A002203 -id:A002203 - cp's OEIS frontend

A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2.

2, 1, 0, 2, 2, 0, 2, 1, 2, 2, 2, 0, 1, 0, 2, 1, 2, 1, 1, 2, 0, 0, 2, 1, 1, 1, 0, 0, 0, 2, 2, 2, 0, 1, 2, 1, 0, 2, 0, 0, 2, 0, 2, 1, 0, 2, 1, 0, 0, 0, 1, 2, 0, 2, 1, 0, 2, 0, 2, 2, 1

Offset: 0

Wolfdieter Lang, Apr 05 2016

This is the scaled first difference sequence of A271222. See the formula.
The digits of the other 3-adic integer sqrt(-2), are given in A271223. See also a comment on A268924 for the two 3-adic numbers sqrt(-2), called there u and -u.
a(n) is the unique solution of the linear congruence 2*A271222(n)*a(n) + A271226(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case.
  a(0) = 2 follows from the formula given below.
For details see the Wolfdieter Lang link under A268992.
The first k digits in the base 3 representation of A002203(3^k) = A006266(k) give the first k terms of the sequence. - Peter Bala, Nov 26 2022

			a(4) = 2 because 2*59*2 + 43 = 279 == 0 (mod 3).
a(4) = - 43*(2*59) (mod 3) = -1*(2*(-1)) (mod 3) = 2.
A271222(5) = 221  = 2*3^0 + 1*3^1 + 0*3^2 + 2*3^3 + 2*3^4.

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.

Cf. A268924, A268992, A271222, A271226, A271223 (companion).

a(n) = truncate(-sqrt(-2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016

a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A271222(n), n >= 0.
a(n) = - A271226(n)*2*A271222(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.
A271222(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.

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

A056236 a(n) = (2 + sqrt(2))^n + (2 - sqrt(2))^n.

Original entry on oeis.org

2, 4, 12, 40, 136, 464, 1584, 5408, 18464, 63040, 215232, 734848, 2508928, 8566016, 29246208, 99852800, 340918784, 1163969536, 3974040576, 13568223232, 46324811776, 158162800640, 540001579008, 1843680714752, 6294719700992

Offset: 0

Henry Bottomley, Aug 11 2000

First differences give A060995. - Jeremy Gardiner, Aug 11 2013
Binomial transform of A002203 [Bhadouria].
The binomial transform of this sequence is 2, 6, 22, 90, 386, .. = 2*A083878(n). - R. J. Mathar, Nov 10 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Pooja Bhadouria, Deepika Jhala, and Bijendra Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92, sequence B_2.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 9.
Takao Komatsu, Asymmetric Circular Graph with Hosoya Index and Negative Continued Fractions, arXiv:2105.08277 [math.CO], 2021.
Youngwoo Kwon, Binomial transforms of the modified k-Fibonacci-like sequence, arXiv:1804.08119 [math.NT], 2018.
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Cf. A006012, A007052, A020727.

LinearRecurrence[{4,-2},{2,4},30] (* Harvey P. Dale, Jan 18 2013 *)

PARI
```
a(n) = 2*real((2+quadgen(8))^n);
```

[lucas_number2(n,4,2) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

a(n) = 4*a(n-1) - 2*a(n-2).
a(n) = a(n-2) - a(n-1) + 2*A020727(n-1).
a(n) = 2*A006012(n) = 4*A007052(n-1).
For n>2, a(n) = floor((2+sqrt(2))*a(n-1)).
G.f.: 2*(1-2*x)/(1-4*x+2*x^2).
From L. Edson Jeffery, Apr 08 2011: (Start)
a(n) = 2^(2*n)*(cos(Pi/8)^(2*n) + cos(3*Pi/8)^(2*n)).
a(n) = 3*a(n-1) + Sum_{k=1..(n-2)} a(k), for n>1, with a(0)=2, a(1)=4. (End)
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 8*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015

More terms from James Sellers, Aug 25 2000

A135532 a(n) = 2*a(n-1) + a(n-2), with a(0)= -1, a(1)= 3.

