A002249 -id:A002249 - cp's OEIS frontend

A107920 Lucas and Lehmer numbers with parameters (1 +- sqrt(-7))/2.

0, 1, 1, -1, -3, -1, 5, 7, -3, -17, -11, 23, 45, -1, -91, -89, 93, 271, 85, -457, -627, 287, 1541, 967, -2115, -4049, 181, 8279, 7917, -8641, -24475, -7193, 41757, 56143, -27371, -139657, -84915, 194399, 364229, -24569, -753027, -703889, 802165, 2209943, 605613, -3814273

Offset: 0

Michael Somos, May 28 2005

The sequences A001607, A077020, A107920, A167433, A169998 are all essentially the same except for signs.
This is an example of a sequence of Lehmer numbers. In this case, the two parameters, alpha and beta, are (1 +- i*sqrt(7))/2. Bilu, Hanrot, Voutier and Mignotte show that all terms of a Lehmer sequence a(n) have a primitive factor for n > 30. Note that for this sequence, a(30) = 24475 = 5*5*11*89 has no primitive factors. - T. D. Noe, Oct 29 2003
Row sums of Riordan array (1/(1+2*x^2), x/(1+2*x^2)). - Paul Barry, Sep 10 2005
Pisano period lengths: 1, 1, 8, 2, 24, 8, 21, 2, 24, 24, 10, 8, 168, 21, 24, 4, 144, 24, 360, 24, ... - R. J. Mathar, Aug 10 2012
This is the Lucas Sequence U_n(P, Q) = U_n(1, 2). V_n(1, 2) = A002249(n). - Raphie Frank, Dec 25 2013
Note that (A002249(n)/2)^2 + 7*(a(n)/2)^2 = 2^n for all n in N. This is a specific case of the Lucas sequence identity (V_n/2)^2 - D*(U_n/2)^2 = Q^n where V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), Q = (a*b) = 2 and D = (a - b)^2 = -7; a = (1 + sqrt(-7))/2 and b = (1 - sqrt(-7))/2. - Raphie Frank, Nov 26 2015
For the special case where |a(n)| = 1, true for n if and only if n is in {1, 2, 3, 5, 13} = {A215795(n) + 1} = {A060728(n) - 2}, then, additionally, by the Lucas sequence identity (U_2n = U_n*V_n), we have (a(2n)/2)^2 + 7*(a(n)/2)^2 = 2^n. - Raphie Frank, Nov 26 2015

			G.f. = x + x^2 - x^3 - 3*x^4 - x^5 + 5*x^6 + 7*x^7 - 3*x^8 - 17*x^9 - 11*x^10 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Christian Ballot, Lucasnomial Fuss-Catalan Numbers and Related Divisibility Questions, J. Int. Seq., Vol. 21 (2018), Article 18.6.5.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
F. Beukers, The multiplicity of binary recurrences, Compositio Mathematica, Tome 40 (1980) no. 2 , p. 251-267. See Theorem 2 p. 259.
Y. Bilu, G. Hanrot, P. M. Voutier and M. Mignotte, Existence of primitive divisors of Lucas and Lehmer numbers, [Research Report] RR-3792, INRIA. 1999, pp.41, HAL Id : inria-00072867.
M. Mignotte, Propriétés arithmétiques des suites récurrentes, Besançon, 1988-1989, see p. 14. In French.
Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
Eric Weisstein's World of Mathematics, Lehmer Number
Wikipedia, Lucas Sequence
Index entries for linear recurrences with constant coefficients, signature (1,-2).
Index entries for sequences related to Chebyshev polynomials.

Similar sequences: A001607, A077020, A107920, A167433, A169998.

Cf. A002249, A049310, A060728, A109466, A215795, A227078.

[0] cat [n le 2 select 1 else Self(n-1)-2*Self(n-2): n in [1..45]]; // Vincenzo Librandi, Nov 27 2015

a:= n-> (Matrix([[1,1],[ -2,0]])^n)[1,2]: seq(a(n), n=0..45); # Alois P. Heinz, Sep 03 2008

LinearRecurrence[{1, -2}, {0, 1}, 50] (* Vincenzo Librandi, Nov 27 2015 *)
a[ n_] := Im[ ((1 + Sqrt[-7]) / 2)^n // FullSimplify] 2 / Sqrt[7]; (* Michael Somos, Jan 19 2017 *)
a[n_] := If[n < 2, n, Hypergeometric2F1[1 - n/2, (1 - n)/2, 1 - n, 8]];
Table[a[n], {n, 0, 45}] (* Peter Luschny, Feb 23 2018 *)

