A038761 -id:A038761 - cp's OEIS frontend

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

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

A054488 Expansion of (1+2x)/(1-6x+x^2).

Original entry on oeis.org

1, 8, 47, 274, 1597, 9308, 54251, 316198, 1842937, 10741424, 62605607, 364892218, 2126747701, 12395593988, 72246816227, 421085303374, 2454265004017, 14304504720728, 83372763320351, 485932075201378, 2832219687887917

Offset: 0

Barry E. Williams, May 04 2000

Bisection (even part) of Chebyshev sequence with Diophantine property.
b(n)^2 - 8*a(n)^2 = 17, with the companion sequence b(n)= A077240(n).
The odd part is A077413(n) with Diophantine companion A077239(n).

			8 = a(1) = sqrt((A077240(1)^2 - 17)/8) = sqrt((23^2 - 17)/8)= sqrt(64) = 8.

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

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.
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A002203, A002315, A038761, A077239, A077240, A077413.

Cf. A077241 (even and odd parts).

a:=[1,8];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 19 2020

I:=[1,8]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 19 2020

R:=PowerSeriesRing(Integers(), 21); Coefficients(R!( (1+2*x)/(1-6*x+x^2))); // Marius A. Burtea, Jan 20 2020

a[0]:=1: a[1]:=8: 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},{1,8},30] (* Harvey P. Dale, Oct 09 2017 *)
Table[(LucasL[2*n+1, 2] + Fibonacci[2*n, 2])/2, {n,0,30}] (* G. C. Greubel, Jan 19 2020 *)

my(x='x+O('x^30)); Vec((1+2*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jan 19 2020

apply( {A054488(n)=[1,8]*([0,-1;1,6]^n)[,1]}, [0..30]) \\ M. F. Hasler, Feb 27 2020

[(lucas_number2(2*n+1,2,-1) + lucas_number1(2*n,2,-1))/2 for n in (0..30)] # G. C. Greubel, Jan 19 2020

a(n) = 6*a(n-1) - a(n-2), a(0)=1, a(1)=8.
a(n) = ((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1) + 2*((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n))/(4*sqrt(2)).
a(n) = S(n, 6) + 2*S(n-1, 6), with S(n, x) Chebyshev's polynomials of the second kind, A049310. S(n, 6) = A001109(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-9)^k. - Philippe Deléham, Mar 05 2014
a(n) = (Pell(2*n+2) + 2*Pell(2*n))/2 = (Pell-Lucas(2*n+1) + Pell(2*n))/2. - G. C. Greubel, Jan 19 2020
E.g.f.: (1/4)*exp(3*x)*(4*cosh(2*sqrt(2)*x) + 5*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Jan 27 2020

More terms from James Sellers, May 05 2000

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

Original entry on oeis.org

1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321, 3551015162, 9884647086, 20696833093, 57611945197, 120629983398, 335787024098

Offset: 1

John W. Layman, Nov 29 2006

First differences are apparently in A143608. [R. J. Mathar, Jul 17 2009]

Alternative definition: T_n and (T_n - 1)/2 are triangular numbers. - Raphie Frank, Sep 06 2012

Michael De Vlieger, Table of n, a(n) for n = 1..2613
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.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001108, A001652, A001653, A006452, A008844, A046090, A046172, A077442, A098790, A216134.

A124124 := proc(n)
coeftayl(x*(1+x-2*x^2+x^3+x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)), x=0, n);
end proc:
seq(A124124(n), n=1..20); # Wesley Ivan Hurt, Aug 04 2014
# Alternative:
a[1]:= 1: a[2]:= 2: a[3]:= 6:
for n from 4 to 1000 do
a[n]:= (3 + 2*(n mod 2))*(a[n-1]-a[n-2])+a[n-3]
od:
seq(a[n],n=1..100); # Robert Israel, Aug 13 2014

LinearRecurrence[{1,6,-6,-1,1},{1,2,6,13,37},40] (* Harvey P. Dale, Nov 05 2011 *)
CoefficientList[Series[(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)*(x^2 - 2*x - 1)*(x^2 + 2*x - 1)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 04 2014 *)

for(n=1,10^10,if(issquare(2*n^2+2*n-3),print1(n,", "))) \\ Derek Orr, Aug 13 2014

