OFFSET

1,3

COMMENTS

Square Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 3 we can write n^2 as x^2 + 2*y^2 + 3*2^z, where x,y,z are nonnegative integers with y even and z > 1.

It is known that a positive integer n has the form x^2 + 2*y^2 with x and y integers if and only if the p-adic order of n is even for any prime p == 5 or 7 (mod 8).

See also A301472 for the list of positive integers not of the form x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.

If n^2 = x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers, then it is easy to see that x is not divisible by 3.

The Square Conjecture implies that for each n = 1,2,3,... we can write 3*n^2 as x^2 + 2*y^2 + 2^z with x,y,z nonnegative integers. In fact, if (3*n)^2 = u^2 + 2*v^2 + 3*2^z with u,v,z integers and z >= 0, then u^2 == v^2 (mod 3) and thus we may assume u == v (mod 3) without loss of generality, hence 3*n^2 = (u^2+2*v^2)/3 + 2^z = x^2 + 2*y^2 + 2^z with x = (u+2*v)/3 and y = (u-v)/3 integers.

On March 25, 2018 Qing-Hu Hou at Tianjin Univ. finished his verification of the Square Conjecture for n <= 4*10^8. Then I used Hou's program to verify the conjecture for n <= 5*10^9. - Zhi-Wei Sun, Apr 10 2018

I have found a counterexample to the Square Conjecture, namely a(5884015571) = 0. Note that 5884015571 is the product of the three primes 7, 17 and 49445509. - Zhi-Wei Sun, Apr 15 2018

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, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

EXAMPLE

a(2) = 1 with 2^2 = 1^2 + 2*0^2 + 3*2^0.

a(3) = 2 with 3^2 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.

a(4) = 1 with 4^2 = 2^2 + 2*0^2 + 3*2^2.

a(1131599953) = 1 with 1131599953^2 = 316124933^2 + 2*768304458^2 + 3*2^6.

a(5884015571) = 0 since there are no nonnegative integers x,y,z such that x^2 + 2*y^2 + 3*2^z = 5884015571^2.

MATHEMATICA

f[n_]:=f[n]=FactorInteger[n];

g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n], i], 1], 8]==5||Mod[Part[Part[f[n], i], 1], 8]==7)&&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]);

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

tab={}; Do[r=0; Do[If[QQ[n^2-3*2^k], Do[If[SQ[n^2-3*2^k-2x^2], r=r+1], {x, 0, Sqrt[(n^2-3*2^k)/2]}]], {k, 0, Log[2, n^2/3]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]

CROSSREFS

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Mar 21 2018

STATUS

approved