PARI
```
{a(n) = imag(quadgen(-7)^n)};
```

my(x='x+O('x^100)); concat(0, Vec(x/(1-x+2*x^2))) \\ Altug Alkan, Dec 04 2015

[lucas_number1(n,1,2) for n in range(0, 46)] # Zerinvary Lajos, Apr 22 2009

G.f.: x / (1 - x + 2*x^2).
a(n) = a(n-1) - 2*a(n-2).
a(n) = -(-1)^n*A001607(n).
From Paul Barry, Sep 10 2005: (Start)
a(n+1) = Sum_{k=0..n} C((n+k)/2, k)*(-2)^((n-k)/2)*(1+(-1)^(n-k))/2.
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-2)^k. (End)
a(n+1) = Sum_{k=0..n} A109466(n,k)*2^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = ((1 - i*sqrt(7))^n - (1 + i*sqrt(7))^n)*i/(2^n*sqrt(7)), where i=sqrt(-1). - Bruno Berselli, Jul 01 2011
(a(2*(A060728(n)) - 4))^2 = (A002249(A060728(n) - 2))^2 = 2^(A060728(n)) - 7 = A227078(n), the Ramanujan-Nagell squares. - Raphie Frank, Dec 25 2013
a(n) = -a(-n) * 2^n for all n in Z. - Michael Somos, Jan 19 2017
G.f.: x / (1 - x / (1 + 2*x / (1 - 2*x))). - Michael Somos, Jan 19 2017
a(n) = S(n-1, 1/sqrt(2))*(sqrt(2))^(n-1), n >= 0, with the Chebyshev S polynomials (coefficients in A049310), and S(-1, x) = 0. - Wolfdieter Lang, Feb 22 2018
a(n) = hypergeom([1-n/2, (1-n)/2], [1-n], 8) for n >= 2. - Peter Luschny, Feb 23 2018

A001607 a(n) = -a(n-1) - 2*a(n-2).

Original entry on oeis.org

0, 1, -1, -1, 3, -1, -5, 7, 3, -17, 11, 23, -45, -1, 91, -89, -93, 271, -85, -457, 627, 287, -1541, 967, 2115, -4049, -181, 8279, -7917, -8641, 24475, -7193, -41757, 56143, 27371, -139657, 84915, 194399, -364229, -24569, 753027, -703889

Offset: 0

N. J. A. Sloane

The sequences A001607, A077020, A107920, A167433, A169998 are all essentially the same except for signs.

Apart from the sign, this is an example of a sequence of Lehmer numbers. In this case, the two parameters, alpha and beta, are (1 +- i*sqrt(7))/2. Bilu, Hanrot, Voutier and Mignotte show that all terms of a Lehmer sequence a(n) have a primitive factor for n > 30. Note that for this sequence, a(30) = 24475 = 5*5*11*89 has no primitive factors. - T. D. Noe, Oct 29 2003

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
Y. Bilu, G. Hanrot, P. M. Voutier, and M. Mignotte, Existence of primitive divisors of Lucas and Lehmer numbers, [Research Report] RR-3792, INRIA. 1999, pp. 41, HAL Id : inria-00072867.
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 14.
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 82.
Erwin Just, Problem E2367, Amer. Math. Monthly, 79 (1972), 772.
G. P. Michon, Never Back to -1.
M. Mignotte, Propriétés arithmétiques des suites récurrentes, Besançon, 1988-1989, see p. 14. In French.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (-1,-2).

Apart from signs, same as A077020.
Cf. A001607, A002249, A077021, A107920, A167433, A169998.
Cf. A172250.

[n eq 1 select 0 else n eq 2 select 1 else -Self(n-1)-2*Self(n-2): n in [1..50]]; // Vincenzo Librandi, Aug 22 2011

LinearRecurrence[{-1,-2},{0,1},60] (* Harvey P. Dale, Aug 21 2011 *)

a(n)=if(n<0,0,polcoeff(x/(1+x+2*x^2)+x*O(x^n),n))

a(n)=if(n<0,0,2*imag(((-1+quadgen(-28))/2)^n))

A001607=BinaryRecurrenceSequence(-1,-2,0,1)
[A001607(n) for n in range(51)] # G. C. Greubel, Mar 24 2024