It appears that a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) if n is even, a(n) = 5*a(n-1)-5*a(n-2)+a(n-3) if n is odd. Can anyone confirm this?
Corrected and confirmed (using the g.f.) by Robert Israel, Aug 27 2014
2*a(n) = sqrt(7+2*A077442(n-1)^2)-1. - R. J. Mathar, Dec 03 2006
a(n) = a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [R. J. Mathar, Jul 17 2009]
For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - Charlie Marion, Jan 03 2012
From Raphie Frank, Sep 06 2012: (Start)
If y = A006452(n), then a(n) = 2y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n))).
Also see A216134 [a(n) = y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n)))].
(End)
From Hermann Stamm-Wilbrandt, Aug 27 2014: (Start)
a(2*n+2) = A098586(2*n).
a(2*n+1) = A098790(2*n).
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>6. (End)
a(2*n+1)^2 + (a(2*n+1)+1)^2 = A038761(n)^2 + 2^2. - Hermann Stamm-Wilbrandt, Aug 31 2014

More terms from Harvey P. Dale, Feb 07 2011

More terms from Wesley Ivan Hurt, Aug 04 2014

A054489 Expansion of (1+4x)/(1-6x+x^2).

Original entry on oeis.org

1, 10, 59, 344, 2005, 11686, 68111, 396980, 2313769, 13485634, 78600035, 458114576, 2670087421, 15562409950, 90704372279, 528663823724, 3081278570065, 17959007596666, 104672767009931, 610077594462920

Offset: 0

Barry E. Williams, May 04 2000

Numbers n such that 8*n^2 + 41 is a square.

(x, y) = (a(n), a(n+1)) are solutions to x^2 + y^2 - 6*x*y = 41. - John O. Oladokun, Mar 17 2021

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

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.
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).

Cf. A000129, A002203, A038761, A054488.

a:=[1,10];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 19 2020

I:=[1,10]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 19 2020

a[0]:=1: a[1]:=10: 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

Table[(LucasL[2*n+1, 2] + 3*Fibonacci[2*n, 2])/2, {n,0,30}] (* G. C. Greubel, Jan 19 2020 *)
LinearRecurrence[{6,-1},{1,10},20] (* Harvey P. Dale, Jun 11 2024 *)

vector(31, n, polchebyshev(n-1,2,3) +4*polchebyshev(n-2,2,3) ) \\ G. C. Greubel, Jan 19 2020

[chebyshev_U(n,3) +4*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Jan 19 2020

a(n) = 6*a(n-1) - a(n-2), a(0)=1, a(1)=10.
a(n) = (10*((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)).
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = ChebyshevU(n,3) + 4*ChebyshevU(n-1,3).
a(n) = (Pell(2*n+2) + 4*Pell(2*n))/2 = (Pell-Lucas(2*n+1) + 3*Pell(2*n))/2.
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 7*sinh(2*sqrt(2)*x)/(2*sqrt(2)) ). (End)

More terms from James Sellers, May 05 2000

A164640 a(n) = 8*a(n-2) for n > 2; a(1) = 1, a(2) = 6.

Original entry on oeis.org

1, 6, 8, 48, 64, 384, 512, 3072, 4096, 24576, 32768, 196608, 262144, 1572864, 2097152, 12582912, 16777216, 100663296, 134217728, 805306368, 1073741824, 6442450944, 8589934592, 51539607552, 68719476736, 412316860416

Offset: 1

Klaus Brockhaus, Aug 20 2009

Interleaving of A001018 and A083233 without initial term 1.

Binomial transform is A164544. Third binomial transform is A038761.

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (0, 8).

Cf. A001018 (powers of 8), A083233 ((3*8^n+(0)^n)/4), A164544, A038761.

[ n le 2 select 5*n-4 else 8*Self(n-2): n in [1..26] ];

LinearRecurrence[{0,8},{1,6},40] (* Harvey P. Dale, Nov 06 2013 *)

a(n)=(7-(-1)^n)*2^(1/4*(6*n-15+3*(-1)^n)) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (7-(-1)^n)*2^(1/4*(6*n -15+3*(-1)^n)).

G.f.: x*(1+6*x)/(1-8*x^2).

G.f. corrected by Klaus Brockhaus, Sep 18 2009

A100525 Bisection of A048654.

