A101386 -id:A101386 - cp's OEIS frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

1, 9, 53, 309, 1801, 10497, 61181, 356589, 2078353, 12113529, 70602821, 411503397, 2398417561, 13979001969, 81475594253, 474874563549, 2767771787041, 16131756158697, 94022765165141, 548004834832149, 3194006243827753, 18616032628134369, 108502189524978461

Offset: 0

Barry E. Williams, May 02 2000

Bisection of A048654. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

This gives part of the (increasingly sorted) positive solutions y to the Pell equation x^2 - 2*y^2 = +7. For the x solutions see A038762. For the other part of solutions see A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			A038762(3)^2 - 2*a(4)^2 = 2547^2 - 2*1801^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001542, A001652, A001653, A038762, A046090, A048654, A055997, A101386, A124124, A253811.

I:=[1, 9]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

a[0]:=1: a[1]:=9: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,9},40] (* Vincenzo Librandi, Nov 16 2011 *)

a(n)=([0,1; -1,6]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

a(n) = (9*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = sqrt(2*(A038762(n))^2-14)/2.
For n>1, a(n)-4a(n-1)=A001541(n)-A001542(n-2); e.g. 309-4*53=97=99-2. - Charlie Marion, Nov 12 2003
For n>0, a(n)=A046090(n)+A001653(n)+A001652(n-1)=A055997(n+1)+A001652(n-1); e.g., 309=120+169+20. - Charlie Marion, Oct 11 2006
G.f.: (1+3*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = third binomial transform of 1,6,8,48,64,384. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n)^2 + 2^2 = A124124(2*n+1)^2 + (A124124(2*n+1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
a(n) = irrational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only part of the general positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference in A038762 on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
a(n) = S(n, 6) + 3*S(n-1, 6), n >= 0, with the Chebyshev S-polynomials evaluated at x=6. See S(n-1, 6) = A001109(n). - Wolfdieter Lang, Mar 30 2015
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

Edited: Replaced the unspecific Pell comment. Moved a formula from the comment section to the formula section. - Wolfdieter Lang, Feb 05 2015

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573, 8072799748093144, 47051725024430291, 274237550398488602

Offset: 0

Barry E. Williams, May 02 2000

This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015

			n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
       65^2 - 2*(2*23)^2 = -7,
      (4*a(2))^2 - 2*A054490(2)^2 =
      (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001653, A054490.

A038725(n) = a(-n).

a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,4},30] (* Harvey P. Dale, Aug 06 2020 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
a(n) = A001653(n+1) - A001109(n). - Antonio Alberto Olivares, Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008
G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - Philippe Deléham, Mar 05 2014
a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015
a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - Wolfdieter Lang, Mar 19 2015
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020

More terms from James Sellers, May 03 2000

A054490 Expansion of (1+5x)/(1-6x+x^2).

Original entry on oeis.org

1, 11, 65, 379, 2209, 12875, 75041, 437371, 2549185, 14857739, 86597249, 504725755, 2941757281, 17145817931, 99933150305, 582453083899, 3394785353089, 19786259034635, 115322768854721, 672150354093691, 3917579355707425, 22833325780150859

Offset: 0

Barry E. Williams, May 04 2000

A Pellian-related second-order recursive sequence.
Third binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
Binomial transform of A164607. - R. J. Mathar, Oct 26 2011
Pisano period lengths: 1, 1, 4, 2, 6, 4, 3, 2, 12, 6, 12, 4, 14, 3, 12, 2, 8, 12, 20, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = x1 = a(n) of the first class of the (generalized) Pell equation x^2 - 2*y^2 = -7. For the corresponding y1 terms see 2*A038723(n). All positive solutions of the second class are given by (x2(n), y2(n)) = (A255236(n), A038725(n+1)), n >= 0. See (A254938(1), 2*A255232(1)) for the fundamental solution (1, 2) of the first class. See the Nagell reference, Theorem 111, p. 210, Theorem 110, p. 208, Theorem 108a, pp. 206-207.
This sequence also gives all positive solutions y = y1 of the first class of the Pell equation x^2 - 2*y^2 = 14. The corresponding solutions x1 are given in 4*A038723. This follows from the preceding comment. (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(0) = -(2*A038761(0) - A038762(0)), a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1.
This follows from the general trivial fact that if X^2 - D*Y^2 = N (X, Y positive integers, D > 1, not a square, and N a non-vanishing integer) then x:= D*Y +/- X and y:= Y +/- X (correlated signs) satisfy x^2 - D*y^2 = -(D-1)*N. with integers x and y. Here D = 2 and N = 7. (End)

