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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.

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A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 10, 108, 1078, 10680, 105730, 1046628, 10360558, 102558960, 1015229050, 10049731548, 99482086438, 984771132840, 9748229241970, 96497521286868, 955226983626718, 9455772314980320
Offset: 0

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Author

Charlie Marion, Mar 12 2009

Keywords

Comments

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=3, then first(2) = 312 = 14*21 - 0 + 18.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=3 and n=2, first(2) = 312 = 7*21 + 6*27 + 3 and last(2) = 363 = 8*21 + 7*27 + 6. a(n) = (2^(n+1)((1+sqrt(3/2))^(2n+1) + (1-sqrt(3/2))^(2n+1)) - 2*2)/4; e.g., 108 = (2^3((1+sqrt(3/2))^5 + (1-sqrt(3/2))^5) - 2*2)/4.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=3 and n=2, then first(2) = 312 = (3^3((1+sqrt((4/3))^(5) + (1-sqrt(4/3)^5) - 2*3)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 1078 = 2*11*49 and a(4) = 10680 = 3*40*89.
In general, if first(n) is the first term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim n->inf first(n+1)/first(n) = k*(1 + sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

Examples

			a(2) = 108 since 108^2 + 109^2 + 110^2 = 133^2 + 134^2.

References

  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
  • W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

Links

Crossrefs

Cf. A001652, A157088, A157092, A157096.

Programs

  • Magma
    [(1/2 + Sqrt(6)/4)*(5 + 2*Sqrt(6))^n - (Sqrt(6)/4 - 1/2)*(5 - 2*Sqrt(6))^n - 1: n in [0..50]]; // G. C. Greubel, Nov 04 2017
  • Mathematica
    CoefficientList[Series[2*x*(-5 + x)/((x - 1)*(x^2 - 10*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(2*(x - 5)/((x-1)*(x^2 - 10*x + 1)))) \\ G. C. Greubel, Nov 04 2017

Formula

For n > 1, a(n) = 10*a(n-1) - a(n-2) + 8.
For n > 0, a(n) = 5*a(n-1) + 4*A157085(n-1) + 2.
Lim_{n->inf} a(n+1)/a(n) = 2(1 + sqrt(3/2))^2 = 5 + 2*sqrt(6).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 2*x*(x - 5)/((x-1)*(x^2 - 10*x + 1)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3).
a(n) = 2*A087125(n) = A054320(n) - 1. (End)
From Sergei N. Gladkovskii, Jan 12 2012: (Start)
G.f.: 1/(x-1) + (x+1)/(x^2-10*x+1).
a(n) = (1/2 + sqrt(6)/4)*(5 + 2*sqrt(6))^n - (sqrt(6)/4 - 1/2)*(5 - 2*sqrt(6))^n - 1. (End)
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