A334567 Least value m > 0 such that Diophantine equation z^2 - y^2 - x^2 = m, when the positive integers, x, y and z are consecutive terms of an arithmetic progression, has exactly n solutions.

1, 3, 27, 15, 63, 135, 384, 315, 960, 1995, 1155, 1575, 2835, 3840, 5775, 4095, 6720, 14400, 14175, 10395, 13440, 20475, 20160, 36855, 48384, 26880, 46080, 108675, 57600, 51975, 40320, 190575, 100800, 193536, 107520, 172800, 126720, 80640, 174720, 120960, 744975

Offset: 0

Author

Bernard Schott, May 19 2020

Keywords

• nonn

Comments

Equivalently, if d is the common difference of the arithmetic progression (x, y, z), then a(n) is the smallest integer m such that the Diophantine equation y * (4d-y) = m with y>0, d>0 and y-d >0 has exactly n solutions (see A334566).

This sequence is not increasing: a(2) = 27 > a(3) = 15.

Examples

			a(4) = 63 because 11^2-7^2-3^2 = 13^2-9^2-5^2 = 27^2-21^2-15^2 = 79^2-63^2-47^2 = 63 and there is no term m < 63 in the context such that z^2 - y^2 - x^2 = m has 4 solutions.
		

Links

• Project Euler, Problem 135: Same differences
• Project Euler, Problem 136: Singleton difference

Crossrefs

Cf. A334566.

Programs

Maple

g:= proc(y,m) local d;
  d:= m/(4*y)+y/4;
  d::posint and y > d
end proc:
f:= proc(m) local L;
  nops(select(g, numtheory:-divisors(m),m));
end proc:
V:= Array(0..50): count:= 0:
for x from 1 while count < 51 do
  v:= f(x);
  if v <= 50 and V[v] = 0 then V[v]:= x; count:= count+1;
  fi
od:
convert(V,list); # Robert Israel, May 19 2020

Mathematica

ok[n_, x_] := Block[{d = (x + n/x)/4}, IntegerQ[d] && x > d]; t = Table[ Length@ Select[ Divisors[n], ok[n, #] &], {n, 21000}]; k=0; Reap[ While[ (v = Position[ t, k++]) != {}, Sow[v[[1, 1]]]]][[2, 1]] (* Giovanni Resta, May 19 2020 *)

Extensions

More terms from Giovanni Resta, May 19 2020