G.f.: x/(1+x+2*x^2).
a(n) = Sum_{k=0..n-1} (-1)^(n-k-1)*binomial(n-k-1, k)*2^k = -2/sqrt(7)*(-sqrt(2))^n*(sin(n*arctan(sqrt(7)))). - Vladeta Jovovic, Feb 05 2003
x/(x^2+x+2) = Sum_{n>=0} a(n)*(x/2)^n. - Benoit Cloitre, Mar 12 2002
4*2^n = A002249(n)^2 + 7*A001607(n)^2. See A077020, A077021.
a(n+1) = Sum_{k=0..n} A172250(n,k)*(-1)^k. - Philippe Deléham, Feb 15 2012
G.f.: x - 2*x^2 + 2*x^2/(G(0)+1) where G(k) = 1 + x/(1 - x/(x - 1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 16 2012
a(n) = 2^((n-1)/2)*ChebyshevU(n-1, -1/(2*sqrt(2))). - G. C. Greubel, Mar 24 2024
a(n) = (i*(((-1 - i*sqrt(7))/2)^n - ((-1 + i*sqrt(7))/2)^n))/sqrt(7). - Alan Michael Gómez Calderón, Jul 09 2024; after T. D. Noe, Oct 29 2003

A060728 Numbers n such that Ramanujan's equation x^2 + 7 = 2^n has an integer solution.

Original entry on oeis.org

3, 4, 5, 7, 15

Offset: 1

Lekraj Beedassy, Apr 25 2001

See A038198 for corresponding x. - Lekraj Beedassy, Sep 07 2004
Also numbers such that 2^(n-3)-1 is in A000217, i.e., a triangular number. - M. F. Hasler, Feb 23 2009
With respect to M. F. Hasler's comment above, all terms 2^(n-3) - 1 are known as the Ramanujan-Nagell triangular numbers (A076046). - Raphie Frank, Mar 31 2013
Interestingly enough, all the solutions correspond to noncomposite x, i.e., x = 1 for the first term, and primes 3, 5, 11, 181 for the following terms. - M. F. Hasler, Mar 11 2024

			The fifth and ultimate solution to Ramanujan's equation is obtained for the 15th power of 2, so that we have x^2 + 7 = 2^15 yielding x = 181.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 181, p. 56, Ellipses, Paris 2008.
J. Roberts, Lure of the Integers. pp. 90-91, MAA 1992.
Ian Stewart & David Tall, Algebraic Number Theory and Fermat's Last Theorem, 3rd Ed. Natick, Massachusetts (2002): 96-98.

T. Skolem, S. Chowla and D. J. Lewis, The Diophantine Equation 2^(n+2)-7=x^2 and Related Problems. Proc. Amer. Math. Soc. 10 (1959) 663-669. [_M. F. Hasler_, Feb 23 2009]
Anonymous, Developing a general 2nd degree Diophantine Equation x^2 + p = 2^n
M. Beeler, R. W. Gosper and R. Schroeppel, HAKMEM: item 31: A Ramanujan Problem (R. Schroeppel)
Curtis Bright, Solving Ramanujan's Square Equation Computationally
Spencer De Chenne, The Ramanujan-Nagell Theorem: Understanding the Proof
T. Do, Developing A General 2nd Degree Diophantine Equation x^2 + p = 2^n
A. Engel, Problem-Solving Strategies. p. 126.
Gerry Myerson, Bibliography
T. Nagell, The Diophantine equation x^2 + 7 = 2^n, Ark. Mat. 4 (1961), no. 2-3, 185-187.
S. Ramanujan, Journal of the Indian Mathematical Society, Question 464(v,120)
Eric Weisstein's World of Mathematics, Ramanujan's Square Equation
Eric Weisstein's World of Mathematics, Diophantine Equation 2nd Powers
Wikipedia, Carmichael's Theorem
Wikipedia, Diophantine equation

Cf. A002249, A038198, A076046, A077020, A077021, A107920, A215795, A227078

[n: n in [0..100] | IsSquare(2^n-7)]; // Vincenzo Librandi, Jan 07 2014

ramaNagell[n_] := Reduce[x^2 + 7 == 2^n, x, Integers] =!= False; Select[ Range[100], ramaNagell] (* Jean-François Alcover, Sep 21 2011 *)

is(n)=issquare(2^n-7) \\ Anders Hellström, Dec 12 2015

a(n) = log_2(8*A076046(n) + 8) = log_2(A227078(n) + 7)

Empirically, a(n) = Fibonacci(c + 1) + 2 = ceiling[e^((c - 1)/2)] + 2 where {c} is the complete set of positive solutions to {n in N | 2 cos(2*Pi/n) is in Z}; c is in {1,2,3,4,6} (see A217290).

