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 A157088 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values. 8
 0, 21, 312, 4365, 60816, 847077, 11798280, 164328861, 2288805792, 31878952245, 444016525656, 6184352406957, 86136917171760, 1199732487997701, 16710117914796072, 232741918319147325 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=4, then first(2) = 680 = 18*36 - 0 + 32. 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, first(2) = 680 = 9*36 + 8*44 + 4 and last(2) = 764 = 10*36 + 9*44 + 8. In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=4 and n=2, then first(2) = 680 = (4^3((1+sqrt(5/4)^5 + (1-sqrt(5/4))^5)-2*4)/4. 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 first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 4365 = 3*15*97 and a(4) = 60816 = 4*84*181. In general, 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 G. C. Greubel, Table of n, a(n) for n = 0..870 Tanya Khovanova, Recursive Sequences Ron Knott, Pythagorean Triples and Online Calculators Index entries for linear recurrences with constant coefficients, signature (15, -15, 1). FORMULA For n > 1, a(n) = 14*a(n-1) - a(n-2) + 18. For n > 0, a(n) = 7*a(n-1) + 6*A157089(n-1) + 3. Lim_{n->inf} a(n+1)/a(n) = 3(1+sqrt(4/3))^2 = 7 + 2*sqrt(12). a(n) = (3^(n+1)((1+sqrt(4/3))^(2n+1) + (1-sqrt(4/3))^(2n+1)) - 2*3)/4. From R. J. Mathar, Mar 19 2009: (Start) G.f.: 3*x*(-7+x)/((x-1)*(x^2-14*x+1)). a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) = 3*A001921(n). (end) EXAMPLE a(2)=312 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 361^2 + 363^2. MATHEMATICA CoefficientList[Series[3*x*(-7 + x)/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *) PROG (PARI) x='x+O('x^50); concat(, Vec(3*x*(-7+x)/((x-1)*(x^2-14*x+1)))) \\ G. C. Greubel, Nov 04 2017 (MAGMA) [Round((3^(n+1)((1+Sqrt(4/3))^(2n+1)+(1-Sqrt(4/3))^(2n+1))-2*3)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017 CROSSREFS Cf. A001652, A157084, A157092, A157096. Sequence in context: A306428 A317201 A281254 * A226990 A016321 A019041 Adjacent sequences:  A157085 A157086 A157087 * A157089 A157090 A157091 KEYWORD nonn AUTHOR Charlie Marion, Mar 12 2009 STATUS approved

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Last modified May 21 00:57 EDT 2019. Contains 323429 sequences. (Running on oeis4.)