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%I M1413 N0552 #1017 Mar 31 2024 14:03:08
%S 0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461,80782,195025,
%T 470832,1136689,2744210,6625109,15994428,38613965,93222358,225058681,
%U 543339720,1311738121,3166815962,7645370045,18457556052,44560482149,107578520350,259717522849
%N Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).
%C Sometimes also called lambda numbers.
%C Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
%C Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - _Emeric Deutsch_, Oct 27 2002
%C a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.
%C Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - _Wolfdieter Lang_, Jan 10 2003
%C Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - _Amarnath Murthy_, Mar 22 2003
%C This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - _Ross La Haye_, Aug 18 2003
%C Number of 132-avoiding two-stack sortable permutations.
%C From _Herbert Kociemba_, Jun 02 2004: (Start)
%C For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.
%C Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)
%C Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
%C Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - _David W. Wilson_
%C Sums of antidiagonals of A038207 [Pascal's triangle squared]. - _Ross La Haye_, Oct 28 2004
%C The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - _Jack Brennen_, Nov 13 2004
%C a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - _Reinhard Zumkeller_, Dec 03 2004
%C Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - _Charlie Marion_, Apr 01 2006
%C (0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - _Tom Copeland_, Oct 29 2007
%C Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - _Gary W. Adamson_, Dec 21 2007
%C A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - _Gary W. Adamson_, Mar 16 2008
%C Prime Pell numbers with an odd index gives the RMS value (A141812) of prime RMS numbers (A140480). - _Ctibor O. Zizka_, Aug 13 2008
%C From _Clark Kimberling_, Aug 27 2008: (Start)
%C Related convergents (numerator/denominator):
%C lower principal convergents: A002315/A001653
%C upper principal convergents: A001541/A001542
%C intermediate convergents: A052542/A001333
%C lower intermediate convergents: A005319/A001541
%C upper intermediate convergents: A075870/A002315
%C principal and intermediate convergents: A143607/A002965
%C lower principal and intermediate convergents: A143608/A079496
%C upper principal and intermediate convergents: A143609/A084068. (End)
%C Equals row sums of triangle A143808 starting with offset 1. - _Gary W. Adamson_, Sep 01 2008
%C Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - _Philippe Deléham_, Oct 28 2008
%C a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
%C Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - _Gary W. Adamson_, Dec 29 2008
%C Starting with offset 1 = row sums of triangle A153345. - _Gary W. Adamson_, Dec 24 2008
%C From _Charlie Marion_, Jan 07 2009: (Start)
%C In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
%C let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
%C and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
%C let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)
%C and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
%C For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
%C In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
%C k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
%C b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
%C for example, if k=1 and n=3, then a(1,n) = a(n+1) and
%C 1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
%C 1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
%C b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
%C b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
%C Cf. A001333, A142238, A142239, A153313, A153314, A153315, A153316, A153317, A153318.
%C (End)
%C Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - _Gary W. Adamson_, Jan 18 2009
%C It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - _Gary W. Adamson_, Feb 21 2009
%C Equals eigensequence of triangle A154325. - _Gary W. Adamson_, Feb 12 2009
%C Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - _Martin Griffiths_, Apr 25 2009
%C a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - _Sture Sjöstedt_, May 29 2009
%C The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - _Sarah-Marie Belcastro_, Jul 04 2009
%C As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - _Mark Dols_, May 18 2010
%C Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - _Gary W. Adamson_, Jul 16 2010
%C Numbers k such that 2*k^2 +- 1 is a square. - _Vincenzo Librandi_, Jul 18 2010
%C Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - _Gary W. Adamson_, Aug 06 2010
%C [u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - _James R. Buddenhagen_, Aug 14 2010
%C An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - _Johannes W. Meijer_, Aug 15 2010
%C Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - _Carmine Suriano_, Jan 14 2011
%C Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - _Charlie Marion_, Jan 18 2012
%C A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - _Sture Sjöstedt_, Oct 20 2012
%C Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - _Vladimir Shevelev_, Feb 22 2013
%C a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - _Bob Selcoe_, Jun 21 2013
%C Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./_., ._\., where . is an empty square. - _Jean M. Morales_, Sep 18 2013
%C Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - _Charlie Marion_, Nov 26 2013
%C An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - _Matthew Lehman_, Dec 25 2013
%C Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - _Charlie Marion_, Jan 30 2014
%C a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - _R. J. Mathar_, Feb 03 2014
%C a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - _David Neil McGrath_, Oct 28 2014
%C For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015
%C This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - _Tom Edgar_, Jan 28 2015
%C A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - _Michael Somos_, Jan 03 2017
%C a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from _Bob Selcoe_ dated Jun 21 2013.) - _Gregory L. Simay_, Sep 07 2017
%C Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - _J. Devillet_, Sep 28 2017
%C Also the number of matchings in the (n-1)-centipede graph. - _Eric W. Weisstein_, Sep 30 2017
%C Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - _Gregory L. Simay_, May 25 2018
%C (1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - _Gary W. Adamson_, Jul 17 2019
%C Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - _Brian Hopkins_, Aug 17 2020
%C Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - _Michael A. Allen_, Jan 23 2023
%C Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - _Amiram Eldar_, Oct 02 2023
%C a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - _Enrique Navarrete_, Dec 15 2023
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%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
%D Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
%D J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 53.
