cp's OEIS Frontend

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2

Offset: 0

Peter Luschny, Jul 22 2024

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |

Paolo Xausa, Rows n = 0..150 of the triangle, flattened
Peter Luschny, Illustrating the polynomials.

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).
Cf. A039961, A065941 (qStirling2), A103631, A108299, A131259. A133607, A194005, A333143, A374438 (m=3).

function T(n,k) // T = A374439
  if k lt 0 or k gt n then return 0;
  elif k le 1 then return k+1;
  else return T(n-1,k) + T(n-2,k-2);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025

A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
# Alternative, using the function qStirling2 from A333143:
T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
seq(seq(T(n, k), k = 0..n), n = 0..10);

A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k > n: return 0
    if k < 2: return k + 1
    return T(n - 1, k) + T(n - 2, k - 2)

from math import comb as binomial
def T(n: int, k: int) -> int:
    o = k & 1
    return binomial(n - o - (k - o) // 2, (k - o) // 2) << o

def P(n, x):
    if n < 0: return P(n, x)
    return sum(T(n, k)*x**k for k in range(n + 1))
def sgn(x: int) -> int: return (x > 0) - (x < 0)
# Table of interpolated sequences
print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))

from sage.combinat.q_analogues import q_stirling_number2
def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A221174 a(0)=-4, a(1)=5; thereafter a(n) = 2*a(n-1) + a(n-2).

-4, 5, 6, 17, 40, 97, 234, 565, 1364, 3293, 7950, 19193, 46336, 111865, 270066, 651997, 1574060, 3800117, 9174294, 22148705, 53471704, 129092113, 311655930, 752403973, 1816463876, 4385331725, 10587127326, 25559586377, 61706300080, 148972186537, 359650673154

Offset: 0

N. J. A. Sloane, Jan 04 2013

From Greg Dresden, May 08 2023: (Start)
For n >= 3, 2*a(n) is the number of ways to tile this figure of length n-1 with two colors of squares and one color of domino. For n=8, we have here the figure of length n-1=7, and it has 2*a(8) = 2728 different tilings.
. 
|||_   _ _
|||_|||_|_|
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012-2014.
Index entries for linear recurrences with constant coefficients, signature (2,1).

Cf. A000129, A078343, A221172, A221173, A221175.

a221174 n = a221174_list !! n
a221174_list = -4 : 5 : zipWith (+)
                        (map (* 2) $ tail a221174_list) a221174_list
-- Reinhard Zumkeller, Jan 04 2013

LinearRecurrence[{2, 1}, {-4, 5}, 50] (* Paolo Xausa, Sep 02 2024 *)

Vec(-(13*x-4)/(x^2+2*x-1) + O(x^50)) \\ Colin Barker, Jul 10 2015

a(n) = 13*A000129(n) - 4*A000129(n+1). - R. J. Mathar, Jan 14 2013
G.f.: -(13*x-4) / (x^2+2*x-1). - Colin Barker, Jul 10 2015
a(n) is the numerator of the continued fraction [4, 2, ..., 2, 4] with n-3 2's in the middle. For denominators, see A048654. - Greg Dresden and Tongjia Rao, Sep 02 2021

A048694 Generalized Pellian with second term equal to 7.

1, 7, 15, 37, 89, 215, 519, 1253, 3025, 7303, 17631, 42565, 102761, 248087, 598935, 1445957, 3490849, 8427655, 20346159, 49119973, 118586105, 286292183, 691170471, 1668633125, 4028436721, 9725506567

Offset: 0

Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, ... . - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1)

with(combinat): a:=n->5*fibonacci(n, 2)+fibonacci(n+1, 2): seq(a(n), n=0..26); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{6},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,7},40] (* Harvey P. Dale, Jul 22 2011 *)

a[0]:1$
a[1]:7$
a[n]:=2*a[n-1]+a[n-2]$
A048694(n):=a[n]$
makelist(A048694(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = ((6+sqrt(2))(1+sqrt(2))^n - (6-sqrt(2))(1-sqrt(2))^n)/2*sqrt(2).
a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=7.
G.f.: (1+5*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008
a(n) = ((1+sqrt(18))(1+sqrt(2))^n+(1-sqrt(18))(1-sqrt(2))^n)/2 offset 0. a(n) = first binomial transform of 1,6,2,12,4,24. - Al Hakanson (hawkuu(AT)gmail.com), Aug 01 2009

A048696 Generalized Pellian with second term equal to 9.

1, 9, 19, 47, 113, 273, 659, 1591, 3841, 9273, 22387, 54047, 130481, 315009, 760499, 1836007, 4432513, 10701033, 25834579, 62370191, 150574961, 363520113, 877615187, 2118750487, 5115116161, 12348982809, 29813081779, 71975146367

Offset: 0

Binomial transform of 5,6,10,12,20,24,40. - Al Hakanson (hawkuu(AT)gmail.com), Aug 12 2009
Binomial transform of A164587. Inverse binomial transform of A164298. - Klaus Brockhaus, Aug 17 2009
For n > 0: a(n) = A105082(n) - A105082(n-1). - Reinhard Zumkeller, Dec 15 2013

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).

a048696 n = a048696_list !! n
a048696_list = 1 : 9 : zipWith (+)
               a048696_list (map (2 *) $ tail a048696_list)
-- Reinhard Zumkeller, Dec 15 2013

[ n le 2 select 8*n-7 else 2*Self(n-1)+Self(n-2): n in [1..28] ]; // Klaus Brockhaus, Aug 17 2009

with(combinat): a:=n->7*fibonacci(n, 2)+fibonacci(n+1, 2): seq(a(n), n=0..25); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{8},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,9},30] (* Harvey P. Dale, Apr 20 2012 *)

a[0]:1$
a[1]:9$
a[n]:=2*a[n-1]+a[n-2]$
A048696(n):=a[n]$
makelist(A048696(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=9.
a(n) = ((4*sqrt(2)+1)(1+sqrt(2))^n - (4*sqrt(2)-1)(1-sqrt(2))^n)/2.
G.f.: (1+7*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008

A164073 a(n) = 2*a(n-2) for n > 2; a(1) = 1, a(2) = 3.

