A135532 -id:A135532 - cp's OEIS frontend

A162255 a(n) = 2*a(n-2) for n > 2; a(1) = 3, a(2) = 2.

3, 2, 6, 4, 12, 8, 24, 16, 48, 32, 96, 64, 192, 128, 384, 256, 768, 512, 1536, 1024, 3072, 2048, 6144, 4096, 12288, 8192, 24576, 16384, 49152, 32768, 98304, 65536, 196608, 131072, 393216, 262144, 786432, 524288, 1572864, 1048576, 3145728, 2097152

Offset: 1

Klaus Brockhaus, Jun 29 2009

Apparently a(n) = A074323(n+1). a(n) = A072946(n-1) for n > 1.
Partial sums are in A164053.
Binomial transform is A135532 without initial term -1. Second binomial transform is A161938.

Index entries for linear recurrences with constant coefficients, signature (0, 2).

Cf. A074323, A072946, A164053, A135532, A161938.

LinearRecurrence[{0,2},{3,2},50] (* Harvey P. Dale, Aug 28 2012 *)

m=42; u=concat([3, 2], vector(m-2)); for(n=3, m, u[n]=2*u[n-2]); u

a(n) = (2^(1/4))^(3+2*n+(-1)^n) * (2-(-1)^n)/2.
G.f.: x*(3+2*x)/(1-2*x^2).
E.g.f.: cosh(sqrt(2)*x) + 3*sinh(sqrt(2)*x)/sqrt(2) - 1. - Stefano Spezia, May 26 2024

G.f. corrected, comments and cross-references added by Klaus Brockhaus, Aug 08 2009

Corrected by Harvey P. Dale, Aug 28 2012

A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

Original entry on oeis.org

1, 2, 6, 15, 37, 90, 218, 527, 1273, 3074, 7422, 17919, 43261, 104442, 252146, 608735, 1469617, 3547970, 8565558, 20679087, 49923733, 120526554, 290976842, 702480239, 1695937321, 4094354882, 9884647086, 23863649055, 57611945197

Offset: 0

Creighton Dement, Oct 30 2004

nonn

Previous name was: a(n) = A048739(n) - A000129(n).

Partial sums of Pell numbers A000129 except omit next-to-last Pell number. E.g., 37 = 0+1+2+5+12+29 - 12.

M. Bicknell-Johnson and G. E. Bergum, The Generalized Fibonacci Numbers {C(n)}, C(n)=C(n-1)+C(n-2)+K, Applications of Fibonacci Numbers, 1986, pp. 193-205.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3, -1, -1).

Cf. A006451, A000129, A048739, A124124, A024537, A074323, A052542.

a[0] = 1; a[1] = 2; a[n_] := a[n] = 2a[n - 1] + a[n - 2] + 1; Table[ a[n], {n, 0, 28}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{3,-1,-1},{1,2,6},31] (* Harvey P. Dale, Oct 15 2011 *)
CoefficientList[Series[(x^2 - x + 1)/((1 - x) (1 - 2 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 14 2014 *)

a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.
G.f.: (x^2-x+1)/((1-x)(1-2x-x^2)).
a(n+1) = - A024537(n+1) + 2*A048739(n+1) - 2*A048739(n).
a(n) = - A024537(n) + A052542(n+1).
Partial sums of A074323. - Paul Barry, Mar 11 2007
a(n) = (sqrt(2)+1)^n*(3/4+sqrt(2)/4)+(sqrt(2)-1)^n*(3/4-sqrt(2)/4)*(-1)^n-1/2; - Paul Barry, Mar 11 2007
a(0)=1, a(1)=2, a(2)=6, a(n)=3*a(n-1)-a(n-2)-a(n-3). [Harvey P. Dale, Oct 15 2011]
a(2*n) = A124124(2*n+1). - Hermann Stamm-Wilbrandt, Aug 03 2014
a(2*n+1) = A006451(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
2*a(n) = A135532(n+1)-1. - R. J. Mathar, Jan 13 2023

More terms from Robert G. Wilson v, Nov 04 2004
Definition edited by N. J. A. Sloane, Aug 03 2014
New name from existing formula by Joerg Arndt, Aug 13 2014

A164682 a(n) = 2*a(n-2) for n > 2; a(1) = 5, a(2) = 8.