Original entry on oeis.org

-1, 3, 5, 13, 31, 75, 181, 437, 1055, 2547, 6149, 14845, 35839, 86523, 208885, 504293, 1217471, 2939235, 7095941, 17131117, 41358175, 99847467, 241053109, 581953685, 1404960479, 3391874643, 8188709765, 19769294173, 47727298111, 115223890395, 278175078901, 671574048197

Offset: 0

Paul Curtz, Feb 21 2008

Double binomial transform of [1, 3, -5, 13, -31, 75, -181, ...] = the Pell-like sequence A048655: (1, 5, 11, 27, 65, 157, ...). - Gary W. Adamson, Jul 23 2008

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

Cf. A000129, A002203, A048655.

I:=[-1,3]; [n le 2 select I[n] else 2*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, May 22 2021

LinearRecurrence[{2,1},{-1,3},25] (* G. C. Greubel, Oct 17 2016 *)

a(n)=([0,1; 1,2]^n*[-1;3])[1,1] \\ Charles R Greathouse IV, Oct 17 2016

[(lucas_number2(n,2,-1) + 2*lucas_number2(n-1,2,-1))/2 for n in (0..30)] # G. C. Greubel, May 22 2021

From R. J. Mathar, Feb 23 2008: (Start)
O.g.f.: (-1 + 5*x)/(1 - 2*x - x^2).
a(n) = 5*A000129(n) - A000129(n+1). (End)
a(n) = ((3+sqrt(2))*(1+sqrt(2))^n + (3-sqrt(2))*(1-sqrt(2))^n)/2 with offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Jun 17 2009
a(n) = (1/2)*(A002203(n) + 2*A002203(n-1)). - G. C. Greubel, May 22 2021

More terms from R. J. Mathar, Feb 23 2008

A164298 a(n) = ((1+4sqrt(2))(2+sqrt(2))^n + (1-4sqrt(2))(2-sqrt(2))^n)/2.

Original entry on oeis.org

1, 10, 38, 132, 452, 1544, 5272, 18000, 61456, 209824, 716384, 2445888, 8350784, 28511360, 97343872, 332352768, 1134723328, 3874187776, 13227304448, 45160842240, 154188760064, 526433355776, 1797355902976, 6136556900352, 20951515795456, 71532949381120

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 12 2009

Binomial transform of A048696. Second binomial transform of A164587. Inverse binomial transform of A164299.

This sequence is part of a class of sequences defined by the recurrence a(n,m) = 2*(m+1)*a(n-1,m) - ((m+1)^2 - 2)*a(n-2,m) with a(0) = 1 and a(1) = m+9. The generating function is Sum_{n>=0} a(n,m)*x^n = (1 - (m-7)*x)/(1 - 2*(m+1)*x + ((m+1)^2 - 2)*x^2) and has a series expansion in terms of Pell-Lucas numbers defined by a(n, m) = (1/2)*Sum_{k=0..n} binomial(n,k)*m^(n-k)*(5*Q(k) + 4*Q(k-1)). - G. C. Greubel, Mar 12 2021

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

Sequences in the class a(n, m): this sequence (m=1), A164299 (m=2), A164300 (m=3), A164301 (m=4), A164598 (m=5), A164599 (m=6), A081185 (m=7), A164600 (m=8).
Cf. A007070, A016921, A048696, A056236, A112032, A164587, A241204.
Cf. A016116(n+1).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+4*r)*(2+r)^n+(1-4*r)*(2-r)^n)/2: n in [0..27] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 17 2009

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+6*x)/(1-4*x+2*x^2) )); // G. C. Greubel, Dec 14 2018

a:=n->((1+4*sqrt(2))*(2+sqrt(2))^n+(1-4*sqrt(2))*(2-sqrt(2))^n)/2: seq(floor(a(n)),n=0..25); # Muniru A Asiru, Dec 15 2018

LinearRecurrence[{4,-2}, {1,10}, 50] (* or *) CoefficientList[Series[(1 + 6*x)/(1 - 4*x + 2*x^2), {x,0,50}], x] (* G. C. Greubel, Sep 12 2017 *)

my(x='x+O('x^50)); Vec((1+6*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Sep 12 2017