Added keyword "full", M. F. Hasler, Feb 23 2009

A038198 Numbers n such that n^2 + 7 is a power of 2.

Original entry on oeis.org

1, 3, 5, 11, 181

Offset: 1

N. J. A. Sloane

The exponents of the corresponding powers of 2 are 3, 4, 5, 7, 15 (see Ramanujan). - N. J. A. Sloane, Jun 01 2014

The terms lead to identities resembling Machin's Pi/4 = arctan(1/1) = 4*arctan(1/5) - arctan(1/239), for example, arctan(sqrt(7)/1) = 5*arctan(sqrt(7)/11) + 2*arctan(sqrt(7)/181), which can also be expressed as arcsin(sqrt(7/2^3)) = 5*arcsin(sqrt(7/2^7)) + 2*arcsin(sqrt(7/2^15)) (cf. A168229). - Joerg Arndt, Nov 09 2012

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 181, p. 56, Ellipses, Paris 2008.
L. J. Mordell, Diophantine Equations, Academic Press, NY, 1969, p. 205.
S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962. See Question 464, p. 327. - N. J. A. Sloane, Jun 01 2014

Spencer De Chenne, The Ramanujan-Nagell Theorem: Understanding the Proof
Eric Weisstein's World of Mathematics, Ramanujan's Square Equation
Wikipedia, Lucas Sequence

Cf. A060728, A002249, A107920.

ok[n_] := Reduce[k>0 && n^2 + 7 == 2^k, k, Integers] =!= False; Select[Range[1000], ok] (* Jean-François Alcover, Sep 21 2011 *)

[x | n<-[0..99], issquare(2^n-7,&x)] \\ M. F. Hasler, Mar 11 2024

A077021 a(n) is the unique odd positive solution y of 2^n = 7x^2 + y^2.

Original entry on oeis.org

1, 3, 5, 1, 11, 9, 13, 31, 5, 57, 67, 47, 181, 87, 275, 449, 101, 999, 797, 1201, 2795, 393, 5197, 5983, 4411, 16377, 7555, 25199, 40309, 10089, 90707, 70529, 110885, 251943, 30173, 473713, 534059, 413367, 1481485, 654751, 2308219, 3617721

Offset: 3

Ed Pegg Jr, Oct 17 2002

nonn

Restate the formula and divide both sides by 4, then y^2 - (-7)*x^2 = 2^n and (y/2)^2 - (-7)*(x/2)^2 = 2^(n-2). Let y = V_n, x = U_n, -7 = D, and 2^(n-2) = Q^n. We then have this sequence as the absolute values for V_n = A002249(n)(excluding a(0) = 2) in relation to the Lucas sequence identity: (V_n/2)^2 - D*(U_n/2)^2 = Q^n with V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), D = (a - b)^2 = P^2 - 4*Q = -7, Q = (a*b) = 2; a = (1 + sqrt(-7))/2, b = (1 - sqrt(-7))/2, P = 1. By the Ramanujan-Nagell Theorem, iff y is in +- {1, 3, 5, 11, 181} = +-A038198, then |x| = 1 and we are left with 2^n = 7 + y^2. See A060728 and note that a(A060728(n) - 3) = A038198(n). - Raphie Frank, Dec 05 2015

A. Engel, Problem-Solving Strategies, p. 126.

T. D. Noe, Table of n, a(n) for n=3..500
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Eric Weisstein's World of Mathematics, Diophantine Equations 2nd Powers

Cf. A077020 (x).

Cf. A002249, A038198, A060728.

a = {}; Do[k = Expand[((1 + I Sqrt[7])/2)^n + ((1 - I Sqrt[7])/2)^n]; AppendTo[a, Abs[k]], {n, 1, 50}]; a (* Artur Jasinski, Oct 05 2008 *)

a(n+2) = abs(A002249(n)). - Artur Jasinski, Oct 05 2008 [With correction by Jianing Song, Nov 21 2018]

A105578 a(n+3) = 2a(n+2) - 3a(n+1) + 2a(n); a(0) = 1, a(1) = 1, a(2) = 0.

