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A157089
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Consider all Consecutive Integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.
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4
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3, 27, 363, 5043, 70227, 978123, 13623483, 189750627, 2642885283, 36810643323, 512706121227, 7141075053843, 99462344632563, 1385331749802027, 19295182152595803, 268747218386539203, 3743165875258953027
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OFFSET
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0,1
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COMMENTS
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For n > 1, a(n) = 14*a(n-1) - a(n-2) - 12; e.g., 5043 = 14*363 - 27 - 12. 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=4, then last(2) = 764 = 18*44 - 4 - 24.
For n > 0, a(n) = 8*a(n-1) + 7*A157088(n-1)+6; e.g., 5043 = 8*312 + 7*363 + 6.
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=4 and n=2, then first(2) = 680 = 9*36 + 8*44 + 4 and last(2) = 764 = 10*36 + 9*44 + 8.
a(n) = 3^n*4((1+sqrt(4/3))^(2n+1) - (1-sqrt(4/3))^(2n+1))/(4*sqrt(4/3)) + 2/2; e.g., 363 = 3^2*4((1+sqrt((4/3))^5 - (1-sqrt(4/3))^5)/(4*sqrt(4/3)) + 2/2. 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=4 and n=2, then last(2) = 764 = 4^2*5((1+sqrt((5/4))^5 - (1-sqrt(5/4))^5)/(4*sqrt(5/4)) + 3/2.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction fraction 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) = 5043 = (84^2 + 3^2*13^2 + 3*2*84*13)/3 and a(4) = 70227 = (3*(209^2 + 97^2 + 2*209*97))/4.
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+25+26+27+361+362+363 = 1170 = 6*195.
Lim_{n->infinity} a(n+1)/a(n) = 3*(1+sqrt(4/3))^2 = 7 + 2*sqrt(12).
In general, if first(n) is the first term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).
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REFERENCES
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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.
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LINKS
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FORMULA
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Empirical g.f.: (3-18*x+3*x^2)/(1-15*x+15*x^2-x^3). - Colin Barker, Jan 01 2012
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EXAMPLE
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a(3)=363 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 362^2 + 363^2.
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MATHEMATICA
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LinearRecurrence[{15, -15, 1}, {3, 27, 363}, 20] (* Harvey P. Dale, May 14 2022 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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