%D John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
%D S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
%D Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
%D R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
%D Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
%D Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
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%D Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.
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%H Dorota Bród, <a href="https://doi.org/10.2478/amsil-2019-0011">On a New One Parameter Generalization of Pell Numbers</a>, Annales Mathematicae Silesianae 33 (2019), 66-76.
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%H <a href="/index/Cor#core">Index entries for "core" sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,1).
%F G.f.: x/(1 - 2*x - x^2). - _Simon Plouffe_ in his 1992 dissertation.
%F a(2n+1)=A001653(n). a(2n)=A001542(n). - _Ira Gessel_, Sep 27 2002
%F G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - _Peter Bala_, Jan 04 2015
%F a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
%F a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
%F For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - _Shahreer Al Hossain_, Aug 18 2019
%F a(n) = integer nearest a(n-1)/(sqrt(2) - 1), where a(0) = 1. - _Clark Kimberling_
%F a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
%F a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
%F a(2n) = 2*a(n)*A001333(n). - _John McNamara_, Oct 30 2002
%F a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1.
%F Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - _Paul Barry_, May 09 2003
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - _Paul Barry_, May 13 2003
%F a(n-2) + a(n) = (1 + sqrt(2))^(n-1) + (1 - sqrt(2))^(n-1) = A002203(n-1). (A002203(n))^2 - 8(a(n))^2 = 4(-1)^n. - _Gary W. Adamson_, Jun 15 2003
%F Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
%F a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
%F Apart from initial terms, inverse binomial transform of A052955. - _Paul Barry_, May 23 2004
%F a(n)^2 + a(n+2k+1)^2 = A001653(k)*A001653(n+k); e.g., 5^2 + 70^2 = 5*985. - _Charlie Marion_ Aug 03 2005
%F a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - _Paul Barry_, Aug 28 2005
%F a(n) = a(n-1) + A001333(n-1) = A001333(n) - a(n-1) = A001109(n)/A001333(n) = sqrt(A001110(n)/A001333(n)^2) = ceiling(sqrt(A001108(n)/2)). - _Henry Bottomley_, Apr 18 2000
%F a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - _T. D. Noe_, Jan 19 2006
%F Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) = d(2n+1) = A001652(n); then ((-1)^n)*(c(n) + d(n)) = a(n). [Proof given by _Max Alekseyev_.] - _Creighton Dement_, Jul 21 2005
%F a(r+s) = a(r)*a(s+1) + a(r-1)*a(s). - _Lekraj Beedassy_, Sep 03 2006
%F a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A006645. - _Sergio Falcon_, Nov 22 2006
%F From _Miklos Kristof_, Mar 19 2007: (Start)
%F Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers (A002203):
%F For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
%F For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
%F For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
%F For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
%F F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
%F F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
%F F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
%F F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
%F F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
%F F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
%F a(n+1)*a(n) = 2*Sum_{k=0..n} a(k)^2 (a similar relation holds for A001333). - _Creighton Dement_, Aug 28 2007
%F a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - _Tom Copeland_, Nov 30 2007
%F Equals row sums of unsigned triangle A133156. - _Gary W. Adamson_, Apr 21 2008
%F a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - _Emeric Deutsch_, Aug 29 2008
%F a(n) = A000045(n) + Sum_{k=1..n-1} A000045(k)*a(n-k). - _Roger L. Bagula_ and _Gary W. Adamson_, Sep 07 2008
%F From _Hieronymus Fischer_, Jan 02 2009: (Start)
%F fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
%F See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x).
