A005667 -id:A005667 - cp's OEIS frontend

A084133 Duplicate of A005667.

1, 3, 19, 117, 721, 4443, 27379, 168717, 1039681, 6406803, 39480499, 243289797

Offset: 0

dead

Tanya Khovanova, Non Recursions

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803

Offset: 0

N. J. A. Sloane, May 24 1994

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd  b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd  b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3)  = 2*L(n - 2);
L(n) - L(n - 4)  = 5*F(n - 2);
L(n) - L(n - 6)  = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

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Index entries for sequences related to Chebyshev polynomials..
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for two-way infinite sequences.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011

Magma
```
[Lucas(n): n in [0..120]];
```

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

{a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

{a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

{a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022

from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023

[(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025

[lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A002386 Primes (lower end) with record gaps to the next consecutive prime: primes p(k) where p(k+1) - p(k) exceeds p(j+1) - p(j) for all j < k.

Original entry on oeis.org

2, 3, 7, 23, 89, 113, 523, 887, 1129, 1327, 9551, 15683, 19609, 31397, 155921, 360653, 370261, 492113, 1349533, 1357201, 2010733, 4652353, 17051707, 20831323, 47326693, 122164747, 189695659, 191912783, 387096133, 436273009, 1294268491

Offset: 1

N. J. A. Sloane

See the links by Jens Kruse Andersen et al. for very large gaps.

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 133.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, Sect 6.1, Table 1.
M. Kraitchik, Recherches sur la Théorie des Nombres. Gauthiers-Villars, Paris, Vol. 1, 1924, Vol. 2, 1929, see Vol. 1, p. 14.
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).

Brian Kehrig, Table of n, a(n) for n = 1..83 (first 75 terms from M. F. Hasler and N. J. A. Sloane, terms n = 76..77 added by Charles R Greathouse IV)
R. K. Guy, Letter to N. J. A. Sloane, Aug 1986
R. K. Guy, Letter to N. J. A. Sloane, 1987
Lutz Kämmerer, A fast probabilistic component-by-component construction of exactly integrating rank-1 lattices and applications, arXiv:2012.14263 [math.NA], 2020.
Jens Kruse Andersen and Norman Luhn, Record Prime Gaps
Alexei Kourbatov, Tables of record gaps between prime constellations, arXiv preprint arXiv:1309.4053 [math.NT], 2013.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, arXiv:1506.03042 [math.NT], 2015; and J. Int. Seq. 18 (2015) #15.11.2.
Alexei Kourbatov and Marek Wolf, Predicting maximal gaps in sets of primes, arXiv preprint arXiv:1901.03785 [math.NT], 2019.
Thomas R. Nicely, First occurrence prime gaps [For local copy see A000101]
Thomas R. Nicely, New maximal prime gaps and first occurrences, Math. Comput. 68,227 (1999) 1311-1315.
Tomás Oliveira e Silva, Gaps between consecutive primes
D. Shanks, On maximal gaps between successive primes, Math. Comp., 18 (1964), 646-651.
Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.
Eric Weisstein's World of Mathematics, Prime Gaps
Wikipedia, Prime gap
Robert G. Wilson v, Notes (no date)
J. Young and A. Potler, First occurrence prime gaps, Math. Comp., 52 (1989), 221-224.
Index entries for primes, gaps between

Cf. A000040, A001223, A000101 (upper ends), A005250 (record gaps), A000230, A111870, A111943.
Cf. A005669, A134266, A070866.
See also A205827(n) = A000040(A214935(n)), A182514(n) = A000040(A241540(n)).

s = {2}; gm = 1; Do[p = Prime[n]; g = Prime[n + 1] - p; If[g > gm, Print[p]; AppendTo[s, p]; gm = g], {n, 2, 1000000}]; s   (* Jean-François Alcover, Mar 31 2011 *)
Module[{nn=10^7,pr,df},pr=Prime[Range[nn]];df=Differences[pr];DeleteDuplicates[ Thread[ {Most[ pr],df}],GreaterEqual[#1[[2]],#2[[2]]]&]][[All,1]] (* The program generates the first 26 terms of the sequence. *) (* Harvey P. Dale, Sep 24 2022 *)

a(n)=local(p,g);if(n<2,2*(n>0),p=a(n-1);g=nextprime(p+1)-p;while(p=nextprime(p+1),if(nextprime(p+1)-p>g,break));p) /* Michael Somos, Feb 07 2004 */

p=q=2;g=0;until( g<(q=nextprime(1+p=q))-p && print1(q-g=q-p,","),) \\ M. F. Hasler, Dec 13 2007

a(n) = A000101(n) - A005250(n) = A008950(n-1) - 1. - M. F. Hasler, Dec 13 2007
A000720(a(n)) = A005669(n).
a(n) = A000040(A005669(n)). - M. F. Hasler, Apr 26 2014

