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A279612 Number of ways to write n = x^2 + y^2 + z^2 + w^2 with x + 2*y - 2*z a power of 4 (including 4^0 = 1), where x,y,z,w are nonnegative integers. 11
1, 1, 1, 2, 3, 1, 1, 1, 3, 5, 2, 1, 3, 4, 1, 1, 3, 5, 5, 4, 3, 2, 3, 2, 4, 5, 1, 3, 4, 4, 1, 1, 5, 7, 7, 2, 3, 7, 3, 2, 4, 3, 4, 2, 8, 5, 1, 1, 6, 8, 3, 6, 7, 8, 2, 3, 3, 6, 8, 4, 6, 5, 2, 2, 9, 7, 7, 7, 7, 12, 3, 1, 9, 10, 7, 1, 10, 10, 2, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*q (k = 0,1,2,... and q = 1, 2, 3, 6, 7, 8, 12, 15, 27, 31, 47, 72, 76, 92, 111, 127).

(ii) Let a and b be positive integers with gcd(a,b) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y a power of two (including 2^0 = 1) if and only if (a,b) = (1,1), (2,1), (2,3).

(iii) Let a,b,c be positive integers with a <= b and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z a power of two if and only if (a,b,c) is among the triples (1,1,1), (1,1,2), (1,2,1), (1,2,2), (1,3,1), (1,3,2), (1,3,3), (1,3,4), (1,3,5), (1,4,1), (1,4,2), (1,4,3), (1,4,4), (1,5,1), (1,5,2), (1,5,4), (1,5,5,), (1,6,3), (1,7,4), (1,7,7), (1,8,1), (1,9,2), (2,3,1), (2,3,3), (2,3,4), (2,5,1), (2,5,3), (2,5,4), (2,5,5), (2,7,1), (2,7,3), (2,7,7), (2,9,3), (2,11,5), (3,4,3), (7,8,7).

(iv) Let a,b,c be positive integers with b <= c and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z a power of two if and only if (a,b,c) is among the triples (2,2,1), (4,2,1), (4,3,1), (4,4,3).

(v) Let a,b,c be positive integers with a <= b, c <= d, and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,2,1,1), (1,2,1,2), (1,2,1,3), (1,3,1,2), (1,3,2,3), (1,3,2,4), (1,4,1,2), (1,7,2,6), (1,9,1,4), (2,2,2,3), (2,3,1,2), (2,3,1,3), (2,3,2,3), (2,3,6,1), (2,4,1,2), (2,5,1,2), (2,5,2,3), (2,5,3,4), (3,4,1,2), (3,4,1,3), (3,4,1,5), (3,4,2,5), (3,4,3,4), (3,8,1,10), (3,8,2,3), (4,5,1,5).

(vi) Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,1,2,2), (1,1,2,3), (1,1,2,4), (1,2,2,3), (1,2,3,4), (1,2,4,3), (1,2,6,7), (1,3,4,4), (1,4,6,5), (2,3,5,4).

(vii) For any positive integers a,b,c,d, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z -d*w a power of two.

(viii) Let a and b be positive integers, and c and d be nonnegative integers. Then, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z + d*w a power of two.

We have verified a(n) > 0 for all n = 1..2*10^7. The conjecture that a(n) > 0 for all n > 0 appeared in arXiv:1701.05868.

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. Also available from arXiv:1604.06723 [math.NT].

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

EXAMPLE

a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 2*1 - 2*1 = 4^0.

a(15) = 1 since 15 = 3^2 + 1^2 + 2^2 + 1^2 with 3 + 2*1 - 2*2 = 4^0.

a(27) = 1 since 27 = 4^2 + 1^2 + 1^2 + 3^2 with 4 + 2*1 - 2*1 = 4.

a(31) = 1 since 31 = 3^2 + 2^2 + 3^2 + 3^2 with 3 + 2*2 - 2*3 = 4^0.

a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + 5^2 with 3 + 2*2 - 2*3 = 4^0.

a(72) = 1 since 72 = 8^2 + 0^2 + 2^2 + 2^2 with 8 + 2*0 - 2*2 = 4.

a(76) = 1 since 76 = 1^2 + 5^2 + 5^2 + 5^2 with 1 + 2*5 - 2*5 = 4^0.

a(92) = 1 since 92 = 4^2 + 6^2 + 6^2 + 2^2 with 4 + 2*6 - 2*6 = 4.

a(111) = 1 since 111 = 9^2 + 1^2 + 5^2 + 2^2 with 9 + 2*1 - 2*5 = 4^0.

a(127) = 1 since 127 = 7^2 + 2^2 + 5^2 + 7^2 with 7 + 2*2 - 2*5 = 4^0.

MATHEMATICA

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

FP[n_]:=FP[n]=n>0&&IntegerQ[Log[4, n]];

Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&FP[x+2y-2z], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]

CROSSREFS

Cf. A000079, A000118, A000290, A271518, A275656, A275675, A275676, A275738, A278560, A279616.

Sequence in context: A180050 A275738 A202603 * A261019 A140737 A108756

Adjacent sequences:  A279609 A279610 A279611 * A279613 A279614 A279615

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 15 2016

STATUS

approved

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Last modified July 15 00:36 EDT 2020. Contains 335762 sequences. (Running on oeis4.)