Original entry on oeis.org

4, 22, 128, 746, 4348, 25342, 147704, 860882, 5017588, 29244646, 170450288, 993457082, 5790292204, 33748296142, 196699484648, 1146448611746, 6681992185828, 38945504503222, 226991034833504, 1323000704497802, 7711013192153308, 44943078448422046

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

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

Cf. A001109, A038761, A048654.

I:=[4,22,128]; [n le 3 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 13 2015

CoefficientList[Series[(4-2x)/(1-6x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 13 2015 *)
LinearRecurrence[{6,-1},{4,22},30] (* Harvey P. Dale, Mar 25 2016 *)

Vec((4-2*x)/(1-6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

[2*(2*chebyshev_U(n,3) - chebyshev_U(n-1,3)) for n in (0..30)] # G. C. Greubel, Jun 28 2022

G.f.: 2*(2-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(0)=4, a(1)=22, a(n) = 6*a(n-1) - a(n-2) for n>1. - Philippe Deléham, Sep 19 2009
a(n) = 2*A038725(n+1). - R. J. Mathar, Sep 27 2014
a(n) = ( (5 + 4*sqrt(2))*(3 + 2*sqrt(2))^n - (5 - 4*sqrt(2))*(3 - 2*sqrt(2))^n )/(2*sqrt(2)). - Colin Barker, Oct 13 2015
From G. C. Greubel, Jun 28 2022: (Start)
a(n) = 2*( 2*ChebyshevU(n, 3) - ChenyshevU(n-1, 3) ).
E.g.f.: 2*exp(3*x)*( 2*cosh(2*sqrt(2)*x) + (5/(2*sqrt(2)))*sinh(2*sqrt(2)*x) ). (End)

A255236 All positive solutions x of the second class of the Pell equation x^2 - 2*y^2 = -7.

Original entry on oeis.org

5, 31, 181, 1055, 6149, 35839, 208885, 1217471, 7095941, 41358175, 241053109, 1404960479, 8188709765, 47727298111, 278175078901, 1621323175295, 9449763972869, 55077260661919, 321013799998645, 1871005539329951, 10905019435981061, 63559111076556415

Offset: 0

Wolfdieter Lang, Feb 26 2015

For the corresponding y = y2 terms see 2*A038725(n+1).
The Pell equation x^2 - 2*y^2 = 7 has two classes of solutions. See, e.g., the Nagell reference and comments under A254938 and A255233. Here the positive solutions based on the fundamental solution (5, 4) (the second largest positive solution) are considered.
The positive solutions of the first class are given in (A054490(n), 2*A038723(n)), n >= 0.
The combined solutions of both classes are given in (A077446, 4*A077447).
The solutions (x(n), y(n)) of x^2 - 2*y^2 = -7 translate to the solutions (X(n), Y(n)) = (2*y(n) , x(n)) of the Pell equation X^2 - 2*Y^2 = 14.

			n = 2: 181^2 - 2*(2*64)^2  = -7; (4*64)^2 - 2*181^2 = 14.
n = 2: 2*53 + 75 = 181. - _Wolfdieter Lang_, Mar 19 2015

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

Cf. A038725, A054490, A038723, A077446, 4*A077447, A254938, A255233.

I:=[5,31]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

CoefficientList[Series[(5 + x) / (1 - 6 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)

Vec((5 + x)/(1 - 6*x + x^2) + O(x^30)) \\ Michel Marcus, Mar 20 2015

a(n) = 5*S(n, 6) + 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).
G.f.: (5 + x)/(1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2), n >= 2, with a(-1) = -1 and a(0) = 5.
a(n) = 2*A038761(n) + A038762(n), n >= 0. See the Mar 19 comment on A054490. - Wolfdieter Lang, Mar 19 2015
a(n) = ((3-2*sqrt(2))^n*(-8+5*sqrt(2)) + (3+2*sqrt(2))^n*(8+5*sqrt(2))) / (2*sqrt(2)). - Colin Barker, Oct 13 2015

A082981 Start with the sequence S(0)={1,1} and for k>0 define S(k) to be I(S(k-1)) where I denotes the operation of inserting, for i=1,2,3..., the term a(i)+a(i+1) between any two terms for which 4a(i+1)<=5a(i). The listed terms are the initial terms of the limit of this process as k goes to infinity.

