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
Keywords
Examples
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.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Project Euler, Problem 135: Same differences
- Project Euler, Problem 136: Singleton difference
Crossrefs
Programs
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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
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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 *)
Formula
Extensions
More terms from Giovanni Resta, May 06 2020
Comments