Definition clarified by Harvey P. Dale, Sep 24 2022

A005250 Record gaps between primes.

Original entry on oeis.org

1, 2, 4, 6, 8, 14, 18, 20, 22, 34, 36, 44, 52, 72, 86, 96, 112, 114, 118, 132, 148, 154, 180, 210, 220, 222, 234, 248, 250, 282, 288, 292, 320, 336, 354, 382, 384, 394, 456, 464, 468, 474, 486, 490, 500, 514, 516, 532, 534, 540, 582, 588, 602, 652

Offset: 1

N. J. A. Sloane, R. K. Guy, May 20 1991

Here a "gap" means prime(n+1) - prime(n), but in other references it can mean prime(n+1) - prime(n) - 1.
a(n+1)/a(n) <= 2, for all n <= 80, and a(n+1)/a(n) < 1 + f(n)/a(n) with f(n)/a(n) <= epsilon for some function f(n) and with 0 < epsilon <= 1. It also appears, with the small amount of data available, for all n <= 80, that a(n+1)/a(n) ~ 1. - John W. Nicholson, Jun 08 2014, updated Aug 05 2019
Equivalent to the above statement, A053695(n) = a(n+1) - a(n) <= a(n). - John W. Nicholson, Jan 20 2016
Conjecture: a(n) = O(n^2); specifically, a(n) <= n^2. - Alexei Kourbatov, Aug 05 2017
Conjecture: below the k-th prime, the number of maximal gaps is about 2*log(k), i.e., about twice as many as the expected number of records in a sequence of k i.i.d. random variables (see arXiv:1709.05508 for a heuristic explanation). - Alexei Kourbatov, Mar 16 2018

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 133.
R. K. Guy, Unsolved Problems in Number Theory, A8.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Brian Kehrig, Table of n, a(n) for n = 1..83 (first 77 terms from John W. Nicholson, terms n=78..80 added by Rodolfo Ruiz-Huidobro)
Jens Kruse Andersen and Norman Luhn, Record Prime Gaps
Alex Beveridge, Table giving known values of A000101(n), A005250(n), A107578(n)
R. P. Brent, J. H. Osborn and W. D. Smith, Lower bounds on maximal determinants of +-1 matrices via the probabilistic method, arXiv preprint arXiv:1211.3248 [math.CO], 2012.
C. K. Caldwell, Table of prime gaps
C. K. Caldwell, Gaps up to 1132
R. K. Guy, Letter to N. J. A. Sloane, Aug 1986
R. K. Guy, Letter to N. J. A. Sloane, 1987
Lutz Kämmerer, A fast probabilistic component-by-component construction of exactly integrating rank-1 lattices and applications, arXiv:2012.14263 [math.NA], 2020.
Alexei Kourbatov, Maximal gaps between prime k-tuples: a statistical approach, arXiv preprint arXiv:1301.2242 [math.NT], 2013 and J. Int. Seq. 16 (2013) #13.5.2.
Alexei Kourbatov, Tables of record gaps between prime constellations, arXiv preprint arXiv:1309.4053 [math.NT], 2013.
Alexei Kourbatov, The distribution of maximal prime gaps in Cramer's probabilistic model of primes, arXiv preprint arXiv:1401.6959 [math.NT], 2014.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, arXiv:1506.03042 [math.NT], 2015; and J. Int. Seq. 18 (2015) #15.11.2.
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015; and Int. Math. Forum, 10 (2015), 283-288.
Alexei Kourbatov, On the distribution of maximal gaps between primes in residue classes, arXiv preprint arXiv:1610.03340 [math.NT], 2016.
Alexei Kourbatov, On the nth record gap between primes in an arithmetic progression, arXiv:1709.05508 [math.NT], 2017; and Int. Math. Forum, 13 (2018), 65-78.
Alexei Kourbatov and Marek Wolf, Predicting maximal gaps in sets of primes, arXiv:1901.03785 [math.NT], 2019.
Ya-Ping Lu and Shu-Fang Deng, An upper bound for the prime gap, arXiv:2007.15282 [math.GM], 2020.
Thomas R. Nicely, Some Results of Computational Research in Prime Numbers [See local copy in A007053]
Thomas R. Nicely, First occurrence prime gaps [For local copy see A000101]
Tomás Oliveira e Silva, Gaps between consecutive primes
D. Shanks, On maximal gaps between successive primes, Mathematics of Computation, 18(88), 646-651. (1964).
Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.
Eric Weisstein's World of Mathematics, Prime Gaps
Wikipedia, Prime gap
Robert G. Wilson v, Notes (no date)
Marek Wolf, A Note on the Andrica Conjecture, arXiv:1010.3945 [math.NT], 2010.
J. Young and A. Potler, First occurrence prime gaps, Math. Comp., 52 (1989), 221-224.
Index entries for primes, gaps between