Original entry on oeis.org

5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, 2560, 4096, 5120, 8192, 10240, 16384, 20480, 32768, 40960, 65536, 81920, 131072, 163840, 262144, 327680, 524288, 655360, 1048576, 1310720, 2097152, 2621440, 4194304

Offset: 1

Klaus Brockhaus, Aug 21 2009

Interleaving of A020714 and A000079 without initial terms 1, 2, 4.
First differences are in A162255.
Binomial transform is A135532 without initial terms -1, 3. Fourth binomial transform is A164537.

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

Equals A094958 (numbers of the form 2^n or 5*2^n) without initial terms 1, 2, 4.

Cf. A020714 (5*2^n), A000079 (powers of 2), A162255, A135532, A164537.

[ n le 2 select 2+3*n else 2*Self(n-2): n in [1..40] ];

LinearRecurrence[{0,2},{5,8},60] (* Harvey P. Dale, Jul 20 2022 *)

a(n) = (9-(-1)^n)*2^(1/4*(2*n-5+(-1)^n)).

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

A048745 Partial sums of A048654.

Original entry on oeis.org

1, 5, 14, 36, 89, 217, 526, 1272, 3073, 7421, 17918, 43260, 104441, 252145, 608734, 1469616, 3547969, 8565557, 20679086, 49923732, 120526553, 290976841, 702480238, 1695937320, 4094354881, 9884647085, 23863649054, 57611945196, 139087539449, 335787024097

Offset: 0

Barry E. Williams

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A000129, A005409, A098790.

I:=[1,5,14]; [n le 3 select I[n] else 3*Self(n-1) -Self(n-2) -Self(n-3): n in [1..31]]; // G. C. Greubel, May 23 2021

t={1,5}; Do[AppendTo[t, t[[-2]] + 2*t[[-1]] + 3], {n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Jan 27 2012 *)
Accumulate[LinearRecurrence[{2,1},{1,4},30]] (* or *) LinearRecurrence[{3,-1,-1},{1,5,14},30] (* Harvey P. Dale, Aug 03 2020 *)

a(n)=polcoeff((1+2*x)/(1-3*x+x^2+x^3)+x*O(x^n),n) \\ Paul D. Hanna

[(5*lucas_number1(n+1,2,-1) + 3*lucas_number1(n,2,-1) -3)/2 for n in (0..30)] # G. C. Greubel, May 23 2021

a(n) = 2*a(n-1) + a(n-2) + 3, a(0)=1, a(1)=5.
a(n) = ( ((4+(5/2)*sqrt(2))*(1+sqrt(2))^n - (4-(5/2)*sqrt(2))*(1-sqrt(2))^n)/ 2*sqrt(2) ) - 3/2.
G.f.: (1+2*x)/((1-x)*(1-2*x-x^2)). - Paul D. Hanna, Feb 22 2005
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n>2, a(0)=1, a(1)=5, a(2)=14. - Philippe Deléham, Dec 16 2008
2*a(n) = A135532(n+2) - 3. - R. J. Mathar, Mar 06 2013
a(n) = (1/2)*( 5*P(n+1) + 3*P(n) - 3), where P(n) = A000129(n). - G. C. Greubel, May 23 2021

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).

A161938 a(n) = ((3+sqrt(2))(2+sqrt(2))^n + (3-sqrt(2))(2-sqrt(2))^n)/2.

Original entry on oeis.org

3, 8, 26, 88, 300, 1024, 3496, 11936, 40752, 139136, 475040, 1621888, 5537472, 18906112, 64549504, 220385792, 752444160, 2569005056, 8771131904, 29946517504, 102243806208, 349082189824, 1191841146880, 4069200207872, 13893118537728, 47434073735168

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jun 22 2009, Jun 29 2009

Second binomial transform of A162255.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Cf. A162255, A135532, A083878.

a:=[3,8];; for n in [3..25] do a[n]:=4*a[n-1]-2*a[n-2]; od; a; # Muniru A Asiru, Sep 28 2018

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((3+r)*(2+r)^n+(3-r)*(2-r)^n)/2: n in [0..23] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 01 2009

I:=[3,8]; [n le 2 select I[n] else 4*Self(n-1) - 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Sep 28 2018

seq(coeff(series((3-4*x)/(1-4*x+2*x^2),x,n+1), x, n), n = 0..25); # Muniru A Asiru, Sep 28 2018

CoefficientList[Series[(3-4*x)/(1-4*x+2*x^2), {x,0,50}], x] (* G. C. Greubel, Sep 28 2018 *)

my(x='x+O('x^50)); Vec((3-4*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Sep 28 2018

a(n) = 4*a(n-1) - 2*a(n-2) for n>1; a(0) = 3; a(1) = 8.
G.f.: (3-4*x)/(1-4*x+2*x^2).
From G. C. Greubel, Sep 28 2018: (Start)
a(2*n) = 2^(n-1) * (Q(2*n +1) + 2*Q(2*n)), Q(m) = Pell-Lucas numbers = A002203(m).
a(2*n+1) = 2^(n-1) * (P(2*n+2) + 2*P(2*n+1)), P(m) = Pell numbers = A000129(m). (End)

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 01 2009

A137232 a(n) = -a(n-1) + 7a(n-2) + 3a(n-3) with a(0) = a(1) = 0, a(2) = 1.

