 1

Author

Zhi-Wei Sun, Jun 13 2019

Keywords

• nonn

Comments

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

Conjecture 2: If f(x) is one of the polynomials x^2, x*(5x+1), x*(5x+1)/2, x*(7x+1)/2, x*(7x+5)/2, then any positive integers n can be written as (2^a*5^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.

Conjecture 3: Any positive integers n can be written as (2^a*7^b)^2 + c*(7c+5)/2 + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.

See also A308640 for similar conjectures.

Examples

			a(12) = 1 with 12 = (2^1*5^0)^2 + (-1)*(3*(-1)+1)/2 + 2*(3*2+1)/2.
a(22) = 1 with 22 = (2^2*5^0)^2 + (-1)*(3*(-1)+1)/2 + (-2)*(3*(-2)+1)/2.
a(50) = 1 with 50 = (2^2*5^0)^2 + (-3)*(3*(-3)+1)/2 + (-4)*(3*(-4)+1)/2.
a(330) = 1 with 330 = (2^2*5^0)^2 + (-8)*(3*(-8)+1)/2 + 12*(3*12+1)/2.
a(8650) = 1 with 8650 = (2^5*5^0)^2 + 8*(3*8+1)/2 + (-71)*(3*(-71)+1)/2.
a(29440) = 1 with 29440 = (2*5)^2 + (-80)*(3*(-80)+1)/2 + (-115)*(3*(-115)+1)/2.
a(48459) = 1 with 48459 = (2^7*5^0)^2 + 20*(3*20+1)/2 + (-145)*(3*(-145)+1)/2.
a(153035) = 1 with 153035 = (2*5^2)^2 + 35*(3*35+1)/2 + (-315)*(3*(-315)+1)/2.
a(164043) = 1 with 164043 = (2^2*5^2)^2 + (-46)*(3*(-46)+1)/2 + 317*(3*317+1)/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, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640.

Programs

Mathematica

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={};Do[r=0;Do[If[PenQ[n-4^a*25^b-c(3c+1)/2],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{c,-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]