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 A157093 Consider all Consecutive Integer Pythagorean 9-tuples (X,X+1,X+2,X+3,X+4,Z-3,Z-2,Z-1,Z) ordered by increasing Z; sequence gives Z values. 4
 4, 44, 764, 13684, 245524, 4405724, 79057484, 1418628964, 25456263844, 456794120204, 8196837899804, 147086288076244, 2639356347472564, 47361327966429884, 849864547048265324, 15250200518902345924 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS For n>1, a(n) = 18*a(n-1)-a(n-2)-24; e.g., 13684=18*764-44-24. 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=5, then last(2)=1385=22*65-5-40. For n>0, a(n)=10*A157092(n-1)+9*a(n-1)+8; e.g., 13684=10*680+9*764+8. 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+10*65+10. a(n)=4^n*5((1+sqrt(5/4))^(2n+1)-(1-sqrt(5/4))^(2n+1))/(4*sqrt(5/4))+3/2; e.g., 764=4^2*5((1+sqrt((5/4))^5-(1-sqrt(5/4))^5)/(4*sqrt(5/4))+3/2; 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=5 and n=2, then last(2)=1385=5^2*6((1+sqrt((6/5))^5-(1-sqrt(6/5))^5)/(4*sqrt(6/5))+4/2. 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)=13684=(144^2+4^2*17^2+4*3*144*17)/4 and a(4)=245524=(4(341^2+161^2+3*341*161))/5. 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+41+42+43+44+761+762+763+764=3230=10*323. Lim n->inf a(n+1)/a(n)=4(1+sqrt(5/4))^2=9+2sqrt(20). In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim n->inf lasf(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 Tanya Khovanova, Recursive Sequences Ron Knott, Pythagorean Triples and Online Calculators Index entries for linear recurrences with constant coefficients, signature (19, -19, 1). FORMULA Empirical G.f.: 4*(1-8*x+x^2)/((1-x)*(1-18*x+x^2)). [Colin Barker, Mar 27 2012] EXAMPLE a(2)=764 since 680^2+681^2+682^2+683^2+684^2=761^2+762^2+763^2+764^2. CROSSREFS Cf. A001653, A157085, A157089, A157097. Sequence in context: A291198 A053332 A276369 * A088594 A144827 A144004 Adjacent sequences:  A157090 A157091 A157092 * A157094 A157095 A157096 KEYWORD nonn AUTHOR Charlie Marion, Mar 12 2009 STATUS approved

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Last modified November 13 01:32 EST 2018. Contains 317118 sequences. (Running on oeis4.)