%I #19 Jun 15 2019 19:32:47
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%C 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.
%C We have verified Conjectures 1-2 for all n = 1..10^6.
%C See also A308640, A308641 and A308656 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A308644/b308644.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-017-9354-4">Universal sums of three quadratic polynomials</a>, Sci. China Math., in press.
%e a(152) = 1 with 152 = (3^0*5^0)^2 + (-4)*(3*(-4)+1)/2 + 6*(7*6+1)/2.
%e a(129894) = 1 with 129894 = (3^0*5^1)^2 + 154*(3*154+1)/2 + 164*(7*164+1)/2.
%e a(200963) = 1 with 200963 = (3^1*5^0)^2 + 364*(3*364+1)/2 + 24*(7*24+1)/2.
%e a(371278) = 1 with 371278 = (3^3*5^1)^2 + (-382)*(3*(-382)+1)/2 + (-196)*(7*(-196)+1)/2.
%e a(534699) = 1 with 534699 = (3^2*5^2)^2 + 543*(3*543+1)/2 + (-109)*(3*(-109)+1)/2.
%t PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
%t 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]
%Y Cf. A000244, A000290, A000351, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308656.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Jun 13 2019