cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A377726 Lengths of the long leg of the unique primitive Pythagorean triple (x,y,z) such that (x-y+z)/2 is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

84, 3280, 113764, 3878112, 131820084, 4478459440, 152138450884, 5168244315840, 175568258308884, 5964153062868112, 202605638937276964, 6882627588628286880, 233806732478308836084, 7942546277279354556400, 269812766698548756220804, 9165691521493946935370112
Offset: 1

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Author

Keywords

Examples

			Triangles begins:
  n=1:     13,        84,        85;
  n=2:     81,      3280,      3281;
  n=3:    477,    113764,    113765;
  ...
This sequence gives the middle column.
		

References

  • Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Crossrefs

Cf. A002315, A377016, A377017, A377725, A362545 (short legs).

Programs

  • Mathematica
    ra[n_]:=ra[n]=Module[{ra},ra=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;{2ra-1,2ra^2-2ra,2ra^2-2ra+1}];exradio={};Do[exradio=Join[exradio,FullSimplify[ra[n]]],{n,0,10}];exradio

Formula

a(n) = 2 * A002315(n) * (A002315(n) - 1).

A362540 Number of chordless cycles of length >= 4 in the n-flower graph.

Original entry on oeis.org

3, 23, 63, 127, 273, 583, 1287, 2975, 6993, 16535, 39525, 95071, 229029, 552199, 1332375, 3215807, 7762611, 18739607, 45240309, 109217983, 263673699, 636563527, 1536798717, 3710157407, 8957109801, 21624374039, 52205854257, 126036078751, 304278008331, 734592089095
Offset: 2

Views

Author

Eric W. Weisstein, Apr 24 2023

Keywords

Comments

The n-flower graph can be defined for n >= 3 without multiple edges. It is a snark for odd n >= 5. The sequence has been extended to n=2 using the recurrence. - Andrew Howroyd, Apr 26 2023

Crossrefs

Cf. A362545.

Programs

  • Mathematica
    LinearRecurrence[{4, -5, 4, -2, -4, 7, -8, 6, 4, -6, 0, 1}, {3, 23, 63, 127, 273, 583, 1287, 2975, 6993, 16535, 39525, 95071}, 20]
    CoefficientList[Series[(-3 - 11 x + 14 x^2 + 22 x^3 + 6 x^4 + 68 x^5 - 9 x^6 - 19 x^7 + 25 x^8 - 13 x^9 - 9 x^10 + x^11)/((-1 + x)^3 (1 - x - x^2 - 3 x^3 - 5 x^4 - 3 x^5 - 4 x^6 + 3 x^8 + x^9)), {x, 0, 20}], x]
    Table[(1 + (-1)^n)/2 + 2 (-I)^n ChebyshevT[n, I] + 3 (n - 2) n + RootSum[-1 + # + #^3 &, #^n &] + RootSum[1 + # + #^3 &, #^n &], {n, 2, 20}]

Formula

From Andrew Howroyd, Apr 26 2023: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - 2*a(n-4) - 4*a(n-5) + 7*a(n-6) - 8*a(n-7) + 6*a(n-8) + 4*a(n-9) - 6*a(n-10) + a(n-12).
G.f.: x^2*(3 + 11*x - 14*x^2 - 22*x^3 - 6*x^4 - 68*x^5 + 9*x^6 + 19*x^7 - 25*x^8 + 13*x^9 + 9*x^10 - x^11)/((1 - x)^3*(1 + x)*(1 - 2*x - x^2)*(1 + x^2 - x^3)*(1 + x^2 + x^3)). (End)
2*a(n) = -2*(-1)^n*A112455(n) +1+6*n^2-12*n+(-1)^n-2*A112455(n)+4*A001333(n). - R. J. Mathar, Feb 18 2024

Extensions

a(2)-a(4) and a(17) and beyond from Andrew Howroyd, Apr 26 2023
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