A157088 -id:A157088 - cp's OEIS frontend

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.

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

Offset: 0

Charlie Marion, Mar 12 2009

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).

			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.

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.

G. C. Greubel, Table of n, a(n) for n = 0..740 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Cf. A001652, A157084, A157088, A157092.

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

CoefficientList[Series[5*x*(x-11)/((x-1)*(x^2-22*x+1)),{x,0,20}],x] (* Vincenzo Librandi, Jun 09 2012 *)

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

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

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

A157092 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 X values.

Original entry on oeis.org

0, 36, 680, 12236, 219600, 3940596, 70711160, 1268860316, 22768774560, 408569081796, 7331474697800, 131557975478636, 2360712083917680, 42361259535039636, 760141959546795800, 13640194012307284796, 244763350261984330560, 4392100110703410665316, 78813038642399407645160

Offset: 0

Charlie Marion, Mar 12 2009

nonn

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=5, then first(2) = 1260 = 22*55 - 0 + 50.
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 + 11*65 + 10.
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=5 and n=2, then first(2) = 1260 = (5^3((1+sqrt((6/5))^5 + (1-sqrt(6/5))^5) - 2*5)/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) = 36 = 4*1*9 and a(2) = 680 = 5*8*17.
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).

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

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.

G. C. Greubel, Table of n, a(n) for n = 0..790
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A001652, A157084, A157088, A157096.

[Round((4^(n+1)*((1+Sqrt(5/4))^(2*n+1) + (1-Sqrt(5/4))^(2*n+1)) - 2*4)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017

RecurrenceTable[{a[0]==0,a[1]==36,a[n]==18a[n-1]-a[n-2]+32},a,{n,20}] (* or *) LinearRecurrence[{19,-19,1},{0,36,680},20] (* Harvey P. Dale, Oct 09 2012 *)

x='x+O('x^50); concat([0], Vec(4*x*(-9+x)/((x-1)*(x^2-18*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 18*a(n-1) - a(n-2) + 32.
For n > 0, a(n) = 9*a(n-1) + 8*A157093(n-1) + 4.
a(n) = (4^(n+1)((1+sqrt(5/4))^(2n+1) + (1-sqrt(5/4))^(2n+1)) - 2*4)/4.
Lim_{n->inf} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 4*x*(-9+x)/((x-1)*(x^2-18*x+1)).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3).
a(n) = 4*A119032(n+1). (End)
For n > 0, 1/a(n) = Sum_{k>=1} F(3*k)/phi^(6*k*n + 3*k), where F(n) = A000045(n) and phi = A001622 = (sqrt(5)+1)/2. - Diego Rattaggi, Dec 28 2019
E.g.f.: (1/2)*((2 + sqrt(5))*exp((9+4*sqrt(5))*x) + (2 - sqrt(5))*exp((9-4*sqrt(5))*x) - 4*exp(x)). - Stefano Spezia, Dec 29 2019

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

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

Charlie Marion, Mar 12 2009

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).

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

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.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

Cf. A001652, A157088, A157092, A157096.

[(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

CoefficientList[Series[2*x*(-5 + x)/((x - 1)*(x^2 - 10*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)

x='x+O('x^50); concat([0], Vec(2*(x - 5)/((x-1)*(x^2 - 10*x + 1)))) \\ G. C. Greubel, Nov 04 2017

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)

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

Charlie Marion, Mar 12 2009

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

			a(3) = 363 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 362^2 + 363^2.

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.

Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001653, A157085, A157093, A157097.

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

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)

A261972 The first of three 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

25, 361, 5041, 70225, 978121, 13623481, 189750625, 2642885281, 36810643321, 512706121225, 7141075053841, 99462344632561, 1385331749802025, 19295182152595801, 268747218386539201, 3743165875258953025, 52135575035238803161, 726154884618084291241

Offset: 1

Colin Barker, Sep 07 2015

For the first of the corresponding four consecutive positive integers, see A157088.

			25 is in the sequence because 25^2 + 26^2 + 27^2 = 2030 = 21^2 + 22^2 + 23^2 + 24^2.

