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A379509
Sum of the legs of the unique primitive Pythagorean triple whose inradius is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.
5
7, 127, 3527, 115199, 3886471, 131868799, 4478743367, 152140105727, 5168253960967, 175568314524799, 5964153390518471, 202605640846963199, 6882627599758753927, 233806732543181952127, 7942546277657462785607, 269812766700752532479999, 9165691521506791484696071, 311363698964290393026435199
OFFSET
0,1
COMMENTS
For all n: a(n) == 7 (mod 8).
REFERENCES
Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.
LINKS
Miguel-Ángel Pérez García-Ortega, El Libro de las Ternas Pitagóricas.
FORMULA
a(n) = A377725(n) + A385977(n).
a(n) = 2*A002315(n)^2 + 4*A002315(n) + 1.
G.f.: (7 - 153*x - 111*x^2 + x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)). - Andrew Howroyd, Nov 16 2025
EXAMPLE
For n=1, the short leg is A377725(1) = 15 the long leg is A385977(1) = 112 so the sum of the legs is then a(1) = 15 + 112 = 127.
MATHEMATICA
s[n_]:=s[n]=Module[{r}, r=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2; {2r^2+4r+1}]; sumas={}; Do[semis=Join[sumas, FullSimplify[s[n]]], {n, 0, 17}]; sumas
PROG
(PARI) a(n)=my(t=polcoef((1 + x)/(1 - 6*x + x^2) + O(x*x^n), n)); 2*t^2 + 4*t + 1; \\ Andrew Howroyd, Nov 16 2025
CROSSREFS
KEYWORD
nonn,easy
STATUS
approved