[( (1+6*x)/(1-4*x+2*x^2) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Dec 14 2018; Mar 12 2021

a(n) = 4*a(n-1) - 2*a(n-2) for n > 1; a(0)=1, a(1)=10.
G.f.: (1+6*x)/(1-4*x+2*x^2).
E.g.f.: (cosh(sqrt(2)*x) + 4*sqrt(2)*sinh(sqrt(2)*x))*exp(2*x). - G. C. Greubel, Sep 12 2017
From G. C. Greubel, Mar 12 2021: (Start)
a(n) = A056236(n) + 8*A007070(n-1).
a(n) = (1/2)*Sum_{k=0..n} binomial(n,k)*(5*Q(k) + 4*Q(k-1)), where Q(n) = Pell-Lucas(n) = A002203(n). (End)

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 17 2009

A054458 Convolution triangle based on A001333(n), n >= 1.

Original entry on oeis.org

1, 3, 1, 7, 6, 1, 17, 23, 9, 1, 41, 76, 48, 12, 1, 99, 233, 204, 82, 15, 1, 239, 682, 765, 428, 125, 18, 1, 577, 1935, 2649, 1907, 775, 177, 21, 1, 1393, 5368, 8680, 7656, 4010, 1272, 238, 24, 1, 3363, 14641, 27312, 28548, 18358, 7506, 1946, 308, 27, 1

Offset: 0

Wolfdieter Lang, Apr 26 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The G.f. for the row polynomials p(n,x) (increasing powers of x) is LPell(z)/(1-x*z*LPell(z)) with LPell(z) given in 'Formula'.
Column sequences are A001333(n+1), A054459(n), A054460(n) for m=0..2.
Mirror image of triangle in A209696. - Philippe Deléham, Mar 24 2012
Subtriangle of the triangle given by (0, 3, -2/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012
Riordan array ((1+x)/(1-2*x-x^2), (x+x^2)/(1-2*x-x^2)). - Philippe Deléham, Mar 25 2012

			Fourth row polynomial (n=3): p(3,x)= 17+23*x+9*x^2+x^3.
Triangle begins :
  1
  3, 1
  7, 6, 1
  17, 23, 9, 1
  41, 76, 48, 12, 1
  99, 233, 204, 82, 15, 1
  239, 682, 765, 428, 125, 18, 1. - _Philippe Deléham_, Mar 25 2012
(0, 3, -2/3, -1/3, 0, 0, 0, ...) DELTA (1, 0, 0, 0, ...) begins :
  1
  0, 1
  0, 3, 1
  0, 7, 6, 1
  0, 17, 23, 9, 1
  0, 41, 76, 48, 12, 1
  0, 99, 233, 204, 82, 15, 1
  0, 239, 682, 765, 428, 125, 15, 1. - _Philippe Deléham_, Mar 25 2012

Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A002203(n+1)/2. Row sums: A055099(n).
Cf. A001333, A054459, A054460.
Cf. A040000, A001333, A055099, A126473, A126501, A126528.

a(n, m) := ((n-m+1)*a(n, m-1) + (2n-m)*a(n-1, m-1) + (n-1)*a(n-2, m-1))/(4*m), n >= m >= 1; a(n, 0)= A001333(n+1); a(n, m) := 0 if n

G.f. for column m: LPell(x)*(x*LPell(x))^m, m >= 0, with LPell(x)= (1+x)/(1-2*x-x^2) = g.f. for A001333(n+1).

G.f.: (1+x)/(1-2*x-y*x-x^2-y*x^2). - Philippe Deléham, Mar 25 2012

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0) = T(1,1) = 1, T(1,0) = 3 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2012

Sum_{k=0..n} T(n,k)*x^k = A040000(n), A001333(n+1), A055099(n), A126473(n), A126501(n), A126528(n) for x = -1, 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Mar 25 2012

A164600 a(n) = 18a(n-1) - 79a(n-2) for n > 1; a(0) = 1, a(1) = 17.