Colin Barker, Table of n, a(n) for n = 1..873
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A157088, A261973, A261974.

LinearRecurrence[{15,-15,1},{25,361,5041},20] (* Harvey P. Dale, Jul 16 2025 *)

Vec(-x*(x^2-14*x+25)/((x-1)*(x^2-14*x+1)) + O(x^40))

a(n) = 15*a(n-1)-15*a(n-2)+a(n-3) for n>3.
G.f.: -x*(x^2-14*x+25) / ((x-1)*(x^2-14*x+1)).
a(n) = (-2-(7-4*sqrt(3))^n*(-2+sqrt(3))+(2+sqrt(3))*(7+4*sqrt(3))^n)/2. - Colin Barker, Mar 05 2016

A261973 The first of three consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.

Original entry on oeis.org

137, 6341, 291593, 13406981, 616429577, 28342353605, 1303131836297, 59915722116101, 2754820085504393, 126661808211086021, 5823688357624452617, 267763002642513734405, 12311274433198007330057, 566050860924465823448261, 26026028328092229871289993

Offset: 1

Colin Barker, Sep 07 2015

For the first of the corresponding eleven consecutive positive integers, see A261974.
From Zak Seidov, Sep 07 2015: (Start)
Positive values x of solutions (x, y) to the Diophantine equation 380 + 110x + 11x^2 - 6y - 3y^2 = 0, with values of y in A261974.
Note that there are also solutions with negative x: (x,y) = (-77,137), (-3317, 6341), (-152285, 291593), (-7001573, 13406981), ... with values of y in A261974. (End)

			137 is in the sequence because 137^2 + 138^2 + 139^2 = 57134 = 67^2 + ... + 77^2.

Colin Barker, Table of n, a(n) for n = 1..601
Index entries for linear recurrences with constant coefficients, signature (47,-47,1).

Cf. A157088, A261972, A261974.

I:=[137,6341,291593]; [n le 3 select I[n] else 47*Self(n-1)-47*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Sep 08 2015

LinearRecurrence[{47, -47, 1}, {137, 6341, 291593}, 20] (* Vincenzo Librandi, Sep 08 2015 *)

Vec(-x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)) + O(x^40))

a(n) = 47*a(n-1)-47*a(n-2)+a(n-3) for n>3.
G.f.: -x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)).
a(n) = -1+3*(23+4*sqrt(33))^(-n)+3*(23+4*sqrt(33))^n. - Colin Barker, Mar 03 2016

A261974 The first of eleven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of three consecutive positive integers.

Original entry on oeis.org

67, 3307, 152275, 7001563, 321919843, 14801311435, 680538406387, 31289965382587, 1438657869192835, 66146972017488043, 3041322054935257363, 139834667555004350875, 6429353385475264883107, 295610421064307180272267, 13591650015572655027641395

Offset: 1

Colin Barker, Sep 07 2015

For the first of the corresponding three consecutive positive integers, see A261973.

			67 is in the sequence because 67^2 + ... + 77^2 = 57134 = 137^2 + 138^2 + 139^2.

Colin Barker, Table of n, a(n) for n = 1..601
Index entries for linear recurrences with constant coefficients, signature (47,-47,1).

Cf. A157088, A261972, A261973.

LinearRecurrence[{47,-47,1},{67,3307,152275},20] (* Harvey P. Dale, Jul 02 2016 *)

Vec(x*(5*x^2-158*x-67)/((x-1)*(x^2-46*x+1)) + O(x^40))

a(n) = 47*a(n-1)-47*a(n-2)+a(n-3) for n>3.
G.f.: x*(5*x^2-158*x-67) / ((x-1)*(x^2-46*x+1)).
a(n) = -5+3*sqrt(3/11)*(23+4*sqrt(33))^(-n)*(-1+(23+4*sqrt(33))^(2*n)). - Colin Barker, Mar 03 2016

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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.

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A157092 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 X values.

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

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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.

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A261972 The first of three consecutive positive integers the sum of the squares of which is equal to the sum of the squares of four consecutive positive integers.

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A261973 The first of three consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.

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