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Number of primitive Pythagorean-like triples a^2+b^2=c^2+k for k=-2 with 0<c<=10^n.
3

%I #10 Oct 17 2015 09:22:38

%S 3,19,182,1779,17697,176794,1768021,17676780,176776851,1767763756

%N Number of primitive Pythagorean-like triples a^2+b^2=c^2+k for k=-2 with 0<c<=10^n.

%C It is conjectured by the first author that a(n)/10^n as n->inf is 1/(4*sqrt(2)) = 0.17677...

%e a(1)=3 because there are 3 solutions (a,b,c) as (1,1,2), (3,5,6), (7,7,10) with 0<c<=10^1.

%t countTriples[m_, k_] := Module[ {c2, c2odd, total = 0, fax, g}, Do[ c2 = c^2 + k; If[c2 < 2, Continue[]]; c2odd = c2; While[EvenQ[c2odd], c2odd /= 2]; If [c2odd==1, If [OddQ[Log[2,c2]], total++ ]; Continue[]]; If[Mod[c2odd, 4] == 3, Continue[]]; g = GCD[c2odd, 100947]; If[g != 1 && g^2 != GCD[c2odd, 10190296809], Continue[]]; fax = Map[{Mod[ #[[1]],4],#[[2]]}&, FactorInteger[c2odd]]; If[Apply[Or, Map[ #[[1]] == 3 && OddQ[ #[[2]]] &, fax]], Continue []]; fax = Cases[fax, {1,aa_}:>aa+1]; fax = Ceiling[Apply[Times,fax]/2]; total += fax;, {c,m}]; total] (* Courtesy of Daniel Lichtblau of Wolfram Research *)

%Y Cf. A101931.

%K more,nonn

%O 1,1

%A _Tito Piezas III_, Aug 11 2006

%E First few terms found by _Tito Piezas III_, James Waldby (j-waldby(AT)pat7.com). Subsequent terms found by Andrzej Kozlowski (akoz(AT)mimuw.edu.pl), Daniel Lichtblau (danl(AT)wolfram.com).

%E a(8)-a(10) from _Hiroaki Yamanouchi_, Oct 17 2015