cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A157096 Consider all consecutive integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 55, 1260, 27715, 608520, 13359775, 293306580, 6439385035, 141373164240, 3103770228295, 68141571858300, 1496010810654355, 32844096262537560, 721074106965172015, 15830786256971246820, 347556223546402258075, 7630406131763878430880, 167521378675258923221335
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

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=6, then first(2) = 2100 = 26*78 - 0 + 72.
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=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.
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=6 and n=2, then first(2) = 2100 = (6^3((1+sqrt((7/6))^5 + (1-sqrt(7/6))^5) - 2*6)/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(1) = 55 = 5*1*11 and a(2) = 1260 = 6*10*21.
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 + 2*sqrt(k^2+k).

Examples

			a(2)=55 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.

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

Crossrefs

Cf. A001652, A157084, A157088, A157092.

Programs

  • Magma
    I:=[0, 55, 1260]; [n le 3 select I[n] else 23*Self(n-1) - 23*Self(n-2) + Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 09 2012
  • Mathematica
    CoefficientList[Series[5*x*(x-11)/((x-1)*(x^2-22*x+1)),{x,0,20}],x] (* Vincenzo Librandi, Jun 09 2012 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(5*x*(x-11)/((x-1)*(x^2-22*x+1)))) \\ G. C. Greubel, Nov 04 2017

Formula

For n > 1, a(n) = 22*a(n-1) - a(n-2) + 50.
For n > 0, a(n) = 11*a(n-1) + 10*A157097(n-1) + 5.
a(n) = (5^(n+1)*((1+sqrt(6/5))^(2n+1) + (1-sqrt(6/5))^(2n+1)) - 2*5)/4.
Lim_{n->inf} a(n+1)/a(n) = 5(1+sqrt(6/5))^2 = 11+2*sqrt(30).
G.f.: 5*x*(x-11)/((x-1)*(x^2-22*x+1)). - Colin Barker, Jun 08 2012
a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). Vincenzo Librandi, Jun 09 2012
a(n) = 5*(-1/2+1/20*(11+2*sqrt(30))^(-n)*(5-sqrt(30)+(5+sqrt(30))*(11+2*sqrt(30))^(2*n))). - Colin Barker, Mar 03 2016

Extensions

Terms a(15) onward added by G. C. Greubel, Nov 06 2017

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.

Original entry on oeis.org

0, 21, 312, 4365, 60816, 847077, 11798280, 164328861, 2288805792, 31878952245, 444016525656, 6184352406957, 86136917171760, 1199732487997701, 16710117914796072, 232741918319147325, 3241676738553266496, 45150732421426583637, 628868577161418904440, 8759009347838438078541
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

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 + 2*sqrt(k^2+k).

Examples

			a(2)=312 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 361^2 + 363^2.

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

Crossrefs

Cf. A001652, A157084, A157092, A157096.

Programs

  • Magma
    [Round((3^(n+1)*((1+Sqrt(4/3))^(2*n+1)+(1-Sqrt(4/3))^(2*n+1))-2*3)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017
  • Mathematica
    CoefficientList[Series[3*x*(-7 + x)/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)
  • PARI
    my(x='x+O('x^50)); concat([0], Vec(3*x*(-7+x)/((x-1)*(x^2-14*x+1)))) \\ G. C. Greubel, Nov 04 2017

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.
Limit_{n->oo} a(n+1)/a(n) = 3*(1+sqrt(4/3))^2 = 7 + 2*sqrt(12).
a(n) = (3^(n+1)*((1+sqrt(4/3))^(2*n+1) + (1-sqrt(4/3))^(2*n+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)

A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 10, 108, 1078, 10680, 105730, 1046628, 10360558, 102558960, 1015229050, 10049731548, 99482086438, 984771132840, 9748229241970, 96497521286868, 955226983626718, 9455772314980320
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

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=3, then first(2) = 312 = 14*21 - 0 + 18.
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=3 and n=2, first(2) = 312 = 7*21 + 6*27 + 3 and last(2) = 363 = 8*21 + 7*27 + 6. a(n) = (2^(n+1)((1+sqrt(3/2))^(2n+1) + (1-sqrt(3/2))^(2n+1)) - 2*2)/4; e.g., 108 = (2^3((1+sqrt(3/2))^5 + (1-sqrt(3/2))^5) - 2*2)/4.
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=3 and n=2, then first(2) = 312 = (3^3((1+sqrt((4/3))^(5) + (1-sqrt(4/3)^5) - 2*3)/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) = 1078 = 2*11*49 and a(4) = 10680 = 3*40*89.
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 + 2*sqrt(k^2+k).

Examples

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

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

Crossrefs

Cf. A001652, A157088, A157092, A157096.