Original entry on oeis.org

0, 0, 1, -1, 8, -12, 65, -125, 544, -1224, 4657, -11593, 40520, -107700, 356561, -988901, 3161728, -9014352, 28179745, -81795025, 252010184, -740036124, 2258722337, -6682944653, 20273892640, -60278338200, 182146752721, -543273442201, 1637465696648, -4893939533892, 14726379083825

Offset: 0

Paul Curtz, Mar 08 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,7,3).

Cf. A000129, A135532.

I:=[0,0,1]; [n le 3 select I[n] else -Self(n-1) +7*Self(n-2) +3*Self(n-3): n in [1..36]]; // G. C. Greubel, Apr 19 2021

Table[((-3)^n + 5*Fibonacci[n,2] -Fibonacci[n+1,2])/14, {n,0,40}] (* G. C. Greubel, Apr 19 2021 *)
LinearRecurrence[{-1,7,3},{0,0,1},40] (* Harvey P. Dale, Apr 26 2022 *)

[((-3)^n +5*lucas_number1(n,2,-1) -lucas_number1(n+1,2,-1))/14 for n in (0..40)] # G. C. Greubel, Apr 19 2021

From R. J. Mathar, Mar 17 2008: (Start)
O.g.f.: x^2/((1+3*x)*(1-2*x-x^2)).
a(n) = ( (-3)^n + A135532(n) )/14. (End)
a(n) = (1/14)*( (-3)^n + 5*Pell(n) - Pell(n+1) ), where Pell(n) = A000129(n). - G. C. Greubel, Apr 19 2021

A227792 Expansion of (1 + 6x + 17x^2 - x^3 - 3x^4)/(1 - 6x^2 + x^4).

Original entry on oeis.org

1, 6, 23, 35, 134, 204, 781, 1189, 4552, 6930, 26531, 40391, 154634, 235416, 901273, 1372105, 5253004, 7997214, 30616751, 46611179, 178447502, 271669860, 1040068261, 1583407981, 6061962064, 9228778026, 35331704123, 53789260175, 205928262674

Offset: 0

Ralf Stephan, Sep 23 2013

Also, values i where A067060(i)/i reaches a new maximum (conjectured).

E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1)

Cf. A041017.

CoefficientList[Series[(1+6x+17x^2-x^3-3x^4)/(1-6x^2+x^4),{x,0,40}],x] (* or *) LinearRecurrence[{0,6,0,-1},{1,6,23,35,134},40] (* Harvey P. Dale, Jun 12 2021 *)

a(n)=polcoeff((-3*x^4-x^3+17*x^2+6*x+1)/(x^4-6*x^2+1)+O(x^100),n)

G.f.: (1+6*x+17*x^2-x^3-3*x^4)/((1+2*x-x^2)*(1-2*x-x^2)).
a(2n) = A038723(n+1), n>0.
a(2n+1) = A001109(n+2).
a(n) = (1/4) * (A135532(n+3) + (-1)^n*A001333(n+2) ).

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A162255 a(n) = 2*a(n-2) for n > 2; a(1) = 3, a(2) = 2.

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A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

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A164682 a(n) = 2*a(n-2) for n > 2; a(1) = 5, a(2) = 8.

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

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A048745 Partial sums of A048654.

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

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A161938 a(n) = ((3+sqrt(2))*(2+sqrt(2))^n + (3-sqrt(2))*(2-sqrt(2))^n)/2.

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A161938 a(n) = ((3+sqrt(2))(2+sqrt(2))^n + (3-sqrt(2))(2-sqrt(2))^n)/2.

A137232 a(n) = -a(n-1) + 7a(n-2) + 3a(n-3) with a(0) = a(1) = 0, a(2) = 1.

A227792 Expansion of (1 + 6x + 17x^2 - x^3 - 3x^4)/(1 - 6x^2 + x^4).