Original entry on oeis.org

1, 17, 227, 2743, 31441, 349241, 3802499, 40854943, 434991553, 4602307457, 48477201539, 509007338599, 5332433173201, 55772217368297, 582637691946467, 6081473282940943, 63438141429166081, 661450156372654961

Offset: 0

Klaus Brockhaus, Aug 17 2009

nonn

Binomial transform of A081185 without initial term 0. Ninth binomial transform of A164587.

This sequence is part of a class of sequences defined by the recurrence a(n,m) = 2*(m+1)*a(n-1,m) - ((m+1)^2 - 2)*a(n-2,m) with a(0) = 1 and a(1) = m+9. The generating function is Sum_{n>=0} a(n,m)*x^n = (1 - (m-7)*x)/(1 - 2*(m+1)*x + ((m+1)^2 - 2)*x^2) and has a series expansion in terms of Pell-Lucas numbers defined by a(n, m) = (1/2)*Sum_{k=0..n} binomial(n,k)*m^(n-k)*(5*Q(k) + 4*Q(k-1)). - G. C. Greubel, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..975 (terms 0..100 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (18, -79).

Sequences in the class a(n, m): A164298 (m=1), A164299 (m=2), A164300 (m=3), A164301 (m=4), A164598 (m=5), A164599 (m=6), A081185 (m=7), this sequence (m=8).

Cf. A147960, A153593, A164587.

[ n le 2 select 16*n-15 else 18*Self(n-1)-79*Self(n-2): n in [1..18] ];

m:=30; S:=series( (1-x)/(1-18*x+79*x^2), x, m+1):
seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Mar 12 2021

LinearRecurrence[{18,-79},{1,17},30] (* Harvey P. Dale, Oct 30 2013 *)

my(x='x+O('x^50)); Vec((1-x)/(1-18*x+79*x^2)) \\ G. C. Greubel, Aug 11 2017

[( (1-x)/(1-18*x+79*x^2) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Mar 12 2021

a(n) = ((1+4*sqrt(2))*(9+sqrt(2))^n + (1-4*sqrt(2))*(9-sqrt(2))^n)/2.
G.f.: (1-x)/(1-18*x+79*x^2).
E.g.f.: exp(9*x)*(cosh(sqrt(2)*x) + 4*sqrt(2)*sinh(sqrt(2)*x)). - G. C. Greubel, Aug 11 2017
From G. C. Greubel, Mar 12 2021: (Start)
a(n) = 2*A147960(n) + 8*A153593(n).
a(n) = (1/2)*Sum_{k=0..n} binomial(n,k)*8^(n-k)*(5*Q(k) + 4*Q(k-1)), where Q(n) = Pell-Lucas(n) = A002203(n). (End)

A271222 One of the two successive approximations up to 3^n for the 3-adic integer sqrt(-2). These are the numbers congruent to 2 mod 3 (except for the initial 0).

Original entry on oeis.org

0, 2, 5, 5, 59, 221, 221, 1679, 3866, 16988, 56354, 174452, 174452, 705893, 705893, 10271831, 24620738, 110714180, 239854343, 627274832, 2951797766, 2951797766, 2951797766, 65713916984, 159857095811, 442286632292

Offset: 0

Wolfdieter Lang, Apr 05 2016

The other approximation for the 3-adic integer sqrt(-2) with numbers 1 (mod 3) is given in A268924.
For the digits of this 3-adic integer sqrt(-2), that is the scaled first differences, see A271224. This 3-adic number has the digits read from the right to the left ... 20020121022200011120021121201022212022012 = -u. For the digits of u see A271223.
For details see the W. Lang link ``Note on a Recurrence or Approximation Sequences of p-adic Square Roots'' given under A268922, also for the Nagell reference and Hensel lifting. Here p = 3,  b = 2, x_2 = 2 and  z_2 = 2.

			n=2: 5^2 + 2 = 27 == 0 (mod 3^2), and 5 is the only solution from {0, 1, ..., 8} which is congruent to 2 modulo 3.
n=3: the only solution of X^2 + 2 == 0 (mod 3^3) with X from {0, ..., 26} and X == 2(mod 3) is 5. The number 22 = A268924(3) also satisfies the first congruence but not the second one: 22  == 1 (mod 3).
n=4: the only solution of X^2 + 2 == 0 (mod 3^4) with X from {0, ..., 80} and X == 2 (mod 3) is 59. The number 22 = A268924(4) also satisfies the first congruence but not the second one: 59  == 1 (mod 3).

Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Seiichi Manyama, Table of n, a(n) for n = 0..2095
Peter Bala, A note on A268924 and A271222, Nov 28 2022.
Wikipedia, Hensel's Lemma.

Cf. A002293, A268922, A268924, A271223, A271224.

with(padic):D2:=op(3,op([evalp(RootOf(x^2+2),3,20)][2])): 0,seq(sum('D2[k]*3^(k-1)','k'=1..n), n=1..20);

a(n) = if (n, 3^n - truncate(sqrt(-2+O(3^(n)))), 0); \\ Michel Marcus, Apr 09 2016

def a271222(n):
      ary=[0]
      a, mod = 2, 3
      for i in range(n):
          b=a%mod
          ary.append(b)
          a=2*b**2 + b + 4
          mod*=3
      return ary
print(a271222(100)) # Indranil Ghosh, Aug 04 2017, after Ruby

def A271222(n)
  ary = [0]
  a, mod = 2, 3
  n.times{
    b = a % mod
    ary << b
    a = 2 * b * b + b + 4
    mod *= 3
  }
  ary
end
p A271222(100) # Seiichi Manyama, Aug 03 2017

a(n)^2 + 2 == 0 (mod 3^n), and a(n) == 2 (mod 3). Representatives of the complete residue system {0, 1, ..., 3^n-1} are taken.
Recurrence for n >= 1: a(n) = modp(a(n-1) +  2*(a(n-1)^2 + 2), 3^n), n >= 2, with a(1) = 2. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.
a(n) = 3^n - A268924(n), n >= 1.
a(n) == A002203(3^n) (mod 3^n). - Peter Bala, Nov 10 2022

A099088 Indices of prime companion Pell numbers, divided by 2 (A001333).

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 16, 19, 29, 47, 59, 163, 257, 421, 937, 947, 1493, 1901, 6689, 8087, 9679, 28753, 79043, 129127, 145969, 165799, 168677, 170413, 172243, 278321, 552283

Offset: 1

T. D. Noe, Sep 24 2004

Note that for A001333(n) to be prime, the index n must be prime or a power of 2. The indices greater than 421 yield probable primes.

Numbers n for which ((1+sqrt(2))^n + (1-sqrt(2))^n)/2 is prime. - Artur Jasinski, Dec 10 2006

F. Le Lionnais, Les Nombres Remarquables. Paris: Hermann, p. 62, 1983.

J. B. Cosgrave and K. Dilcher, Pairs of reciprocal quadratic congruences involving primes, Fib. Quart. 51 (2) (2013) 98, after Theorem 3.
Eric Weisstein's World of Mathematics, Pell Number
Eric Weisstein's World of Mathematics, Integer Sequence Primes

Cf. A002203 (companion Pell numbers), A086395 (primes in A001333), A096650 (indices of prime Pell numbers).

Cf. A005850.

lst={}; a=1; b=1; Do[c=a+2b; a=b; b=c; If[PrimeQ[c], AppendTo[lst, n]], {n, 2, 10000}]; lst
(* Second program: *)
Do[If[PrimeQ[Expand[((1 + Sqrt[2])^n + (1 - Sqrt[2])^n)/2]], Print[n]], {n, 0, 1000}] (* Artur Jasinski, Dec 10 2006 *)

isok(n) = isprime(polchebyshev(n, 1, I)/I^n); \\ Michel Marcus, Dec 07 2018

a(24) from Eric W. Weisstein, May 22 2006
a(25) from Eric W. Weisstein, Aug 29 2006
a(26) from Eric W. Weisstein, Nov 11 2006
a(27) from Eric W. Weisstein, Nov 26 2006
a(28) from Eric W. Weisstein, Dec 10 2006
a(29) from Eric W. Weisstein, Jan 25 2007
a(30) from Robert Price, Dec 07 2018
a(31) from Robert Price, Dec 05 2023

A164299 a(n) = ((1+4sqrt(2))(3+sqrt(2))^n + (1-4sqrt(2))(3-sqrt(2))^n)/2.