1, 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, 6291456

Offset: 1

Klaus Brockhaus, Aug 09 2009

Interleaving of A000079 and A007283.
Binomial transform is A048654. Second binomial transform is A111567. Third binomial transform is A081179 without initial 0. Fourth binomial transform is A164072. Fifth binomial transform is A164031.
Absolute second differences are the sequence itself. - Eric Angelini, Jul 30 2013
Least number having n - 1 Gaussian prime factors, counted with multiplicity, excluding units. See A239628 for a similar sequence. - T. D. Noe, Mar 31 2014
Writing the prime factorizations of the terms of this sequence, the exponents of prime factor 2 give the integers repeated (A004526), while the exponents of prime factor 3 give the sequence of alternating 0's and 1's (A000035). - Alonso del Arte, Nov 30 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (0, 2).

Cf. A000079 (powers of 2), A007283 (3*2^n), A074323, A162255, A048654, A111567, A081179, A164072, A164031, A239628.

[ n le 2 select 2*n-1 else 2*Self(n-2): n in [1..42] ];

terms = 50; CoefficientList[Series[x * (1 + 3 * x)/(1 - 2 * x^2), {x, 0, terms}], x] (* T. D. Noe, Mar 31 2014 *)
Flatten[Table[{2^n, 3 * 2^n}, {n, 0, 31}]] (* Alonso del Arte, Nov 30 2016 *)
CoefficientList[Series[x (1 + 3 x)/(1 - 2 x^2), {x, 0, 44}], x] (* Michael De Vlieger, Dec 13 2016 *)

PARI
```
a(n) = (5 + (-1)^n) * 2^((2*n-9)\/4)
```

Vec(x*(1+3*x)/(1-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Dec 13 2016

a(n) = (5 + (-1)^n) * 2^(1/4 * (2*n - 1 + (-1)^n))/4.
G.f.: x*(1 + 3 * x)/(1 - 2 * x^2).
a(n) = A074323(n), n>=1.
a(n) = A162255(n-1), n>=2.
a(n) = A072946(n-2), n > 2. - R. J. Mathar, Aug 17 2009
a(n+3) = a(n + 2) * a(n + 1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = (2/3)a(n - 1) for odd n > 1, a(n) = 3a(n - 1) for even n. - Alonso del Arte, Nov 30 2016

A048693 Generalized Pellian with 2nd term equal to 6.

1, 6, 13, 32, 77, 186, 449, 1084, 2617, 6318, 15253, 36824, 88901, 214626, 518153, 1250932, 3020017, 7290966, 17601949, 42494864, 102591677, 247678218, 597948113, 1443574444, 3485097001, 8413768446

Offset: 0

Pisano period lengths: 1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, ... (is this A175181?). - R. J. Mathar, Aug 10 2012

			a(n)=[ (5+sqrt(2))(1+sqrt(2))^n-(5-sqrt(2))(1-sqrt(2))^n ]/2*sqrt(2)

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1)

with(combinat): a:=n->4*fibonacci(n-1,2)+fibonacci(n,2): seq(a(n), n=1..26); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{5},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,6},30] (* Harvey P. Dale, Mar 29 2013 *)

a[0]:1$
a[1]:6$
a[n]:=2*a[n-1]+a[n-2]$
A048693(n):=a[n]$
makelist(A048693(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=6.
G.f.: (1+4*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008
a(n) = 4*A000129(n) + A000129(n+1). - R. J. Mathar, Aug 10 2012

A105082 Expansion of (5+4x)/(1-2x-x^2).

5, 14, 33, 80, 193, 466, 1125, 2716, 6557, 15830, 38217, 92264, 222745, 537754, 1298253, 3134260, 7566773, 18267806, 44102385, 106472576, 257047537, 620567650, 1498182837, 3616933324, 8732049485, 21081032294, 50894114073

Offset: 0

Creighton Dement, Apr 06 2005

A floretion-generated, Pellian related sequence.
Floretion Algebra Multiplication Program, FAMP Code: lesloop(infty)-tesforseq[ + .25'i + .25i' - .25'ii' - .25'jj' - .25'kk' + .25'jk' + .25'kj' - .25e ], Fortype: 1A.
For n > 0: A048696(n) = a(n) - a(n-1). - Reinhard Zumkeller, Dec 15 2013

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).

Cf. A001333, A048654, A048696.

Cf. A048772.

a105082 n = a105082_list !! n
a105082_list = scanl (+) 5 $ tail a048696_list
-- Reinhard Zumkeller, Dec 15 2013

a(n+2) = 2*a(n+1) + a(n); FAMP result: a(n) = 2*A001333(n) + 3*A048654(n); SuperSeeker results: a(n+1) - a(n) = A048696(n+1); a(n) + a(n+1) = A048696(n+2)

a(n) = ((9+5*sqrt(2))*(1+sqrt(2))^n - (9-5*sqrt(2))*(1-sqrt(2))^n)/(2*sqrt(2)) - Lambert Herrgesell (zero815(AT)googlemail.com), Jan 26 2007

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

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

A048697 Generalized Pellian with second term equal to 10.

1, 10, 21, 52, 125, 302, 729, 1760, 4249, 10258, 24765, 59788, 144341, 348470, 841281, 2031032, 4903345, 11837722, 28578789, 68995300, 166569389, 402134078, 970837545, 2343809168, 5658455881

Offset: 0

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).