A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 1, 3, 0, 0, 4, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 3, 0, 0, 2, 0

Offset

1,15

Comments

Inspired by the 135th and 136th problems of Project Euler (see links).

If d is the common difference of the arithmetic progression (x, y, z), then the Diophantine equation becomes (y+d)^2 - y^2 - (y-d)^2 = n <==> y^2 - 4dy + n = 0 <==> n = y * (4d-y).

If y is the average term, then y divides n.

Offset is 1 because for n = 0, every (x, y, z)= (3d, 4d, 5d) with d>0 would be solution.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Project Euler, Problem 135: Same differences

Project Euler, Problem 136: Singleton difference

Formula

a(n) = 0 iff n = 4q+1 (A016813), n = 4q+2 (A016825), n = 16q+8 (A051062), q>= 0.

a(n) >= 1 iff n = 4q+3, q >=0 (A004767), n = 16q, q>=1 (A008598), n = 16q+4, q>=0 (A119413), n = 16q+12, q>=0 (A098502).

a(4*q^2) >= 1, for q >= 1, since (q, 2q, 3q) is a solution.

a(p) = 1 for p = 4q+3 prime (A002145).

a(p^2) = 0 for p an odd prime (A065091).

Example

a(3) = 1 because 4^2 - 3^2 - 2^2 = 3.
a(15) = 3 because 5^2 - 3^2 - 1^2 = 7^2 - 5^2 - 3^2 = 19^2 - 15^2 - 11^2 = 15.
If n = 4q+3, q >= 0 then (3q+2, 4q+3, 5q+4) is a solution.
If n = 16q, q >= 1 then (3q-1, 4q, 5q+1) is a solution.
If n = 16q+4, q >= 0 then (6q+1, 8q+2, 10q+3) is a solution.
If n = 16q+12, q >= 0 then (6q+4, 8q+6, 10q+8) is a solution.

Maple

f:= proc(n) local r; r:= floor(sqrt(n/3));
nops(select(t -> n/t + t mod 4 = 0 and t > r, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Jul 31 2020

Mathematica

a[n_] := Length@ Solve[(4 d - x) x == n  && x>0 && x-d>0 && x+d>0, {d, x}, Integers]; Array[a, 90] (* Giovanni Resta, May 06 2020 *)

Crossrefs

Cf. A001248, A002145, A004767, A008598, A016813, A016825, A051062.

Cf. A334567 (least value of n such that a(n) = k>0).

Sequence in context: A361902 A205531 A269246 * A342270 A330248 A247505

Adjacent sequences: A334563 A334564 A334565 * A334567 A334568 A334569

Keyword

nonn

Author

Bernard Schott, May 06 2020

Extensions

More terms from Giovanni Resta, May 06 2020

Status

approved