A227078 -id:A227078 - 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

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

A002249 a(n) = a(n-1) - 2*a(n-2) with a(0) = 2, a(1) = 1.

Original entry on oeis.org

2, 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

Offset: 0

N. J. A. Sloane, Iwan Duursma

4*2^n = A002249(n)^2 + 7*A001607(n)^2. See A077020, A077021.
Among presented initial elements of the sequence a(n), the maximal increasing or decreasing subsequences have length either 3 or 4. - Roman Witula, Aug 21 2012
This is the Lucas Sequence V_n(P, Q) = V_n(1, 2). U_n(1, 2) = A107920(n). - Raphie Frank, Dec 25 2013
The only numbers that occur more than once are 1=a(1)=a(4) and -5=a(3)=a(9).  See Noam D. Elkies's posting in the Mathematics Stack Exchange link. - Robert Israel, Dec 21 2016

			We have a(2)-a(7) = a(5)-a(4) = a(6)+a(4) = a(11)-a(10) = a(12)+a(10)=10. Further the following relations: ((1+i*sqrt(7))/2)^4 + ((1-i*sqrt(7))/2)^4 = 1 and ((1+i*sqrt(7))/2)^8 + ((1-i*sqrt(7))/2)^8 = -31. - _Roman Witula_, Aug 21 2012
G.f. = 2 + x - 3*x^2 - 5*x^3 + x^4 + 11*x^5 + 9*x^6 - 13*x^7 - 31*x^8 + ...
From _Raphie Frank_, Dec 05 2015: (Start)
V_n(1, 2) = a(1*n) = ((a(1) + sqrt(-7))/2)^n + ((a(1) - sqrt(-7))/2)^n; a(1) = 1.
V_n(-3, 4) = a(2*n) = ((a(2) + sqrt(-7))/2)^n + ((a(2) - sqrt(-7))/2)^n; a(2) = -3.
V_n(-5, 8) = a(3*n) = ((a(3) + sqrt(-7))/2)^n + ((a(3) - sqrt(-7))/2)^n; a(3) = -5.
V_n(11, 32) = a(5*n) = ((a(5) + sqrt(-7))/2)^n + ((a(5) - sqrt(-7))/2)^n; a(5) = 11.
V_n(-181, 8192) = a(13*n) = ((a(13) + sqrt(-7))/2)^n + ((a(13) - sqrt(-7))/2)^n; a(13) = -181.
(End)

T. D. Noe, Table of n, a(n) for n = 0..500
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Mathematics Stack Exchange, An exotic sequence, March 2014.
Mihai Prunescu, On other two representations of the C-recursive integer sequences by terms in modular arithmetic, arXiv:2406.06436 [math.NT], 2024. See p. 18.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 17.
Mihai Prunescu and Joseph M. Shunia, On modular representations of C-recursive integer sequences, arXiv:2502.16928 [math.NT], 2025. See p. 6.
Wikipedia, Lucas Sequence.
Index entries for linear recurrences with constant coefficients, signature (1,-2).

Cf. A014551, A107920, A060728.

I:=[2,1]; [n le 2 select I[n] else Self(n-1)-2*Self(n-2): n in [1..50]]; // Vincenzo Librandi, Nov 29 2015

A002249 := proc(n) option remember; >if n = 1 then 1 elif n = 2 then -3 else A002249(n-1>)-2*A002249(n-2); fi; end;

LinearRecurrence[{1,-2}, {2,1}, 50] (* Roman Witula, Aug 21 2012 *)
a[ n_] := 2^(n/2) ChebyshevT[ n, 8^(-1/2)] 2; (* Michael Somos, Jun 02 2014 *)
a[ n_] := 2^Min[0, n] SeriesCoefficient[ (2 - x) / (1 - x + 2 x^2), {x, 0, Abs @ n}]; (* Michael Somos, Jun 02 2014 *)
Table[2 Re[((1 + I Sqrt[7])/2)^n], {n, 0, 40}] (* Jean-François Alcover, Jun 02 2017 *)

{a(n) = if( n<0, 2^n * a(-n), polsym(2 - x + x^2, n)[n+1])}; /* Michael Somos, Jun 02 2014 */

{a(n) = 2 * real( ((1 + quadgen(-28)) / 2)^n )}; /* Michael Somos, Jun 02 2014 */

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

from sympy import sqrt, re, I
def a(n): return 2*re(((1 + I*sqrt(7))/2)**n)
print([a(n) for n in range(40)]) # Indranil Ghosh, Jun 02 2017

[lucas_number2(n,1,2) for n in range(0, 40)] # Zerinvary Lajos, Apr 30 2009

G.f.: (2-x)/(1-x+2x^2). - Michael Somos, Oct 18 2002
a(n) = trace(A^n) for the square matrix A=[1, -2; 1, 0]. - Paul Barry, Sep 05 2003
a(n) = 2^((n+2)/2)*cos(-n*acot(sqrt(7)/7)). - Paul Barry, Sep 06 2003
a(n) = (-1)^n*(2*A110512(n) - A001607(n)) = ((1 + i*sqrt(7))/2)^n + ((1 - i*sqrt(7))/2)^n. - Roman Witula, Aug 21 2012
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(7*k+1)/(x*(7*k+8) + 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
(a(A060728(n) - 2))^2 = (A107920(2*(A060728(n)) - 4))^2 = 2^(A060728(n)) - 7 = A227078(n), the Ramanujan-Nagell squares. - Raphie Frank, Dec 25 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x - 7*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = (A107920(n+1) + 2*A107920(n+2) - A107920(n+3))/2. - Raphie Frank, Nov 28 2015
V_n(P,Q) = a(k*n) = ((a(k) + sqrt(-7))/2)^n + ((a(k) - sqrt(-7))/2)^n for k is in {1, 2, 3, 5, 13} = (A060728(n) - 2), P is in {1, -3, -5, 11, -181} = a(k), and Q is in {2, 4, 8, 32, 8192} = 2^k = (2*A076046(n) + 2) = (A227078(n) - 7)/4. P^2 - 4*Q = -7. - Raphie Frank, Dec 05 2015
From Peter Bala, Nov 16 2022: (Start)
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all positive integers n and k and all primes p.
A268924(n) == a(3^n) (mod 3^n). (End)

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A107920 Lucas and Lehmer numbers with parameters (1 +- sqrt(-7))/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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A002249 a(n) = a(n-1) - 2*a(n-2) with a(0) = 2, a(1) = 1.

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A227077 y solutions to the Diophantine equation 2*x^2*(x^2 - 1) = 3*(y^2 - 1).

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A227079 Solutions n to the Diophantine equation: n = (2*x^2 - 1)^2 = (6*y^2 - 5).

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A227077 y solutions to the Diophantine equation 2x^2(x^2 - 1) = 3*(y^2 - 1).

A227079 Solutions n to the Diophantine equation: n = (2x^2 - 1)^2 = (6y^2 - 5).