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A248980
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One fifth the hypotenuse of primitive Pythagorean triangles (PPT's) whose hypotenuse is the only side that is divisible by primes of the form 4k+1 (sorted and repetitions omitted).
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0
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1, 5, 13, 17, 37, 53, 61, 85, 89, 97, 101, 109, 149, 185, 197, 205, 221, 229, 257, 313, 349, 397, 421, 425, 461, 481, 485, 493, 545, 557, 577, 629, 689, 701, 725, 733, 797, 829, 841, 845, 865, 901, 953, 1021, 1037, 1069, 1073, 1105, 1157, 1165, 1181, 1189, 1193, 1241, 1249, 1301
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OFFSET
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1,2
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COMMENTS
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If only one side of a PPT is divisible by primes of the form 4k+1 that side is the hypotenuse. Furthermore this hypotenuse is divisible by 5 because all PPT's must have at least one side divisible by 5. However this is only a necessary condition. It is not sufficient. For instance, PPT's can share the same hypotenuse divisible by 5, some will have no other side divisible by primes of the form 4k+1 and others will. E.g. PPT's (16,63,65) and (33,56,65) so 13 is in the sequence.
Also an identical sequence is generated from the superset of primitive Heronian triangles if only one side is divisible by primes of the form 4k+1 and this side is divided by 5 (see Yiu link).
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LINKS
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EXAMPLE
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a(1)=1 as (3,4,5) is the first PPT where only the hypotenuse is divisible by primes of the form 4k+1, whereas the PPT (5,12,13) has two sides and the PPT (39,80,89) has three sides divisible by primes of the form 4k+1.
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MATHEMATICA
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pyprimeQ[n0_] := If[Length@Select[FactorInteger[n0], Mod[#[[1]], 4]==1 &] > 0, 1, 0]; lst1 = {}; Do[If[GCD[m, n]==1&&m<n&&OddQ[m+n], AppendTo[lst1, {n^2-m^2, 2m*n, n^2+m^2}]], {n, 1, 100}, {m, 1, n}]; SetAttributes[pyprimeQ, Listable]; lst2=Select[lst1, Total[pyprimeQ[#]]==1 &]; Union@Table[lst2[[k]][[3]]/5, {k, 1, Length[lst2]}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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