Original entry on oeis.org

1, 11, 59, 277, 1249, 5555, 24587, 108637, 479713, 2117819, 9348923, 41268805, 182170369, 804140579, 3549650891, 15668921293, 69165971521, 305313380075, 1347718479803, 5949117218293, 26260673951137, 115920223178771

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 12 2009

nonn

Binomial transform of A164298. Third binomial transform of A164587. Inverse binomial transform of A164300.

This sequence is part of a class of sequences defined by the recurrence a(n,m) = 2*(m+1)*a(n-1,m) - ((m+1)^2 - 2)*a(n-2,m) with a(0) = 1 and a(1) = m+9. The generating function is Sum_{n>=0} a(n,m)*x^n = (1 - (m-7)*x)/(1 - 2*(m+1)*x + ((m+1)^2 - 2)*x^2) and has a series expansion in terms of Pell-Lucas numbers defined by a(n, m) = (1/2)*Sum_{k=0..n} binomial(n,k)*m^(n-k)*(5*Q(k) + 4*Q(k-1)). - G. C. Greubel, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-7).

Sequences in the class a(n, m): A164298 (m=1), this sequence (m=2), A164300 (m=3), A164301 (m=4), A164598 (m=5), A164599 (m=6), A081185 (m=7), A164600 (m=8).

Cf. A081179, A083878, A164587.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+4*r)*(3+r)^n+(1-4*r)*(3-r)^n)/2: n in [0..21] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 17 2009

LinearRecurrence[{6,-7}, {1,11}, 50] (* or *) CoefficientList[Series[(1 + 5*x)/(1 - 6*x + 7*x^2), {x,0,50}], x] (* G. C. Greubel, Sep 12 2017 *)

my(x='x+O('x^50)); Vec((1+5*x)/(1-6*x+7*x^2)) \\ G. C. Greubel, Sep 12 2017

[( (1+5*x)/(1-6*x+7*x^2) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Mar 12 2021

a(n) = 6*a(n-1) - 7*a(n-2) for n > 1; a(0) = 1, a(1) = 11.
G.f.: (1+5*x)/(1-6*x+7*x^2).
E.g.f.: (cosh(sqrt(2)*x) + 4*sqrt(2)*sinh(sqrt(2)*x))*exp(3*x). - G. C. Greubel, Sep 12 2017
From G. C. Greubel, Mar 12 2021: (Start)
a(n) = 2*A083878(n) + 8*A081179(n).
a(n) = (1/2)*Sum_{k=0..n} binomial(n,k)*2^(n-k)*(5*Q(k) + 4*Q(k-1)), where Q(n) = Pell-Lucas(n) = A002203(n). (End)

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 17 2009

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A271224 Digits of one of the two 3-adic integers sqrt(-2). Here the sequence with first digit 2.

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

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A056236 a(n) = (2 + sqrt(2))^n + (2 - sqrt(2))^n.

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A135532 a(n) = 2*a(n-1) + a(n-2), with a(0)= -1, a(1)= 3.

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A164298 a(n) = ((1+4*sqrt(2))*(2+sqrt(2))^n + (1-4*sqrt(2))*(2-sqrt(2))^n)/2.

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A054458 Convolution triangle based on A001333(n), n >= 1.

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

A164298 a(n) = ((1+4sqrt(2))(2+sqrt(2))^n + (1-4sqrt(2))(2-sqrt(2))^n)/2.

A164600 a(n) = 18a(n-1) - 79a(n-2) for n > 1; a(0) = 1, a(1) = 17.

A164299 a(n) = ((1+4sqrt(2))(3+sqrt(2))^n + (1-4sqrt(2))(3-sqrt(2))^n)/2.