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A157085 Consider all Consecutive Integer Pythagorean quintuples (X,X+1,X+2,Z-1,Z) ordered by increasing Z; sequence gives Z values. 4

%I

%S 2,14,134,1322,13082,129494,1281854,12689042,125608562,1243396574,

%T 12308357174,121840175162,1206093394442,11939093769254,

%U 118184844298094,1169909349211682,11580908647818722,114639177128975534

%N Consider all Consecutive Integer Pythagorean quintuples (X,X+1,X+2,Z-1,Z) ordered by increasing Z; sequence gives Z values.

%C "Consecutive Integer Pythagorean quintuple" means that X^2+(X+1)^2+(X+2)^2 = (Z-1)^2+Z^2. - _M. F. Hasler_, Oct 04 2014

%C 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=3, then last(2)=363=14*27-3-12.

%C For n>0, a(n)=6*a(n-1)+5*A157084(n-1)+4; e.g., 1322=6*108+5*134+4.

%C 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=3 and n=2, then first(2)=312=7*21+6*27+3 and last(2)=363=8*21+7*27+6.

%C a(n)=2^n*3((1+sqrt(3/2))^(2n+1)-(1-sqrt(3/2))^(2n+1))/(4*sqrt(3/2))+1/2; e.g., 134=2^2*3((1+sqrt(3/2))^5-(1-sqrt(3/2))^5)/(4*(sqrt(3/2))+1/2.

%C 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=3 and n=2, then last(2)=363=3^2*4((1+sqrt((4/3))^5-(1-sqrt(4/3))^5)/(4*sqrt(4/3))+2/2;

%C 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:

%C 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)=1322=(40^2+2^2*9^2+2*1*40*9)/2 and a(4)=13082=(2(109^2+49^2+1*109*49))/3.

%C 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+13+14+133+134=297=3*99.

%C 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+2sqrt(k^2+k).

%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.

%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.

%D W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11, -11, 1).

%F For n>1, a(n) = 10*a(n-1)-a(n-2)-4; e.g., 1322=10*134-14-4.

%F Lim n->inf a(n+1)/a(n) = 2(1+sqrt(3/2))^2 = 5+2sqrt(6).

%F Empirical G.f.: (2-8*x+2*x^2)/(1-11*x+11*x^2-x^3). - Colin Barker, Jan 01 2012

%e a(3)=134 since 108^2+109^2+110^2=133^2+134^2.

%o (PARI) for(y=0,9e9,yy=(y+2)^2+(y+1)^2;for(x=sqrtint(yy\3),ceil(sqrt(yy/3)),x^2+(x-1)^2+(x-2)^2==yy&&print1(y+2,","))) \\ For illustrative purpose. - _M. F. Hasler_, Oct 04 2014

%Y Cf. A001653, A157089, A157093, A157097.

%K nonn

%O 0,1

%A _Charlie Marion_, Mar 12 2009

%E Edited by _M. F. Hasler_, Oct 04 2014

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Last modified November 14 03:52 EST 2018. Contains 317159 sequences. (Running on oeis4.)