A157096 Consider all consecutive integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

0, 55, 1260, 27715, 608520, 13359775, 293306580, 6439385035, 141373164240, 3103770228295, 68141571858300, 1496010810654355, 32844096262537560, 721074106965172015, 15830786256971246820, 347556223546402258075, 7630406131763878430880, 167521378675258923221335

Offset: 0

Author

Charlie Marion, Mar 12 2009

Keywords

• nonn
• easy

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=6, then first(2) = 2100 = 26*78 - 0 + 72.

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=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.

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=6 and n=2, then first(2) = 2100 = (6^3((1+sqrt((7/6))^5 + (1-sqrt(7/6))^5) - 2*6)/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(1) = 55 = 5*1*11 and a(2) = 1260 = 6*10*21.

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)=55 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.
		

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

• G. C. Greubel, Table of n, a(n) for n = 0..740 (terms 0..200 from Vincenzo Librandi)
• Tanya Khovanova, Recursive Sequences
• Ron Knott, Pythagorean Triples and Online Calculators
• Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Crossrefs

Cf. A001652, A157084, A157088, A157092.

Programs

Magma

I:=[0, 55, 1260]; [n le 3 select I[n] else 23*Self(n-1) - 23*Self(n-2) + Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 09 2012

Mathematica

CoefficientList[Series[5*x*(x-11)/((x-1)*(x^2-22*x+1)),{x,0,20}],x] (* Vincenzo Librandi, Jun 09 2012 *)

PARI

x='x+O('x^50); concat([0], Vec(5*x*(x-11)/((x-1)*(x^2-22*x+1)))) \\ G. C. Greubel, Nov 04 2017

Formula

For n > 1, a(n) = 22*a(n-1) - a(n-2) + 50.

For n > 0, a(n) = 11*a(n-1) + 10*A157097(n-1) + 5.

a(n) = (5^(n+1)*((1+sqrt(6/5))^(2n+1) + (1-sqrt(6/5))^(2n+1)) - 2*5)/4.

Lim_{n->inf} a(n+1)/a(n) = 5(1+sqrt(6/5))^2 = 11+2*sqrt(30).

G.f.: 5*x*(x-11)/((x-1)*(x^2-22*x+1)). - Colin Barker, Jun 08 2012

a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). Vincenzo Librandi, Jun 09 2012

a(n) = 5*(-1/2+1/20*(11+2*sqrt(30))^(-n)*(5-sqrt(30)+(5+sqrt(30))*(11+2*sqrt(30))^(2*n))). - Colin Barker, Mar 03 2016

Extensions

Terms a(15) onward added by G. C. Greubel, Nov 06 2017