A157097 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 Z values.

5, 65, 1385, 30365, 666605, 14634905, 321301265, 7053992885, 154866542165, 3400009934705, 74645352021305, 1638797734533965, 35978904807725885, 789897108035435465, 17341757471971854305, 380728767275345359205, 8358691122585626048165, 183510475929608427700385, 4028871779328799783360265

Offset: 0

Author

Charlie Marion, Mar 12 2009

Keywords

• nonn
• uned
• easy

Comments

In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let last(0)=0, last(1)=k*(2k+3) and, for n > 1, last(n) = (4k+2)*last(n-1) - last(n-2)-2*k*(k-1); e.g., if k=6, then last(2) = 2274 = 26*90 - 6 - 60.

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 last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: if q=(k+1)/k, then last(n) = k^n*(k+1)*((1+sqrt(q))^(2*n+1) - (1-sqrt(q))^(2*n+1))/(4*sqrt(q)) + (k-1)/2; e.g., if k=6 and n=2, then last(2) = 2274 = 6^2*7*((1+sqrt(7/6))^5 - (1-sqrt(7/6))^5)/(4*sqrt(7/6)) + 5/2.

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 last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: last(2n+1) = (e(2n+1)^2 + k^2*e(2n)^2 + k*(k-1)*e(2n+1)*e(n))/k and, for n > 0, last(2n) = (k*(u(2n)^2 + u(2n-1)^2 + (k-1)*u(2n)*u(2n-1)))/(k+1); e.g., a(3) = 30365 = (220^2 + 5^2*21^2 + 5*4*220*21)/5 and a(4) = 666605 = (5*(505^2 + 241^2 + 4*505*241))/6.

In general, if b(0)=1, b(1)=4k+2 and, for n > 1, b(n) = (4k+2)*b(n-1) - b(n-2), and last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple as defined above, then Sum_{i=0..n} (k*last(i) - k(k-1)/2) = k(k+1)/2*b(n); e.g., if n=3, then 1+2+3+4+5+61+62+63+64+65+1381+1382+1383+1384+1385 = 7245 = 15*483.

In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} last(n+1)/last(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

Examples

			a(2)=65 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

• Paolo Xausa, Table of n, a(n) for n = 0..500
• 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. A001653, A157085, A157089, A157093.

Programs

Mathematica

LinearRecurrence[{23, -23, 1}, {5, 65, 1385}, 25] (* Paolo Xausa, May 29 2025 *)

Formula

For n > 1, a(n) = 22*a(n-1) - a(n-2) - 40.

For n > 0, a(n) = 12*A157096(n-1) + 11*a(n-1) + 10.

a(n) = 5^n*6*((1+sqrt(6/5))^(2*n+1) - (1-sqrt(6/5))^(2*n+1))/(4*sqrt(6/5)) + 4/2; e.g., 1385 = 5^2*6*((1+sqrt(6/5))^5 - (1-sqrt(6/5))^5)/(4*sqrt(6/5)) + 4/2.

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

G.f.: 5*(1-10*x+x^2)/((1-x)*(1-22*x+x^2)). - Colin Barker, Mar 27 2012

a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). - Wesley Ivan Hurt, Oct 26 2020

Extensions

a(13), a(15) corrected by Georg Fischer, Oct 26 2020