Records in A001223. For positions of records see A005669.

Cf. A000040, A002386, A000101, A008996, A058320, A107578.

a005250 n = a005250_list !! (n-1)
a005250_list = f 0 a001223_list
   where f m (x:xs) = if x <= m then f m xs else x : f x xs
-- Reinhard Zumkeller, Dec 12 2012

nn=10^7;Module[{d=Differences[Prime[Range[nn]]],ls={1}},Table[If[d[[n]]> Last[ls],AppendTo[ls,d[[n]]]],{n,nn-1}];ls] (* Harvey P. Dale, Jul 23 2012 *)
DeleteDuplicates[Differences[Prime[Range[10^7]]],GreaterEqual] (* The program generates the first 26 terms of the sequence. *) (* Harvey P. Dale, May 12 2022 *)

p=q=2;g=0;until( g<(q=nextprime(1+p=q))-p & print1(g=q-p,","),) \\ M. F. Hasler, Dec 13 2007

p=2; g=0;m=g; forprime(q=3,10^13,g=q-p;if(g>m,print(g", ",p,", ",q);m=g);p=q) \\ John W. Nicholson, Dec 18 2016

a(n) = A000101(n) - A002386(n) = A008996(n-1) + 1. - M. F. Hasler, Dec 13 2007

a(n+1) = 1 + Sum_{i=1..n} A053695(i). - John W. Nicholson, Jan 20 2016

More terms from Andreas Boerner (andreas.boerner(AT)altavista.net), Jul 11 2000
Additional comments from Frank Ellermann, Apr 20 2001
More terms from Robert G. Wilson v, Jan 03 2002, May 01 2006

A005668 Denominators of continued fraction convergents to sqrt(10).

Original entry on oeis.org

0, 1, 6, 37, 228, 1405, 8658, 53353, 328776, 2026009, 12484830, 76934989, 474094764, 2921503573, 18003116202, 110940200785, 683644320912, 4212806126257, 25960481078454, 159975692596981, 985814636660340, 6074863512559021, 37434995712014466, 230684837784645817

Offset: 0

N. J. A. Sloane, Simon Plouffe, R. K. Guy

a(2*n+1) with b(2*n+1) := A005667(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = -1, a(2*n) with b(2*n) := A005667(2*n), n>=1, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n)= 6*S(n-1,2*19) = 6*A078987(n-1), n >= 0 and a(2*n+1) = A097315(n), n >= 0, with S(n,x) Chebyshev's polynomials of the second kind. S(-1,x)=0. See A049310.
Sqrt(10) = 6/2 + 6/37 + 6/(37*1405) + 6/(1405*53353) + ... - Gary W. Adamson, Dec 21 2007
a(p) == 40^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 6's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,6} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 6-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 6 kinds of squares available. (End)

			G.f. = x + 6*x^2 + 37*x^3 + 228*x^4 + 1405*x^5 + 8658*x^6 + 53353*x^7 + ...

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
R. K. Guy, Letter to N. J. A. Sloane, 1987
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 427
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for linear recurrences with constant coefficients, signature (6,1).
Index entries for sequences related to Chebyshev polynomials.

