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

%I #15 Jul 21 2023 06:53:57

%S 1,1,1,2,2,3,1,2,3,1,3,2,2,2,2,4,2,2,5,3,3,3,3,3,4,6,4,3,3,5,4,4,3,6,

%T 5,6,3,2,6,3,6,2,3,4,4,6,5,5,4,4,6,1,4,4,4,6,3,5,2,6,7,3,2,5,5,4,5,6,

%U 8,5,6,5,4,8,3,7,3,3,7,3,6,7,4,4,7,7,4,4,8,7,4,3,6,4,7,7,4,1,6,7

%N Number of ways to write n as (2^a*5^b)^2 + c*(3c+1) + d*(3d+2), where a and b are nonnegative integers, and c and d are integers.

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

%C Conjecture 2: Let r be 1 or 2. Then, any positive integer n can be written as (2^a*5^b)^2 + c*(2c+1) + d*(3d+r), where a and b are nonnegative integers, and c and d are integers.

%C We have verified Conjectures 1 and 2 for all n = 1..10^8.

%H Zhi-Wei Sun, <a href="/A308662/b308662.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58 (2015), No. 7, 1367-1396.

%e a(3) = 1 with 3 = (2^0*5^0)^2 + (-1)*(3*(-1)+1) + 0*(3*0+2).

%e a(7) = 1 with 7 = (2^1*5^0)^2 + (-1)*(3*(-1)+1) + (-1)*(3*(-1)+2).

%e a(10) = 1 with 10 = (2^0*5^0)^2 + 1*(3*1+1) + 1*(3*1+2).

%e a(52) = 1 with 52 = (2^0*5^0)^2 + 3*(3*3+1) + (-3)*(3*(-3)+2).

%e a(98) = 1 with 98 = (2^0*5^1)^2 + 4*(3*4+1) + (-3)*(3*(-3)+2).

%e a(14596) = 1 with 14596 = (2^3*5^0)^2 + (-36)*(3*(-36)+1) + (-60)*(3*(-60)+2).

%e a(22163) = 1 with 22163 = (2^3*5^0)^2 + 66*(3*66+1) + (-55)*(3*(-55)+2).

%e a(150689) = 1 with 150689 = (2^6*5^1)^2 + 117*(3*117+1) + (-49)*(3*(-49)+2).

%t OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]];

%t tab={};Do[r=0;Do[If[OctQ[n-4^a*25^b-x(3x+1)],r=r+1],{a,0,Log[4,n]},{b,0,Log[25,n/4^a]},{x,-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]

%Y Cf. A000079, A000351, A001082, A001318, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308661.

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Jun 15 2019