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A303389 revision #18

A303389
Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.
30
0, 1, 1, 1, 1, 2, 1, 3, 2, 2, 2, 4, 3, 2, 2, 3, 3, 3, 2, 2, 2, 4, 3, 2, 1, 5, 4, 3, 2, 5, 5, 5, 5, 3, 3, 5, 5, 4, 4, 4, 5, 5, 2, 5, 3, 5, 4, 7, 2, 4, 6, 6, 5, 4, 4, 5, 8, 4, 4, 4, 7, 6, 4, 3, 4, 8, 4, 7, 3, 3, 6, 8, 2, 5, 6, 5, 4, 6, 4, 3
OFFSET
1,6
COMMENTS
Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
See also A303401, A303432 and A303540 for similar conjectures.
LINKS
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.
EXAMPLE
a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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], Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1], r=r+1], {x, 0, (Sqrt[4(n-5^j-5^k)+1]-1)/2}]], {j, 0, Log[5, n/2]}, {k, j, Log[5, n-5^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 23 2018
STATUS
editing