Row n=6 of A073133, A172236, A352361.

Cf. A005667, A000045, A000129, A006190, A001076, A052918, A175185 (Pisano periods), A243399.

[n le 2 select n-1 else 6*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

evalf(sqrt(10),200); convert(%,confrac,fractionlist); fractionlist;
A005668:=-z/(-1+6*z+z**2); - Simon Plouffe in his 1992 dissertation.
a := n -> `if`(n<2,n,6^(n-1)*hypergeom([1-n/2,(1-n)/2], [1-n], -1/9)):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Jun 28 2017

LinearRecurrence[{6,1}, {0,1}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
a[ n_] := (-I)^(n - 1) ChebyshevU[ n - 1, 3 I]; (* Michael Somos, May 28 2014 *)
a[ n_] := MatrixPower[ {{0, 1}, {1, 6}}, n + 1][[1, 1]]; (* Michael Somos, May 28 2014 *)
Fibonacci[Range[0,30],6] (* G. C. Greubel, Jun 06 2019 *)
Join[{0},Convergents[Sqrt[10],30]//Denominator] (* Harvey P. Dale, Dec 28 2022 *)

{a(n) = ([0, 1; 1, 6]^(n+1)) [1, 1]}; /* Michael Somos, May 28 2014 */

{a(n) = (-I)^(n-1) * polchebyshev( n-1, 2, 3*I)}; /* Michael Somos, May 28 2014 */

from sage.combinat.sloane_functions import recur_gen3; it = recur_gen3(0,1,6,6,1,0); [next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,6,-1) for n in range(0, 21)]# Zerinvary Lajos, Apr 24 2009

G.f.: x / (1 - 6*x - x^2).
a(n) = 6*a(n-1) + a(n-2).
a(n) = ((-i)^(n-1))*S(n-1, 3*i) with S(n, x) Chebyshev's polynomials of the second kind (see A049310) and i^2=-1.
a(n) = F(n, 6), the n-th Fibonacci polynomial evaluated at x=6. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((3+sqrt(10))^n - (3-sqrt(10))^n)/(2*sqrt(10)).
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*6^(n-1-2*i). (End)
Sum_{n>=1}(-1)^(n-1)/(a(n)*a(n+1)) = sqrt(10) - 3. - Vladimir Shevelev, Feb 23 2013
a(n) = [M^(n+1)]{0,0}, where M = [0,1; 1,6]. - _L. Edson Jeffery, Aug 28 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014
a(n) = 6^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1/9) for n >= 2. - Peter Luschny, Jun 28 2017
G.f.: x/(1 - 6*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 6 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Chebyshev comments from Wolfdieter Lang, Jan 21 2003

A005669 Indices of primes where largest gap occurs.

Original entry on oeis.org

1, 2, 4, 9, 24, 30, 99, 154, 189, 217, 1183, 1831, 2225, 3385, 14357, 30802, 31545, 40933, 103520, 104071, 149689, 325852, 1094421, 1319945, 2850174, 6957876, 10539432, 10655462, 20684332, 23163298, 64955634, 72507380, 112228683, 182837804, 203615628, 486570087

Offset: 1

N. J. A. Sloane

nonn

Conjecture: log a(n) ~ n/2. That is, record prime gaps occur about twice as often as records in an i.i.d. random sequence of comparable length (see arXiv:1709.05508 for a heuristic explanation). - Alexei Kourbatov, Mar 28 2018

H. Riesel, Prime numbers and computer methods for factorization, Progress in Mathematics, Vol. 57, Birkhäuser, Boston, 1985, Chap. 4, see pp. 381-384.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