Programs

  • Magma
    [(1/2 + Sqrt(6)/4)*(5 + 2*Sqrt(6))^n - (Sqrt(6)/4 - 1/2)*(5 - 2*Sqrt(6))^n - 1: n in [0..50]]; // G. C. Greubel, Nov 04 2017
  • Mathematica
    CoefficientList[Series[2*x*(-5 + x)/((x - 1)*(x^2 - 10*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(2*(x - 5)/((x-1)*(x^2 - 10*x + 1)))) \\ G. C. Greubel, Nov 04 2017

Formula

For n > 1, a(n) = 10*a(n-1) - a(n-2) + 8.
For n > 0, a(n) = 5*a(n-1) + 4*A157085(n-1) + 2.
Lim_{n->inf} a(n+1)/a(n) = 2(1 + sqrt(3/2))^2 = 5 + 2*sqrt(6).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 2*x*(x - 5)/((x-1)*(x^2 - 10*x + 1)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3).
a(n) = 2*A087125(n) = A054320(n) - 1. (End)
From Sergei N. Gladkovskii, Jan 12 2012: (Start)
G.f.: 1/(x-1) + (x+1)/(x^2-10*x+1).
a(n) = (1/2 + sqrt(6)/4)*(5 + 2*sqrt(6))^n - (sqrt(6)/4 - 1/2)*(5 - 2*sqrt(6))^n - 1. (End)

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.

Original entry on oeis.org

4, 44, 764, 13684, 245524, 4405724, 79057484, 1418628964, 25456263844, 456794120204, 8196837899804, 147086288076244, 2639356347472564, 47361327966429884, 849864547048265324, 15250200518902345924, 273653744793193961284, 4910517205758588957164, 88115655958861407267644
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

Comments

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.
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.
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.
In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} last(n+1)/last(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

Examples

			a(2)=764 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.

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

Crossrefs

Cf. A001653, A157085, A157089, A157097.

Programs

  • Mathematica
    LinearRecurrence[{19, -19, 1}, {4, 44, 764}, 25] (* Paolo Xausa, May 29 2025 *)

Formula

For n > 1, a(n) = 18*a(n-1) - a(n-2) - 24.
For n > 0, a(n) = 10*A157092(n-1) + 9*a(n-1) + 8.
a(n) = 4^n*5*((1+sqrt(5/4))^(2*n+1) - (1-sqrt(5/4))^(2*n+1))/(4*sqrt(5/4)) + 3/2.
Limit_{n->oo} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
Empirical g.f.: 4*(1-8*x+x^2)/((1-x)*(1-18*x+x^2)). - Colin Barker, Mar 27 2012

A261995 The first of four consecutive positive integers the sum of the squares of which is equal to the sum of the squares of twenty-one consecutive positive integers.

Original entry on oeis.org

42, 123, 315, 1827, 4659, 13650, 34794, 201114, 512610, 1501539, 3827187, 22120875, 56382603, 165155802, 420955938, 2433095298, 6201573882, 18165636843, 46301326155, 267618362067, 682116744579, 1998054897090, 5092724921274, 29435586732234, 75026640329970
Offset: 1

Views

Author

Colin Barker, Sep 08 2015

Keywords

Comments

For the first of the corresponding twenty-one consecutive positive integers, see A261996.

Examples

			42 is in the sequence because 42^2 + ... + 45^2 = 7574 = 8^2 + ... + 28^2.

Links

Crossrefs

Cf. A157092, A246642, A261996.

Programs

  • PARI
    Vec(-3*x*(6*x^8+8*x^6+27*x^5-596*x^4+504*x^3+64*x^2+27*x+14)/((x-1)*(x^8-110*x^4+1)) + O(x^40))

Formula

G.f.: -3*x*(6*x^8+8*x^6+27*x^5-596*x^4+504*x^3+64*x^2+27*x+14) / ((x-1)*(x^8-110*x^4+1)).

A261996 The first of twenty-one consecutive positive integers the sum of the squares of which is equal to the sum of the squares of four consecutive positive integers.

Original entry on oeis.org

8, 44, 128, 788, 2024, 5948, 15176, 87764, 223712, 655316, 1670312, 9654332, 24607376, 72079892, 183720224, 1061889836, 2706588728, 7928133884, 20207555408, 116798228708, 297700153784, 872022648428, 2222647375736, 12846743269124, 32744310328592
Offset: 1

Views

Author

Colin Barker, Sep 08 2015

Keywords

Comments

For the first of the corresponding four consecutive positive integers, see A261995.

Examples

			8 is in the sequence because 8^2 + ... + 28^2 = 7574 = 42^2 + ... + 45^2.

Links

Crossrefs

Cf. A157092, A246642, A261995.

Programs

  • PARI
    Vec(4*x*(x^8+3*x^7+3*x^6+9*x^5-89*x^4-165*x^3-21*x^2-9*x-2)/((x-1)*(x^8-110*x^4+1)) + O(x^40))

Formula

G.f.: 4*x*(x^8+3*x^7+3*x^6+9*x^5-89*x^4-165*x^3-21*x^2-9*x-2) / ((x-1)*(x^8-110*x^4+1)).
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