OFFSET
1,7
COMMENTS
Conjecture: a(n) > 0 for all n > 5.
This has been verified for n up to 5*10^9. Note that 321256731 cannot be written as x^2 + (2*y)^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.
In contrast, Crocker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.
570143 cannot be written as x^2 + y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers, and 2284095 cannot be written as x^2 + y^2 + 2^z + 7*2^w with x,y,z,w nonnegative integers.
Jiao-Min Lin (a student at Nanjing University) found a counterexample to the conjecture: a(18836421387) = 0. - Zhi-Wei Sun, Jul 21 2022
REFERENCES
R. C. Crocker, On the sum of two squares and two powers of k, Colloq. Math. 112(2008), 235-267.
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.
EXAMPLE
a(6) = 1 with 6 = 0^2 + 0^2 + 2^0 + 5*2^0.
a(8) = 2 with 8 = 1^2 + 1^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^1 + 5*2^0.
a(9) = 2 with 9 = 1^2 + 1^2 + 2^1 + 5*2^0 = 0^2 + 0^2 + 2^2 + 5*2^0.
a(10) = 2 with 10 = 0^2 + 2^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^2 + 5*2^0.
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[n-5*2^k-2^m], Do[If[SQ[n-5*2^k-2^m-x^2], r=r+1], {x, 0, Sqrt[(n-5*2^k-2^m)/2]}]], {k, 0, Log[2, n/5]}, {m, 0, If[n/5==2^k, -1, Log[2, n-5*2^k]]}]; tab=Append[tab, r], {n, 1, 60}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 27 2018
STATUS
approved