Original entry on oeis.org

1, 2, 3, 4, 9, 14, 19, 24, 53, 82, 111, 140, 309, 478, 647, 816, 1801, 2786, 3771, 4756, 10497, 16238, 21979, 27720, 61181, 94642, 128103, 161564, 356589, 551614, 746639, 941664, 2078353, 3215042, 4351731, 5488420, 12113529, 18738638, 25363747

Offset: 1

John W. Layman, May 28 2003

nonn

Conjectures:
(1) the section {a(2n+1)}={1,3,9,19,53,111,...} is A077442, the terms of which are solutions of ax^2+7 = a square,
(2) the section {a(4n+1)}={1,9,53,309,1801,...} is A038761,
(3) the section {a(4n+2)}={2,14,82,478,2786,...} is A077444, the terms of which are solutions of 2x^2+8 = a square,
(4) the sequence {a(4n+2)/2}={1,7,41,239,1393,...} is A002315, the terms of which are solutions of 2x^2+2 = a square,
(5) the section {a(4n+4)}={4,24,140,816,4756,...} is A005319, the terms of which are solutions of 2x^2+4=a square,
(6) the sequence {a(4n+4)/4}={1,6,35,204,1189,...} is A001109, the terms of which are solutions of 8x^2+1=a square.

Ivan Neretin, Table of n, a(n) for n = 1..1000
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001109, A002315, A005319, A038761, A077442, A077444.

Most@Nest[If[#[[-2]] >= 4 #[[-1]], Append[Most@#, #[[-1]] + #[[-2]]], Insert[#, #[[-1]] + #[[-2]], -2]] &, {1, 1}, 47] (* Ivan Neretin, Apr 27 2017 *)

It appears that a(n)=6a(n-4)-a(n-8).

Empirical g.f.: x*(x+1)^2*(x^2+1)^2/((x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Nov 06 2014

A164545 a(n) = 4a(n-1) + 4a(n-2) for n > 1; a(0) = 1, a(1) = 8.

Original entry on oeis.org

1, 8, 36, 176, 848, 4096, 19776, 95488, 461056, 2226176, 10748928, 51900416, 250597376, 1209991168, 5842354176, 28209381376, 136206942208, 657665294336, 3175488946176, 15332616962048, 74032423632896, 357460162379776

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009

Binomial transform of A164544. Second binomial transform of A164640. Inverse binomial transform of A038761.

Harvey P. Dale, Table of n, a(n) for n = 0..1000 [extending from n(164) by Vincenzo Librandi]
Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Index entries for linear recurrences with constant coefficients, signature (4,4).

Cf. A038761, A164544, A164640.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((2+3*r)*(2+2*r)^n+(2-3*r)*(2-2*r)^n)/4: n in [0..21] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 19 2009

LinearRecurrence[{4,4},{1,8},30] (* Harvey P. Dale, Dec 25 2011 *)

[(2*i)^n*(chebyshev_U(n, -i) - 2*i*chebyshev_U(n-1, -i)) for n in (0..30)] # G. C. Greubel, Jul 17 2021

a(n) = 4*a(n-1) + 4*a(n-2) for n > 1; a(0) = 1, a(1) = 8.
a(n) = ((2+3*sqrt(2))*(2+2*sqrt(2))^n + (2-3*sqrt(2))*(2-2*sqrt(2))^n)/4.
G.f.: (1 + 4*x)/(1 - 4*x - 4*x^2).
a(n) = (2*i)^n*( ChebyshevU(n, -i) - 2*i*ChebyshevU(n-1, -i) ). - G. C. Greubel, Jul 17 2021

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 19 2009

cp's OEIS Frontend

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

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

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A054488 Expansion of (1+2*x)/(1-6*x+x^2).

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A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

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A054489 Expansion of (1+4*x)/(1-6*x+x^2).

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A164640 a(n) = 8*a(n-2) for n > 2; a(1) = 1, a(2) = 6.

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A054488 Expansion of (1+2x)/(1-6x+x^2).

A054489 Expansion of (1+4x)/(1-6x+x^2).

A164545 a(n) = 4a(n-1) + 4a(n-2) for n > 1; a(0) = 1, a(1) = 8.