A303399 Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.
0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 1, 4, 3, 3, 2, 5, 4, 4, 4, 3, 3, 4, 4, 3, 4, 4, 4, 3, 2, 4, 3, 3, 3, 5, 2, 4, 5, 4, 4, 4, 4, 3, 5, 3, 4, 4, 4, 4, 4, 3, 3, 5, 4, 5, 3, 3, 5, 5, 2, 4, 6, 3, 3, 4, 4, 3
Offset: 1
Keywords
Examples
a(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2. a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2. a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
- Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
- Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.
Crossrefs
Programs
-
Mathematica
f[n_]:=f[n]=FactorInteger[n]; g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0; QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]); tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],r=r+1],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]
Comments