John W. Nicholson, Table of n, a(n) for n = 1..82 (first 77 terms from Charles R Greathouse IV)
Jens Kruse Andersen, The Top-20 Prime Gaps.
Jens Kruse Andersen, New record prime gap.
Jens Kruse Andersen, Maximal gaps.
R. K. Guy, Letter to N. J. A. Sloane, 1987.
Alexei Kourbatov, Tables of record gaps between prime constellations, arXiv preprint arXiv:1309.4053 [math.NT], 2013.
Alexei Kourbatov, On the nth record gap between primes in an arithmetic progression, arXiv:1709.05508 [math.NT], 2017; Int. Math. Forum, 13 (2018), 65-78.
Alexei Kourbatov and Marek Wolf, Predicting maximal gaps in sets of primes, arXiv preprint arXiv:1901.03785 [math.NT], 2019.
Thomas R. Nicely, First occurrence prime gaps. [For local copy see A000101]
Matt Visser, Verifying the Firoozbakht, Nicholson, and Farhadian conjectures up to the 81st maximal prime gap, arXiv:1904.00499 [math.NT], 2019.
Robert G. Wilson v, Notes (no date).
J. Young and A. Potler, First occurrence prime gaps, Math. Comp., 52 (1989), 221-224.
Index entries for primes, gaps between

Cf. A000101, A002386, A005250.

f[n_] := Block[{d, i, m = 0}, Reap@ For[i = 1, i <= n, i++, d = Prime[i + 1] - Prime@ i; If[d > m, m = d; Sow@ i, False]] // Flatten // Rest]; f@ 1000000 (* Michael De Vlieger, Mar 24 2015 *)

a(n) = A000720(A002386(n)).

a(n) = A107578(n) - 1. - Jens Kruse Andersen, Oct 19 2010

A005672 a(n) = Fibonacci(n+1) - 2^floor(n/2).

Original entry on oeis.org

0, 0, 0, 1, 1, 4, 5, 13, 18, 39, 57, 112, 169, 313, 482, 859, 1341, 2328, 3669, 6253, 9922, 16687, 26609, 44320, 70929, 117297, 188226, 309619, 497845, 815656, 1313501, 2145541

Offset: 0

N. J. A. Sloane

R. K. Guy, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. K. Guy, Letter to N. J. A. Sloane, 1987
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992

Gives diagonal sums of triangle A054143.

A005672:=z**3/(z**2+z-1)/(-1+2*z**2); # conjectured by Simon Plouffe in his 1992 dissertation
with (combinat):a[ -1]:=0:a[1]:=0:a[2]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-2] od: seq(a[n-1], n=0..31); # Zerinvary Lajos, Mar 17 2008

a(n) = Fibonacci(n-1) + 2*a(n-2), a(-1)=0, a(1)=0, a(2)=1. - Zerinvary Lajos, Mar 17 2008

A085447 a(n) = 6*a(n-1) + a(n-2), starting with a(0)=2 and a(1)=6.

Original entry on oeis.org

2, 6, 38, 234, 1442, 8886, 54758, 337434, 2079362, 12813606, 78960998, 486579594, 2998438562, 18477210966, 113861704358, 701647437114, 4323746327042, 26644125399366, 164188498723238, 1011775117738794

Offset: 0

Gary W. Adamson, Jul 01 2003

a(n+1)/a(n) converges to 3 + sqrt 10.

a(0)/a(1) = 1/3 = [3]; a(1)/a(2) = 6/38 = [6,3]; a(2)/a(3) = 38/234 = [6,6,3], a(3)/a(4) = 234/1442 = [6,6,6,3]; a(4)/a(5) = 1442/8886 = [6,6,6,6,3];...etc. Lim a(n)/a(n+1) as n approaches infinity = 0.1622776...= 1/(3 + sqrt10) = sqrt(10) - 3.

			a(4) = 6*a(3)+a(2) = 6*234+38 = 1442.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, B. 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 L_{6,n}
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (6,1).

Cf. A005667, A005668.

I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 6, a[n] == 6 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{6,1}, {2,6}, 30] (* G. C. Greubel, Nov 07 2018 *)

x='x+O('x^30); Vec(2*(1-3*x)/(1-6*x-x^2)) \\ G. C. Greubel, Nov 07 2018

a(n) = (3 + sqrt 10)^n + (3 - sqrt 10)^n = A005668(n+1) + A005668(n-1).
O.g.f.: 2*(-1+3*x)/(-1+6*x+x^2). - R. J. Mathar, Dec 02 2007
a(n) = 2*A005667(n). - R. J. Mathar, Nov 10 2013

Edited and extended by Henry Bottomley, Jul 13 2003

A040006 Continued fraction for sqrt(10).