			n = 2: sqrt(8*23^2-7) = 65.
2*19 + 27  = 65. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See y(n) in Lemma 7.9 p. 21.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001109, A038723, A038725, A054488, A054489, A101386, A253811, A255236.

a:=[1,11];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 20 2020

I:=[1,11]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

a[0]:=1: a[1]:=11: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006

CoefficientList[Series[(1+5x)/(1-6x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{6, -1}, {1, 11}, 30] (* G. C. Greubel, Jul 26 2018 *)

my(x='x+O('x^30)); Vec((1+5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

[lucas_number1(2*n+1,2,-1) + 3*lucas_number1(2*n,2,-1) for n in (0..30)] # G. C. Greubel, Jan 20 2020

a(n) = 6*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=11.
a(n) = sqrt(8*A038723(n)^2 - 7).
a(n) = (11*((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) - ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = 11*S(n, 6) + 5*S(n-1, 6), n >= 0, with Chebyshev's polynomials S(n, x) (A049310) evaluated at x=6: S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comments above. - Wolfdieter Lang, Feb 26 2015
a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1. See the Mar 19 2015 comment above. - Wolfdieter Lang, Mar 19 2015
From G. C. Greubel, Jan 20 2020: (Start)
a(n) = Pell(2*n+1) + 3*Pell(2*n).
a(n) = ChebyshevU(n,3) + 5*ChebyshevU(n-1,3).
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 2*sqrt(2)*sinh(2*sqrt(2)*x) ). (End)

More terms from James Sellers, May 05 2000

More terms from Vincenzo Librandi, Mar 20 2015

A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

Original entry on oeis.org

3, 13, 75, 437, 2547, 14845, 86523, 504293, 2939235, 17131117, 99847467, 581953685, 3391874643, 19769294173, 115223890395, 671574048197, 3914220398787, 22813748344525, 132968269668363, 774995869665653, 4517006948325555, 26327045820287677, 153445267973400507

Offset: 0

Barry E. Williams, May 03 2000

This gives part of the (increasingly sorted) positive solutions x to the Pell equation x^2 - 2*y^2 = +7. For the y solutions see A038761. The other part of solutions is found in A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			a(3)^2 - 2*A038761(3)^2 = 437^2 - 2*309^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A038725, A038761, A077443, A101386, A253811.

I:=[3, 13]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

LinearRecurrence[{6,-1},{3,13},40] (* Vincenzo Librandi, Nov 16 2011 *)

x='x+O('x^30); Vec((3-5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

a(n) = sqrt(2*(A038761(n))^2+7).
a(n) = (13*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-3*((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = A077443(2n) = A038725(n)+A038725(n+1).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(n-1)+(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Apr 20 2008
G.f.: (3-5*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008, corrected by R. J. Mathar, Nov 06 2011
a(n) = -5*A001109(n) +3*A001109(n+1). - R. J. Mathar, Nov 06 2011
a(n) = rational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only one part of the positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
E.g.f.: exp(3*x)*(3*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from James Sellers, May 04 2000

Unspecific Pell comment replaced by Wolfdieter Lang, Feb 05 2015

A077442 2*a(n)^2 + 7 is a square.

