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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.
5

%I #14 Jul 30 2022 12:45:57

%S 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,

%T 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,

%U 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

%N 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.

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

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

%C 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.

%C 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.

%C 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.

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

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

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

%H Zhi-Wei Sun, <a href="/A308656/b308656.txt">Table of n, a(n) for n = 1..10000</a>

%e a(13) = 1 with 13 = (2^0*9^0)^2 + 2*(2*2+1) + (-1)*(3*(-1)+1).

%e a(3515) = 1 with 3515 = (2^0*9^1)^2 + 0*(2*0+1) + (-34)*(3*(-34)+1).

%e a(124076) = 1 with 124076 = (2^3*9^1)^2 + 206*(2*206+1) + 106*(3*106+1).

%e a(141518) = 1 with 141518 = (2^1*9^2)^2 + (-188)*(2*(-188)+1) + 122*(3*122+1).

%e a(345402) = 1 with 345402 = (2^7*9^0)^2 + 18*(2*18+1) + (-331)*(3*(-331)+1).

%t PQ[n_]:=PQ[n]=IntegerQ[Sqrt[12n+1]];

%t 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]

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

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Jun 14 2019