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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
2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

"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

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.

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

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.

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.

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;

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)=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.

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.

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).

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..17.

Tanya Khovanova, Recursive Sequences

Ron Knott, Pythagorean Triples and Online Calculators

Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

FORMULA

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

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

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

EXAMPLE

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

PROG

(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

CROSSREFS

Cf. A001653, A157089, A157093, A157097.

Sequence in context: A048990 A089602 A052641 * A073553 A144097 A306081

Adjacent sequences:  A157082 A157083 A157084 * A157086 A157087 A157088

KEYWORD

nonn

AUTHOR

Charlie Marion, Mar 12 2009

EXTENSIONS

Edited by M. F. Hasler, Oct 04 2014

STATUS

approved

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Last modified October 17 01:37 EDT 2018. Contains 316275 sequences. (Running on oeis4.)