Original entry on oeis.org

1, 3, 9, 19, 53, 111, 309, 647, 1801, 3771, 10497, 21979, 61181, 128103, 356589, 746639, 2078353, 4351731, 12113529, 25363747, 70602821, 147830751, 411503397, 861620759, 2398417561, 5021893803, 13979001969, 29269742059, 81475594253

Offset: 0

Gregory V. Richardson, Nov 06 2002

Lim. n -> Inf. a(n)/a(n-2) = 3 + 2*Sqrt(2) = R1*R2. Lim. k -> Inf. a(2*k-1)/a(2*k) = (9 + 4*Sqrt(2))/7 = R1 (ratio #1). Lim. k -> Inf. a(2*k)/a(2*k-1) = (11 + 6*Sqrt(2))/7 = R2 (ratio #2).

a(n) gives for n >= 0 all positive y-values solving the (generalized) Pell equation x^2 - 2*y^2 = 7. A077443(n+1) gives the corresponding x-values. See, e.g., the Nagell reference on how to find all solutions. - Wolfdieter Lang, Feb 05 2015

			a(4)^2 - 2*a(3)^2 = 27^2 - 2*19^2  = +7. - _Wolfdieter Lang_, Feb 05 2015

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077443, A038762, A038761, A101386, A253811.

CoefficientList[Series[(1+3 x+3 x^2+x^3)/ (1-6 x^2+x^4),{x,0,50}],x]  (* Harvey P. Dale, Mar 12 2011 *)
LinearRecurrence[{0, 6, 0, -1},{1,3,9,19},50] (* Sture Sjöstedt, Oct 08 2012 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;3;9;19])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Vec((x+1)^3/(x^2+2*x-1)/(x^2-2*x-1) + O(x^50)) \\ Colin Barker, Mar 27 2016

For n>0, a(2n) = A046090(n) + A001653(n) + A001652(n-1); a(2n+1) = A001652(n+1) - A001652(n-1) - A001653(n-1); e.g. 53=21+29+3; 111=119-3-5. - Charlie Marion, Aug 14 2003
The same recurrences hold for the odd and even indices respectively : a(n+2) = 6*a(n+1) - a(n), a(n+1) = 3*a(n) + 2*(2*a(n)^2+7)^0.5. - Richard Choulet, Oct 11 2007
G.f.: (x+1)^3/(x^2+2*x-1)/(x^2-2*x-1). a(n)= [ -A077985(n)-3*A077985(n-1)+3*A000129(n+1)+A000129(n)]/2. - R. J. Mathar, Nov 16 2007
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=3, a(3)=9, a(4)=19. - Sture Sjöstedt, Oct 08 2012
a(n) = ((-(-1 - sqrt(2))^n*(-2+sqrt(2)) - (-1+sqrt(2))^n*(2+sqrt(2)) + (1-sqrt(2))^n*(-4+3*sqrt(2)) + (1+sqrt(2))^n*(4+3*sqrt(2))))/(4*sqrt(2)). - Colin Barker, Mar 27 2016

Edited: n in Name replaced by a(n). Pell comments moved to comment section. - Wolfdieter Lang, Feb 05 2015

A077443 Numbers k such that (k^2 - 7)/2 is a square.

Original entry on oeis.org

3, 5, 13, 27, 75, 157, 437, 915, 2547, 5333, 14845, 31083, 86523, 181165, 504293, 1055907, 2939235, 6154277, 17131117, 35869755, 99847467, 209064253, 581953685, 1218515763, 3391874643, 7102030325, 19769294173, 41393666187, 115223890395, 241259966797, 671574048197