Original entry on oeis.org

3, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

N. J. A. Sloane

Eventual period is (6). - Zak Seidov, Mar 05 2011
The convergents are given in A005667(n+1)/A005668(n+1), n >= 0. - Wolfdieter Lang, Nov 23 2017
Decimal expansion of 11/30. - Elmo R. Oliveira, Feb 16 2024

			3.162277660168379331998893544... = 3 + 1/(6 + 1/(6 + 1/(6 + 1/(6 + ...)))).

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 276.

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A010467 (decimal expansion), A005667(n+1)/A005668(n+1) (convergents), A248239 (Egyptian fraction).

Cf. A040000.

[6-3*(Binomial(2*n,n) mod 2): n in [0..100]]; // Vincenzo Librandi, Jan 03 2016

Digits := 100: convert(evalf(sqrt(N)),confrac,90,'cvgts'):

ContinuedFraction[Sqrt[10],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
PadRight[{3},120,{6}] (* Harvey P. Dale, Aug 25 2024 *)

contfrac(sqrt(10)) \\ For illustration.

A040006(n)=if(n,6,3) \\ M. F. Hasler, Nov 02 2019

a(n) = 3 + 3*sign(n). a(n) = 6, n > 0. - Wesley Ivan Hurt, Nov 01 2013
From Elmo R. Oliveira, Feb 16 2024: (Start)
G.f.: 3*(1+x)/(1-x).
E.g.f.: 6*exp(x) - 3.
a(n) = 3*A040000(n). (End)

A041006 Numerators of continued fraction convergents to sqrt(6).

Original entry on oeis.org

2, 5, 22, 49, 218, 485, 2158, 4801, 21362, 47525, 211462, 470449, 2093258, 4656965, 20721118, 46099201, 205117922, 456335045, 2030458102, 4517251249, 20099463098, 44716177445, 198964172878, 442644523201, 1969542265682, 4381729054565, 19496458483942

Offset: 0

N. J. A. Sloane

Interspersion of 2 sequences, 2*A054320 and A001079. - Gerry Martens, Jun 10 2015

Hugo Pfoertner, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A041007 (denominators).
Cf. A054320, A001079.
Analog for other sqrt(m): A001333 (m=2), A002531 (m=3), A001077 (m=5), A041008 (m=7), A041010 (m=8), A005667 (m=10), A041014 (m=11), ..., A042936 (m=1000).

I:=[2, 5, 22, 49]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 10 2015

Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[6],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
LinearRecurrence[{0, 10, 0, -1}, {2, 5, 22, 49}, 50] (* Vincenzo Librandi, Jun 10 2015 *)

A41006=contfracpnqn(c=contfrac(sqrt(6)), #c)[1, ][^-1] \\ Discard possibly incorrect last element. NB: a(n)=A41006[n+1]! M. F. Hasler, Nov 01 2019

\\ For correct index & more terms:
A041006(n)={n<#A041006|| A041006=extend(A041006, [2, 10; 4, -1], n\.8); A041006[n+1]}
extend(A, c, N)={for(n=#A+1, #A=Vec(A, N), A[n]=[A[n-i]|i<-c[, 1]]*c[, 2]); A} \\ M. F. Hasler, Nov 01 2019

From M. F. Hasler, Feb 13 2009: (Start)
a(2n) = 2*A142238(2n) = A041038(2n)/2;
a(2n-1) = A142238(2n-1) = A041038(2n-1) = A001079(n). (End)
G.f.: (2 + 5*x + 2*x^2 - x^3)/(1 - 10*x^2 + x^4).
a(n) = ((2 + sqrt(6))^(n+1) + (2 - sqrt(6))^(n+1))/2^(ceiling(n/2) + 1). - Robert FERREOL, Oct 13 2024
E.g.f.: sqrt(2)*sinh(sqrt(2)*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) + cosh(sqrt(2)*x)*(2*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). - Stefano Spezia, Oct 14 2024

More terms from Vincenzo Librandi, Jun 10 2015

cp's OEIS Frontend

A084133 Duplicate of A005667.

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A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

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A002386 Primes (lower end) with record gaps to the next consecutive prime: primes p(k) where p(k+1) - p(k) exceeds p(j+1) - p(j) for all j < k.

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A005250 Record gaps between primes.

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A005668 Denominators of continued fraction convergents to sqrt(10).

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A005669 Indices of primes where largest gap occurs.

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