Original entry on oeis.org

1, 1, 0, -1, 0, 3, 4, -1, -8, -5, 12, 23, 0, -45, -44, 47, 136, 43, -228, -313, 144, 771, 484, -1057, -2024, 91, 4140, 3959, -4320, -12237, -3596, 20879, 28072, -13685, -69828, -42457, 97200, 182115, -12284, -376513, -351944, 401083, 1104972, 302807, -1907136, -2512749, 1301524, 6327023, 3723976

Offset: 0

Creighton Dement, Apr 14 2005

Floretion Algebra Multiplication Program, FAMP Code: ibaseiseq[.5'j + .5'k + .5j' + .5k' + .5'ii' + .5e]

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-3,2).

Cf. A002249, A014551, A078020, A105577, A105579.

Equals (A107920(n) + 1)/2.

-Join[{-1,-1,a=0,b=1},Table[c=1*b-2*a-1;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2011 *)
LinearRecurrence[{2,-3,2},{1,1,0},50] (* Harvey P. Dale, Mar 28 2019 *)

Vec(-(x^2-x+1)/((x-1)*(2*x^2-x+1)) + O(x^100)) \\ Colin Barker, Feb 08 2015

a(n) - a(n+1) = A001607(n); a(n+2) - 2a(n+1) + a(n) = - A078020(n).

G.f.: -(x^2-x+1) / ((x-1)*(2*x^2-x+1)). - Colin Barker, Feb 08 2015

A105577 a(n) = 2a(n-1) - 3a(n-2) + 2*a(n-3) with a(0) = 1, a(1) = 5, a(2) = 6.

Original entry on oeis.org

1, 5, 6, -1, -10, -5, 18, 31, -2, -61, -54, 71, 182, 43, -318, -401, 238, 1043, 570, -1513, -2650, 379, 5682, 4927, -6434, -16285, -3414, 29159, 35990, -22325, -94302, -49649, 138958, 238259, -39654, -516169, -436858, 595483, 1469202, 278239, -2660162, -3216637, 2103690, 8536967

Offset: 0

Creighton Dement, Apr 14 2005

Floretion Algebra Multiplication Program, FAMP Code: 2lesseq[.5'j + .5'k + .5j' + .5k' + .5'ii' + .5e]

Index entries for linear recurrences with constant coefficients, signature (2,-3,2).

Cf. A002249, A014551, A105578.

Equals (1/4) [A107920(n+4) + 2*A107920(n-1) + 3].

LinearRecurrence[{2,-3,2},{1,5,6},50] (* Harvey P. Dale, Apr 13 2019 *)

G.f.: (1+3*x-x^2)/((1-x)*(1-x+2*x^2)). - Colin Barker, Mar 26 2012

E.g.f.: exp(x/2)*(21*exp(x/2) - 7*cos(sqrt(7)*x/2) + 15*sqrt(7)*sin(sqrt(7)*x/2))/14. - Stefano Spezia, May 22 2025

A105579 a(n+3) = 2a(n+2) - 3a(n+1) + 2a(n); a(0) = 1, a(1) = 3, a(2) = 4.

Original entry on oeis.org

1, 3, 4, 1, -4, -3, 8, 17, 4, -27, -32, 25, 92, 45, -136, -223, 52, 501, 400, -599, -1396, -195, 2600, 2993, -2204, -8187, -3776, 12601, 20156, -5043, -45352, -35263, 55444, 125973, 15088, -236855, -267028, 206685, 740744, 327377, -1154108, -1808859, 499360, 4117081, 3118364, -5115795, -11352520

Offset: 0

Creighton Dement, Apr 14 2005

Floretion Algebra Multiplication Program, FAMP Code: famseq[.5'j + .5'k + .5j' + .5k' + .5'ii' + .5e]

Cf. A002249, A014551, A078020, A105577, A105578, A105580.

Cf. Equals (1/2) [A107920(n+4) - 2*A107920(n-1) + 3 ].

Table[(3 - ((1-I*Sqrt[7])^n + (1+I*Sqrt[7])^n)/2^n)/2 // Simplify, {n, 1, 50}] (* Jean-François Alcover, Jun 04 2017 *)

a(n+1) - a(n) = A002249(n).

a(n) = 2*a(n-1)-3*a(n-2)+2*a(n-3). G.f.: (1+x+x^2)/((1-x)*(1-x+2*x^2)). [Colin Barker, Mar 27 2012]

Corrected by T. D. Noe, Nov 07 2006

A272931 a(n) = 2^(n+1)cos(narctan(sqrt(15))).