Offset: 1

Gregory V. Richardson, Nov 06 2002

Lim_{n -> inf} a(n)/a(n-2) = 3 + 2*sqrt(2) = R1*R2. Lim_{k -> inf} a(2*k-1)/a(2*k) = (9 + 4*sqrt(2))/7 = R1 = A156649 (ratio #1). Lim_{k -> inf} a(2*k)/a(2*k-1) = (11 + 6*sqrt(2))/7 = R2 (ratio #2).
Also gives solutions > 3 to the equation x^2-4 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
From Paul Curtz, Dec 15 2012: (Start)
a(n-1) and A006452(n) are companions. Like A000129 and A001333.
Reduced mod 10 this is a sequence of period 12: 3, 5, 3, 7, 5, 7, 7, 5, 7, 3, 5, 3.
(End)
The Pisano periods (periods of the sequence reducing a(n) modulo m) for m>=1 are 1,  1,  8,  4, 12,  8,  6,  4, 24, 12, 24,  8, 28, ... R. J. Mathar, Dec 15 2012
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 56 = 0. - Colin Barker, Feb 08 2014
From Wolfdieter Lang, Feb 05 2015: (Start)
a(n+1) gives for n >= 0 all positive x solutions of the (generalized) Pell equation x^2 - 2*y^2 = +7.
The corresponding y solutions are given in A077442(n), n >= 0. The, e.g., the Nagell reference for finding all solutions.
Because the primitive Pythagorean triangle (3,4,5) is the only one with the sum of legs equal to 7 all positive solutions (x(n),y(n)) = (a(n+1),A077442(n)) of the Pell equation x^2 - 2*y^2 = +7 satisfy x(n) - y(n) < y(n) if n >= 1; only the first solution (x(0),y(0)) = (3,2) satisfies 3-1 > 1. Proof: Primitive Pythagorean triangles are characterized by the positive integer pairs [u,v] with  u+v odd, gcd(u,v) = 1 and u > v. See the Niven et al. reference, Theorem 5.5, p. 232. The leg sum is L = (u+v)^2 - 2*v^2. With L = 7, x = u+v and y = v, every solution (x(n),y(n)) with x(n)-y(n) = u(n) > v(n) = y(n) will correspond to a primitive Pythagorean triangle. Note that because of gcd(x,y) = 1 also gcd(u,v) = 1. But there is only one such triangle with L=7, namely the one with  [u(0),v(0)] = [2,1]. All other solutions with n >= 1 must therefore satisfy x(n)-y(n) < y(n). (End)
For n > 0, a(n+1) is the n-th almost Lucas-cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

			a(3)^2 - 2*A077442(2)^2 = 13^2 - 2*9^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A000129, A001333, A006452, A038761, A038762, A077442, A101386, A124124, A156649, A176981, A216134, A253811.

LinearRecurrence[{0,6,0,-1},{3,5,13,27},50] (* Sture Sjöstedt, Oct 09 2012 *)

a(2n+1) = A038762(n). a(2n) = A101386(n-1).
The same recurrences hold for the odd and the even indices: a(n+2) = 6*a(n) - a(n-2), a(n+1) = 3*a(n) + 2*(2*a(n)^2-14)^0.5 - Richard Choulet, Oct 11 2007
O.g.f.: -x*(x-1)*(3*x^2+8*x+3) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Nov 23 2007
If n is even a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(-n/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(-n/2); if n is odd a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^((n-1)/2) + (1/2)*(3-sqrt(2))*(3-2*sqrt(2))^((n-1)/2). - Antonio Alberto Olivares, Apr 20 2008
a(n) = A000129(n+1) + (-1)^n*A176981(n-1), n>1. - R. J. Mathar, Jul 03 2011
a(n) = A000129(n+1) -(-1)^n*A000129(n-2), rephrasing the formula above. - Paul Curtz, Dec 07 2012
a(n) = sqrt(8*A216134(n)^2 + 8*A216134(n) + 9) = 2*A124124(n) + 1. - Raphie Frank, May 24 2013
E.g.f.: cosh(sqrt(2)*x)*(3*cosh(x) - sinh(x)) + sqrt(2)*(2*cosh(x) - sinh(x))*sinh(sqrt(2)*x) - 3. - Stefano Spezia, Nov 25 2022

More terms from Richard Choulet, Oct 11 2007

Edited: replaced n by a(n) in the name. Moved Pell remarks to the comment section. Added cross references. - Wolfdieter Lang, Feb 05 2015

A253811 Part of the y solutions of the Pell equation x^2 - 2*y^2 = +7.

