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 1-4 of 4 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

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

Original entry on oeis.org

2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

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))^(2*n+1) - (1-sqrt(3/2))^(2*n+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->infinity} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

Examples

			a(3) = 134 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. A001653, A157089, A157093, A157097.

Programs

  • Mathematica
    LinearRecurrence[{11, -11, 1}, {2, 14, 134}, 25] (* Paolo Xausa, May 29 2025 *)
  • PARI
    for(y=0,oo,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

Formula

For n > 1, a(n) = 10*a(n-1) - a(n-2) - 4; e.g., 1322 = 10*134 - 14 - 4.
Limit_{n->oo} a(n+1)/a(n) = 2*(1+sqrt(3/2))^2 = 5 + 2*sqrt(6).
G.f.: 2*(1-4*x+x^2)/((1-x)*(x^2-10*x+1)). - Colin Barker, Jan 01 2012
a(n) = 2*A253175(n+1). - R. J. Mathar, Feb 07 2022

Extensions

Edited by M. F. Hasler, Oct 04 2014

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

A157089 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.

Original entry on oeis.org

3, 27, 363, 5043, 70227, 978123, 13623483, 189750627, 2642885283, 36810643323, 512706121227, 7141075053843, 99462344632563, 1385331749802027, 19295182152595803, 268747218386539203, 3743165875258953027
Offset: 0

Views

Author

Charlie Marion, Mar 12 2009

Keywords

Comments

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.
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.
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.
If u(n) is the numerator and e(n) is the denominator of the n-th continued 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.
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).
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).
The generalization using the binomial formula is : a(n,k)= ((k-1) + (k+1)*k^n*(Sum_{j=0..n} binomial(2*n+1, 2*j+1)*(1+1/k)^j)))/2. - Gerry Martens, Aug 19 2025

Examples

			a(3) = 363 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 362^2 + 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. A001653, A157085, A157093, A157097.

Programs

  • Mathematica
    LinearRecurrence[{15,-15,1},{3,27,363},20] (* Harvey P. Dale, May 14 2022 *)

Formula

For n > 1, a(n) = 14*a(n-1) - a(n-2) - 12.
a(n) = 8*a(n-1) + 7*A157088(n-1) + 6.
G.f.: (3-18*x+3*x^2)/(1-15*x+15*x^2-x^3). - Colin Barker, Jan 01 2012
From Gerry Martens, Aug 19 2025: (Start)
a(n) = ((1 + sqrt(3))^(2*n+1) - (1 - sqrt(3))^(2*n+1))^2/4^(n+1).
a(n) = 1/(3*4^n)*(Sum_{k=0..n} binomial(2*n+1, 2*k+1)*3^(k+1))^2. (End)
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