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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. 4
5, 65, 1385, 30365, 666605, 14634905, 321301265, 7053992885, 154866542165, 3400009934705, 74645352021305, 1638797734533965, 35978904807725885, 78989710803535465, 17341757471971854305, 38072876727534559205 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

For n>1, a(n) = 22*a(n-1)-a(n-2)-40; e.g., 30365=22*1385-65-40.

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.

For n>0, a(n)=12*A157096 (n-1)+11*a(n-1)+10; e.g., 30365=12*1260+11*1385+10.

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.

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

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.

Lim n->inf a(n+1)/a(n)=5(1+sqrt(6/5))^2=11+2sqrt(30).

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

Table of n, a(n) for n=0..15.

Tanya Khovanova, Recursive Sequences

Ron Knott, Pythagorean Triples and Online Calculators

Index entries for linear recurrences with constant coefficients, signature (23, -23, 1).

FORMULA

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

EXAMPLE

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.

CROSSREFS

Cf. A001653, A157085, A157089, A157093.

Sequence in context: A195886 A079482 A147625 * A234295 A251575 A277347

Adjacent sequences:  A157094 A157095 A157096 * A157098 A157099 A157100

KEYWORD

nonn,uned

AUTHOR

Charlie Marion, Mar 12 2009

STATUS

approved

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Last modified December 17 14:50 EST 2018. Contains 318201 sequences. (Running on oeis4.)