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A245648
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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.
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5
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OFFSET
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1,1
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COMMENTS
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Next term comes from a triple with c > 10^5.
The 4 known triples that satisfy the requisite are [3,4,5], [9,12,15], [100, 105, 145], [900, 2625, 2775].
Let po(n) be A176774(n), the least polygonality of a number.
po([3,4,5]) = [3,4,5]; <-----
po([9,12,15]) = [4,5,3];
po([100,105,145]) = [4,3,5]; <-----
po([900,2625,2775]) = [4,5,3].
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?
Let nb(n) be A177025(n) is the number of ways to represent n as a polygonal number.
nb([3,4,5]) = [1,1,1]; <-----
nb([9,12,15]) = [4,5,3];
nb([100,105,145]) = [4,3,5];
nb([900,2625,2775]) = [4,5,3].
So for the highlighted triple, we get [1,1,1]. Are there other Pythagorean triples with the same property? (End)
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
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LINKS
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EXAMPLE
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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).
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MATHEMATICA
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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*)
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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