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A245648 The largest member 'c' of the Pythagorean triples (a,b,c) ordered by increasing c, where the triples consist of a triangular number, a square number and a pentagonal number.

%I A245648 #24 Dec 16 2021 14:08:17
%S A245648 5,15,145,2775
%N A245648 The largest member 'c' of the Pythagorean triples (a,b,c) ordered by increasing c, where the triples consist of a triangular number, a square number and a pentagonal number.
%C A245648 Next term comes from a triple with c > 10^5.
%C A245648 From _Michel Marcus_, Apr 08 2021: (Start)
%C A245648 The 4 known triples that satisfy the requisite are [3,4,5], [9,12,15], [100, 105, 145], [900, 2625, 2775].
%C A245648 Let po(n) be A176774(n), the least polygonality of a number.
%C A245648   po([3,4,5]) = [3,4,5]; <-----
%C A245648   po([9,12,15]) = [4,5,3];
%C A245648   po([100,105,145]) = [4,3,5]; <-----
%C A245648   po([900,2625,2775]) = [4,5,3].
%C A245648 So for the 2 highlighted triples, we have a-gonal^2 + b-gonal^2 = c-gonal^2. Are there other Pythagorean triples with the same property?
%C A245648 Let nb(n) be A177025(n) is the number of ways to represent n as a polygonal number.
%C A245648   nb([3,4,5]) = [1,1,1]; <-----
%C A245648   nb([9,12,15]) = [4,5,3];
%C A245648   nb([100,105,145]) = [4,3,5];
%C A245648   nb([900,2625,2775]) = [4,5,3].
%C A245648 So for the highlighted triple, we get [1,1,1]. Are there other Pythagorean triples with the same property? (End)
%C A245648 Regarding the first question by _Michel Marcus_, if such triple [x,y,z] exists, then z > 10^4. Regarding his second question, if such triple exists, then z > 10^7. - _Ivan N. Ianakiev_, Dec 16 2021
%C A245648 a(5) > 10^11, if it exists. - _Giovanni Resta_, Apr 15 2021
%e A245648 a(1) = 5 as the first such Pythagorean triple is (3,4,5). The next three triples are (9,12,15), (100,105,145), (900,2625,2775).
%t A245648 n=10^3;ppt={};list={};pos=1;t[x_]:=(IntegerPart[Sqrt[2*x]])*(IntegerPart[Sqrt[2*x]]+1)/2;ls[x_]:=Length[Sqrt[x]];lis[x_]:=Length[IntegerPart[Sqrt[x]]];lp[x_]:=Length[(Sqrt[24*x+1]+1)/6];lip[x_]:=Length[IntegerPart[(Sqrt[24*x+1]+1)/6]];Do[y=x+1;z=y+1;While[z<=n,While[z^2<x^2+y^2,z=z+1];If[z^2==x^2+y^2,AppendTo[ppt,{x,y,z}]];y=y+1],{x,1,n}];While[pos<Length[ppt]+1,a=ppt[[pos,1]];b=ppt[[pos,2]];c=ppt[[pos,3]];If[Or[And[t[a]==a,ls[b]==lis[b],lp[c]==lip[c]],And[t[a]==a,ls[c]==lis[c],lp[b]==lip[b]],And[t[b]==b,ls[a]==lis[a],lp[c]==lip[c]],And[t[b]==b,ls[c]==lis[c],lp[a]==lip[a]],And[t[c]==c,ls[a]==lis[a],lp[b]==lip[b]],And[t[c]==c,ls[b]==lis[b],lp[a]==lip[a]]],AppendTo[list,{a,b,c}]];pos++];l=Flatten[Sort[list,#1[[3]]<#2[[3]]&]];Take[l,{3,-1,3}](*Finds the terms through a search within all Pythagorean triples with c <= n*)
%Y A245648 Cf. A000217, A000290, A000326, A245646, A245647.
%Y A245648 Cf. A176774, A177025.
%K A245648 nonn,more,hard
%O A245648 1,1
%A A245648 _Ivan N. Ianakiev_, Jul 28 2014