A308656 Number of ways to write n as (2^a*9^b)^2 + c*(2c+1) + d*(3d+1), where a and b are nonnegative integers, and c and d are integers.

1, 1, 1, 3, 2, 3, 3, 2, 3, 1, 4, 2, 1, 4, 3, 4, 3, 5, 4, 3, 6, 2, 2, 4, 3, 6, 2, 4, 5, 3, 6, 4, 4, 4, 4, 4, 4, 1, 4, 5, 5, 2, 3, 3, 2, 8, 3, 4, 5, 3, 5, 3, 3, 5, 3, 7, 1, 3, 5, 4, 6, 3, 6, 2, 2, 6, 5, 4, 6, 6, 7, 3, 4, 9, 5, 4, 5, 3, 4, 4, 11, 5, 5, 12, 5, 7, 5, 4, 10, 2, 7, 8, 4, 8, 7, 12, 5, 5, 5, 5

Offset

1,4

Comments

Note that {x*(2x+1): x is an integer} = {n*(n+1)/2: n = 0,1,2,...}.

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

Conjecture 2: If f(x) is one of the polynomials x*(4x+1), x*(5x+2), x*(5x+4), x*(7x+3)/2 and x(7x+5)/2, then any positive integer n can be written as (2^a*9^b)^2 + f(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.

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

Conjecture 4: If g(x) is one of the polynomials x*(x+1), x*(4x+3), x*(7x+1)/2, x*(7x+3)/2 and x*(7x+5)/2, then any positive integer n can be written as (2^a*7^b)^2 + g(c) + d*(3d+1)/2, where a and b are nonnegative integers, and c and d are integers.

We have verified a(n) > 0 for all n = 1..10^8, and Conjectures 2-4 for all n = 1..10^6.

See also A308640, A308641, and A308644 for similar conjectures.

Jiao-Min Lin (a student at Nanjing University) has found a counterexample to Conjecture 1: a(2109982225) = 0. - Zhi-Wei Sun, Jul 30 2022

Links

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

Example

a(13) = 1 with 13 = (2^0*9^0)^2 + 2*(2*2+1) + (-1)*(3*(-1)+1).
a(3515) = 1 with 3515 = (2^0*9^1)^2 + 0*(2*0+1) + (-34)*(3*(-34)+1).
a(124076) = 1 with 124076 = (2^3*9^1)^2 + 206*(2*206+1) + 106*(3*106+1).
a(141518) = 1 with 141518 = (2^1*9^2)^2 + (-188)*(2*(-188)+1) + 122*(3*122+1).
a(345402) = 1 with 345402 = (2^7*9^0)^2 + 18*(2*18+1) + (-331)*(3*(-331)+1).

Mathematica

PQ[n_]:=PQ[n]=IntegerQ[Sqrt[12n+1]];
tab={}; Do[r=0; Do[If[PQ[n-81^a*4^b-x(2x+1)], r=r+1], {a, 0, Log[81, n]}, {b, 0, Log[4, n/81^a]}, {x, -Floor[(Sqrt[8(n-81^a*4^b)+1]+1)/4], (Sqrt[8(n-81^a*4^b)+1]-1)/4}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Crossrefs

A000079, A000217, A000420, A001019, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644.

Sequence in context: A291674 A265157 A054263 * A178775 A124874 A230258

Adjacent sequences: A308653 A308654 A308655 * A308657 A308658 A308659

Keyword

nonn

Author

Zhi-Wei Sun, Jun 14 2019

Status

approved