%I #17 Jul 21 2023 12:55:23
%S 1,1,1,2,2,3,1,2,4,1,4,3,3,4,1,7,2,2,7,2,5,2,4,5,1,8,5,2,3,4,6,2,3,4,
%T 2,3,7,6,5,4,7,6,1,7,5,4,6,4,4,1,6,9,2,5,3,3,5,6,7,4,7,5,4,6,6,6,4,4,
%U 5,3,9,7,4,8,2,8,5,4,10,3,9,6,5,6,4,11,7,5,8,4,7,7,8,8,2,14,6,3,8,4
%N Number of ways to write n as (2^a*3^b)^2 + c*(2c+1) + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
%C Conjecture 1: a(n) > 0 for all n > 0.
%C Conjecture 2: Let k be 1 or 2. Then, any positive integer n can be written as (2^a*3^b)^2 + k*c^2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
%C Conjecture 3: Let k be 1 or -1. Then, any positive integer n can be written as (2^a*3^b)^2 + c*(5c+3k)/2 + d*(3d+1)/2, where a,b,c are nonnegative integers and d is an integer.
%C We have verified Conjectures 1-3 for all n = 1..10^6.
%C See also A308641 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A308640/b308640.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="https://www.sciengine.com/SCM/doi/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58 (2015), No. 7, 1367-1396.
%e a(230) = 1 with 230 =(2^3*3^0)^2 + 3*(2*3+1) + 10*(3*10+1)/2.
%e a(2058) = 1 with 2058 = (2^0*3^0)^2 + 25*(2*25+1) + (-23)*(3*(-23)+1)/2.
%e a(26550) = 1 with 26550 = (2^0*3^3)^2 + 14*(2*14+1) + 130*(3*130+1)/2.
%e a(39433) = 1 with 39433 = (2*3^3)^2 + 135*(2*135+1) + 17*(3*17+1)/2.
%e a(505330) = 1 with 505330 = (2*3^2)^2 + 198*(2*198+1) + 533*(3*533+1)/2.
%e a(537830) = 1 with 537830 = (2^5*3^2)^2 + 402*(2*402+1) + (-296)*(3*(-296)+1)/2.
%t PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
%t tab={};Do[r=0;Do[If[PenQ[n-4^a*9^b-c(2c+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[9,n/4^a]},{c,0,(Sqrt[8(n-4^a*9^b)+1]-1)/4}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000079, A000244, A000290, A000566, A001318, A014105, A147875, A308566, A308584, A308621, A308623, A308641.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, Jun 12 2019