Original entry on oeis.org

3, 19, 111, 647, 3771, 21979, 128103, 746639, 4351731, 25363747, 147830751, 861620759, 5021893803, 29269742059, 170596558551, 994309609247, 5795261096931, 33777256972339, 196868280737103, 1147432427450279, 6687726283964571, 38978925276337147, 227185825374058311

Offset: 0

Wolfdieter Lang, Feb 05 2015

All positive solutions y = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding x solutions are given by x(n) = A101386(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See the Nagell reference on how to find all solutions.

			A101386(2)^2 - 2*a(2) = 157^2 - 2*111^2 = +7.

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A001109, A038761, A038762, A077442, A077443, A101386.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x+3)/(x^2-6*x+1))); // G. C. Greubel, Jul 26 2018

LinearRecurrence[{6,-1}, {3,19}, 30] (* or *) CoefficientList[Series[ (x+3)/(x^2-6*x+1), {z, 0, 50}], x]  (* G. C. Greubel, Jul 26 2018 *)

Vec((x+3)/(x^2-6*x+1) + O(x^100)) \\ Colin Barker, Feb 05 2015

a(n) = irrational part of z(n), where z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions.
From Colin Barker, Feb 05 2015: (Start)
a(n) = 6*a(n-1) - a(n-2).
G.f.: (x+3) / (x^2-6*x+1). (End)
a(n) = 3*A001109(n+1) + A001109(n). - R. J. Mathar, Feb 05 2015
E.g.f.: exp(3*x)*(6*cosh(2*sqrt(2)*x) + 5*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

A164737 a(n) = 8*a(n-2) for n > 2; a(1) = 5, a(2) = 12.

Original entry on oeis.org

5, 12, 40, 96, 320, 768, 2560, 6144, 20480, 49152, 163840, 393216, 1310720, 3145728, 10485760, 25165824, 83886080, 201326592, 671088640, 1610612736, 5368709120, 12884901888, 42949672960, 103079215104, 343597383680, 824633720832

Offset: 1

Klaus Brockhaus, Aug 24 2009

nonn

Interleaving of 5*A001018 and 12*A001018.

Binomial transform is A096980 without initial terms 1. Second binomial transform is A164593. Third binomial transform is A101386.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,8).

Cf. A001018 (powers of 8), A067412, A096980, A101386, A164593.

[ n le 2 select 7*n-2 else 8*Self(n-2): n in [1..26] ];

seq(coeff(series( x*(5+12*x)/(1-8*x^2) , x, n+1), x, n), n=1..30); # G. C. Greubel, Apr 16 2020

LinearRecurrence[{0,8}, {5,12}, 30] (* G. C. Greubel, Apr 16 2020 *)

[(13 -7*(-1)^n)*2^((6*n -11 +3*(-1)^n)/4) for n in (1..30)] # G. C. Greubel, Apr 16 2020

a(n) = (13 - 7*(-1)^n)*2^(1/4*(6*n - 11 + 3*(-1)^n)).

G.f.: x*(5 + 12*x)/(1 - 8*x^2).

A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

Original entry on oeis.org

2, 2, 1, 3, 4, 8, 9, 19, 22, 46, 53, 111, 128, 268, 309, 647, 746, 1562, 1801, 3771, 4348, 9104, 10497, 21979, 25342, 53062, 61181, 128103, 147704, 309268, 356589, 746639, 860882, 1802546, 2078353, 4351731, 5017588, 10506008, 12113529, 25363747, 29244646, 61233502

