A303401 Number of ways to write n as a*(3*a-1)/2 + b*(3*b-1)/2 + 3^c + 3^d with a,b,c,d nonnegative integers.

0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 4, 1, 3, 2, 3, 2, 3, 3, 2, 1, 2, 3, 3, 2, 2, 2, 4, 4, 4, 3, 2, 3, 3, 3, 4, 3, 4, 2, 5, 4, 5, 1, 2, 3, 5, 2, 3, 2, 3, 2, 4, 5, 5, 3, 3, 3, 4, 4, 3, 2, 4, 4, 4, 3, 3, 3, 2, 3, 3, 2, 4, 2, 4, 5, 4, 5, 1, 3, 4

Offset: 1

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two pentagonal numbers and two powers of 3.

a(n) > 0 for all n = 2..7*10^6. See A303434 for the numbers of the form x*(3*x-1)/2 + 3^y with x and y nonnegative integers. See also A303389 and A303432 for similar conjectures.

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
• 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.

Crossrefs

Cf. A000244, A000326, A303233, A303338, A303363, A303389, A303393, A303399, A303428, A303432, A303434.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PenQ[n_]:=PenQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0);
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[12(n-3^j-3^k)+1],Do[If[PenQ[n-3^j-3^k-x(3x-1)/2],r=r+1],{x,0,(Sqrt[12(n-3^j-3^k)+1]+1)/6}]],{j,0,Log[3,n/2]},{k,j,Log[3,n-3^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

Formula

a(78) = 1 with 78 = 3*(3*3-1)/2 + 3*(3*3-1)/2 + 3^3 + 3^3.

a(285) = 1 with 285 = 3*(3*1-1)/2 + 11*(3*11-1)/2 + 3^3 + 3^4.

a(711) = 1 with 711 = 9*(3*9-1)/2 + 20*(3*20-1)/2 + 3^0 + 3^1.

a(775) = 1 with 775 = 7*(3*7-1)/2 + 21*(3*21-1)/2 + 3^3 + 3^3.

a(3200) = 1 with 12*(3*12-1)/2 + 44*(3*44-1)/2 + 3^3 + 3^4.

a(13372) = 1 with 13372 = 17*(3*17-1)/2 + 65*(3*65-1)/2 + 3^4 + 3^8.

a(16545) = 1 with 16545 = 0*(3*0-1)/2 + 98*(3*98-1)/2 + 3^0 + 3^7.