Original entry on oeis.org

2, 1, -7, -11, 17, 61, -7, -251, -223, 781, 1673, -1451, -8143, -2339, 30233, 39589, -81343, -239699, 85673, 1044469, 701777, -3476099, -6283207, 7621189, 32754017, 2269261, -128746807, -137823851, 377163377, 928458781, -580194727, -4294029851, -1973250943

Offset: 0

Peter Luschny, May 11 2016

For n >= 1, |a(n)| is the unique odd positive solution y to 4^(n+1) = 15*x^2 + y^2. The value of x is |A106853(n-1)|. - Jianing Song, Jan 22 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (1,-4).

Cf. A087204, A002249, A106853.

seq(simplify(((1-I*sqrt(15))^n + (1+I*sqrt(15))^n)/2^n), n=0..32);

Mathematica
```
LinearRecurrence[{1, -4}, {2, 1}, 33]
```

Vec((2 - x) / (1 - x + 4*x^2) + O(x^40)) \\ Colin Barker, Jan 22 2019

[lucas_number2(i, 1, 4) for i in range(33)]

Let a(x) = x/2 - i*sqrt(15)*x/2 and b(x) = x/2 + i*sqrt(15)*x/2, then:
a(n) = a(1)^n + b(1)^n.
a(n) = n! [x^n] exp(a(x)) + exp(b(x)).
a(n) = [x^n] (2 - x)/(4*x^2 - x + 1).
a(n) = Sum_{k=0..floor(n/2)} (-4)^k*n*(n - k - 1)!/(k!*(n - 2*k)!) for n >= 1.
For n >= 1, 15*a(n)^2 + A106853(n-1)^2 = 4^(n+1). - Jianing Song, Jan 22 2019
a(n) = a(n-1) - 4*a(n-2) for n>1. - Colin Barker, Jan 22 2019
a(n) = 2*A106853(n) - A106853(n-1). - R. J. Mathar, Aug 19 2022

A002248 Number of points on y^2 + xy = x^3 + x^2 + x over GF(2^n).

Original entry on oeis.org

2, 8, 14, 16, 22, 56, 142, 288, 518, 968, 1982, 4144, 8374, 16472, 32494, 65088, 131174, 263144, 525086, 1047376, 2094358, 4193912, 8393806, 16783200, 33550022, 67092488, 134210174, 268460656, 536911222

Offset: 1

N. J. A. Sloane, Iwan Duursma

This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). The point at infinity is counted also. - T. D. Noe, Mar 12 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Hugh Williams, R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-4).

I:=[2, 8, 14, 16]; [n le 4 select I[n] else 4*Self(n-1)-7*Self(n-2)+8*Self(n-3)-4*Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jun 18 2012

Needs["FiniteFields`"]; Table[cnt=1; (* 1 point at infinity *) f=Table[GF[2,n][IntegerDigits[i,2,n]], {i,0,2^n-1}]; Do[If[y^2+x*y-x^3-x^2-x==0, cnt++ ], {x,f}, {y,f}]; cnt, {n,6}] (* T. D. Noe, Mar 12 2009 *)
LinearRecurrence[{4,-7,8,-4},{2,8,14,16},30] (* Vincenzo Librandi, Jun 18 2012 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -4,8,-7,4]^(n-1)*[2;8;14;16])[1,1] \\ Charles R Greathouse IV, Jun 23 2020

a(n) = 2^n + 1 - b(n); b(n) = b(n-1) - 2*b(n-2), b(1)=1, b(2)=-3; b(n) = A002249(n).
G.f.: -2*x*(-1+2*x^2) / ( (x-1)*(2*x-1)*(2*x^2 - x + 1) ).
a(n) = 4*a(n-1) - 7*a(n-2) + 8*a(n-3) - 4*a(n-4). - Vincenzo Librandi, Jun 18 2012

cp's OEIS Frontend

A107920 Lucas and Lehmer numbers with parameters (1 +- sqrt(-7))/2.

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A001607 a(n) = -a(n-1) - 2*a(n-2).

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A060728 Numbers n such that Ramanujan's equation x^2 + 7 = 2^n has an integer solution.

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A038198 Numbers n such that n^2 + 7 is a power of 2.

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A077021 a(n) is the unique odd positive solution y of 2^n = 7x^2 + y^2.

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A105578 a(n+3) = 2a(n+2) - 3a(n+1) + 2a(n); a(0) = 1, a(1) = 1, a(2) = 0.

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A105577 a(n) = 2a(n-1) - 3a(n-2) + 2*a(n-3) with a(0) = 1, a(1) = 5, a(2) = 6.

A272931 a(n) = 2^(n+1)cos(narctan(sqrt(15))).