Offset: 0

Raphie Frank, Dec 30 2015

This sequence gives all x in N | 2*x^2 - 7(-1)^x = y^2. The companion sequence to this sequence, giving y values, is A266505.
A266505(n)/a(n) converges to sqrt(2).
Alternatively, 1/4*(3*A002203(floor[n/2]) - A002203(n-(-1)^n)), where A002203 gives the Companion Pell numbers, or, in Lucas sequence notation, V_n(2, -1).
Alternatively, bisection of A266506.
Alternatively, A048654(n -1) and A078343(n + 1) interlaced.
Alternatively, A100525(n-1), A266507(n), A038761(n) and A253811(n) interlaced.
Let b(n) = (a(n) - a(n)(mod 2))/2, that is b(n) = {1, 1, 0, 1, 2, 4, 4, 9, 11, 23, 26, 55, 64, ...}. Then:
A006452(n) = {b(4n+0) U b(4n+1)} gives n in N such that n^2 - 1 is triangular;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that n^2 + n + 1 is triangular (indices of Sophie Germain triangular numbers);
A216162(n) = {b(4n+0) U b(4n+2) U b(4n+1) U b(4n+3)}, sequences A006452 and A216134 interlaced.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[2,2,1,3]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{0, 2, 0, 1}, {2, 2, 1, 3}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(1 - x) (2 + 4 x + x^2)/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 41}] (* Michael De Vlieger, Dec 31 2015 *)

Vec((1-x)*(2+4*x+x^2)/(1-2*x^2-x^4) + O(x^50)) \\ Colin Barker, Dec 31 2015

a(n) = 1/sqrt(8)*(+sqrt(2)*(1+sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n - 3*(1-sqrt(2))^(floor(n/2)-(-1)^n) + sqrt(2)*(1-sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n + 3*(1+sqrt(2))^(floor(n/2)-(-1)^n)).
a(n) = 1/4*((3*((1+sqrt(2))^floor(n/2)+(1-sqrt(2))^floor(n/2))) - (-1)^n*((1+sqrt(2))^(floor(n/2)-(-1)^n)+(1-sqrt(2))^(floor(n/2)-(-1)^n))).
a(2n) = (+sqrt(2)*(1+sqrt(2))^(n-1) - 3 *(1-sqrt(2))^(n-1) + sqrt(2)*(1-sqrt(2))^(n-1) + 3*(1 + sqrt(2))^(n-1))/sqrt(8) = A048654(n -1).
a(2n) = 1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) - ((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))) = A048654(n -1).
a(2n + 1) = (-sqrt(2)*(1+sqrt(2))^(n+1) - 3 *(1-sqrt(2))^(n+1) - sqrt(2)*(1-sqrt(2))^(n+1) + 3*(1+sqrt(2))^(n+1))/sqrt(8) = A078343(n + 1).
a(2n + 1) =1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) + ((1+sqrt(2))^(n+1)+(1-sqrt(2))^(n+1))) =  A078343(n + 1).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A100525(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A266507(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038761(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A253811(n).
sqrt(2*a(n)^2 - 7(-1)^a(n))*sgn(2*n - 1) = A266505(n).
(a(2n + 1) + a(2n))/2 = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 1) - a(2n))/2 = A000129(n), where A000129 gives the Pell numbers.
(a(2n+2) + a(2n+1))*2 = A002203(n+2)
(a(2n+2) - a(2n+1))*2 = A002203(n-1).
G.f.: (1-x)*(2+4*x+x^2) / (1-2*x^2-x^4). - Colin Barker, Dec 31 2015

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

Original entry on oeis.org

-1, 1, 3, 5, 5, 11, 13, 27, 31, 65, 75, 157, 181, 379, 437, 915, 1055, 2209, 2547, 5333, 6149, 12875, 14845, 31083, 35839, 75041, 86523, 181165, 208885, 437371, 504293, 1055907, 1217471, 2549185, 2939235, 6154277, 7095941, 14857739, 17131117, 35869755, 41358175, 86597249, 99847467

Offset: 0

Raphie Frank, Dec 30 2015

a(n)/A266504(n) converges to sqrt(2).
Alternatively, bisection of A266506.
Alternatively, A135532(n) and A048655(n) interlaced.
Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016

LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* Michael De Vlieger, Dec 31 2015 *)

my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ G. C. Greubel, Jul 26 2018

G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
(sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
(a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
(a(2n + 2) - a(2n + 1))/2 = A000129(n-1).

cp's OEIS Frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

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A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

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A054490 Expansion of (1+5*x)/(1-6*x+x^2).

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A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

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A077442 2*a(n)^2 + 7 is a square.

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A054490 Expansion of (1+5x)/(1-6x+x^2).