OFFSET
0,2
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=5, then first(2) = 1260 = 22*55 - 0 + 50.
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=5 and n=2, then first(2) = 1260 = 11*55 + 10*65 + 5 and last(2) = 1385 = 12*55 + 11*65 + 10.
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=5 and n=2, then first(2) = 1260 = (5^3((1+sqrt((6/5))^5 + (1-sqrt(6/5))^5) - 2*5)/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) = 36 = 4*1*9 and a(2) = 680 = 5*8*17.
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).
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..790
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).
FORMULA
For n > 1, a(n) = 18*a(n-1) - a(n-2) + 32.
For n > 0, a(n) = 9*a(n-1) + 8*A157093(n-1) + 4.
a(n) = (4^(n+1)((1+sqrt(5/4))^(2n+1) + (1-sqrt(5/4))^(2n+1)) - 2*4)/4.
Lim_{n->inf} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 4*x*(-9+x)/((x-1)*(x^2-18*x+1)).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3).
a(n) = 4*A119032(n+1). (End)
For n > 0, 1/a(n) = Sum_{k>=1} F(3*k)/phi^(6*k*n + 3*k), where F(n) = A000045(n) and phi = A001622 = (sqrt(5)+1)/2. - Diego Rattaggi, Dec 28 2019
E.g.f.: (1/2)*((2 + sqrt(5))*exp((9+4*sqrt(5))*x) + (2 - sqrt(5))*exp((9-4*sqrt(5))*x) - 4*exp(x)). - Stefano Spezia, Dec 29 2019
EXAMPLE
a(2)=680 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.
MATHEMATICA
RecurrenceTable[{a[0]==0, a[1]==36, a[n]==18a[n-1]-a[n-2]+32}, a, {n, 20}] (* or *) LinearRecurrence[{19, -19, 1}, {0, 36, 680}, 20] (* Harvey P. Dale, Oct 09 2012 *)
PROG
(PARI) x='x+O('x^50); concat([0], Vec(4*x*(-9+x)/((x-1)*(x^2-18*x+1)))) \\ G. C. Greubel, Nov 04 2017
(Magma) [Round((4^(n+1)*((1+Sqrt(5/4))^(2*n+1) + (1-Sqrt(5/4))^(2*n+1)) - 2*4)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Charlie Marion, Mar 12 2009
EXTENSIONS
Terms a(15) onward added by G. C. Greubel, Nov 06 2017
STATUS
approved