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A301471
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Number of ways to write n^2 as x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.
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25
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0, 1, 2, 1, 3, 4, 3, 1, 5, 4, 4, 4, 5, 4, 10, 1, 4, 7, 4, 4, 10, 4, 3, 4, 6, 6, 11, 4, 7, 10, 6, 1, 9, 5, 7, 7, 7, 6, 12, 4, 6, 12, 7, 4, 14, 4, 8, 4, 3, 8, 10, 6, 8, 13, 6, 4, 16, 8, 7, 10, 7, 6, 14, 1, 7, 11, 6, 5, 16, 9, 5, 7, 7, 7, 18, 6, 7, 14, 6, 4
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OFFSET
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1,3
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301472, A301479, A301579, A301640, A302641.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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