%F a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by _Josh Inman_, Mar 05 2024]
%F a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
%F If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - _Milan Janjic_, May 08 2010
%F a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - _Gary Detlefs_, Sep 09 2010
%F a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
%F a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). - _Charlie Marion_, Apr 13 2011
%F G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Dec 02 2011
%F In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - _Charlie Marion_, Jan 17 2012
%F Sum{n>=1} (-1)^(n-1)/(a(n)*a(n+1)) = sqrt(2) - 1. - _Vladimir Shevelev_, Feb 22 2013
%F From _Vladimir Shevelev_, Feb 24 2013: (Start)
%F (1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
%F (2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
%F (3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
%F (4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
%F a(-n) = -(-1)^n * a(n). - _Michael Somos_, Jun 01 2013
%F G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Aug 10 2013
%F G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 30 2013
%F a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - _Peter Luschny_, Nov 16 2013
%F a(n) = Sum_{k=1,3,5,...<=n} C(n,k)*2^((k-1)/2). - _Vladimir Shevelev_, Feb 06 2014
%F a(2n) = 2*a(n)*(a(n-1) + a(n)). - _John Blythe Dobson_, Mar 08 2014
%F a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - _Charlie Marion_, Mar 27 2014
%F a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - _Charlie Marion_, Mar 30 2014
%F a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - _Art DuPre_, Apr 04 2014
%F a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - _Art DuPre_, Apr 04 2014
%F a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - _Ralf Stephan_, May 23 2014
%F a(n) = round(sqrt(a(2n) + a(2n-1)))/2. - _Richard R. Forberg_, Jun 22 2014
%F a(n) = Product_{k divides n} A008555(k). - _Tom Edgar_, Jan 28 2015
%F a(n+k)^2 - A002203(k)*a(n)*a(n+k) + (-1)^k*a(n)^2 = (-1)^n*a(k)^2. - _Alexander Samokrutov_, Aug 06 2015
%F a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - _Peter Luschny_, Dec 17 2015
%F a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - _Tony Foster III_, May 07 2017
%F a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - _Peter Luschny_, Mar 07 2018
%F From _Rogério Serôdio_, Mar 30 2018: (Start)
%F Some properties:
%F (1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319);
%F (2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
%F (3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
%F (4) a(n) - a(n-2*k+1) = (A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2);
%F (5) Sum_{k=n..n+9} a(k) = 41*A001333(n+5). (End)
%F From _Kai Wang_, Dec 30 2019: (Start)
%F a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
%F a(m+r)*a(n+s) + a(m+s)*a(n+r) = (2*A002203(m+n+r+s) - (-1)^(n+s)*A002203(m-n)*A002203(r-s))/8.
%F A002203(m+r)*A002203(n+s) - A002203(m+s)*A002203(n+r) = (-1)^(n+s)*8*a(m-n)*a(r-s).
%F A002203(m+r)*A002203(n+s) - 8*a(m+s)*a(n+r) = (-1)^(n+s)*A002203(m-n)*A002203(r-s).
%F A002203(m+r)*A002203(n+s) + 8*a(m+s)*a(n+r) = 2*A002203(m+n+r+s)+ (-1)^(n+s)*8*a(m-n)*a(r-s). (End)
%F From _Kai Wang_, Jan 12 2020: (Start)
%F a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
%F a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
%F a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
%F a(m-n) = (-1)^n (a(m)*A002203(n) - A002203(m)*a(n))/2.
%F a(m+n) = (a(m)*A002203(n) + A002203(m)*a(n))/2.
%F A002203(n)^2 - A002203(n+r)*A002203(n-r) = (-1)^(n-r-1)*8*a(r)^2.
%F A002203(m)*A002203(n+1) - A002203(m+1)*A002203(n) = (-1)^(n-1)*8*a(m-n).
%F A002203(m-n) = (-1)^(n)*(A002203(m)*A002203(n) - 8*a(m)*a(n) )/2.
%F A002203(m+n) = (A002203(m)*A002203(n) + 8*a(m)*a(n) )/2. (End)
%F From _Kai Wang_, Mar 03 2020: (Start)
%F Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
%F Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
%F In general, for n > 0,
%F Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
%F a(n) = (A001333(n+3*k) + (-1)^(k-1)*A001333(n-3*k)) / (20*A041085(k-1)) for any k>=1. - _Paul Curtz_, Jun 23 2021
%F Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - _Greg Dresden_, Jan 05 2022
%F From _Peter Bala_, Aug 20 2022: (Start)
%F Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
%F Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319.
%F Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
%F Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
%F G.f. = 1/(1 - Sum_{k>=1} Fibonacci(k)*x^k). - _Enrique Navarrete_, Dec 17 2023
%F Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - _R. J. Mathar_, Feb 05 2024
%e G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
%e From _Enrique Navarrete_, Dec 15 2023: (Start)
%e From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
%e The following table gives the number of compositions of n=6 with Fibonacci number of parts:
%e Composition, number of such compositions, number of compositions of this type:
%e 6, 1, 8;
%e 5+1, 2, 10;
%e 4+2, 2, 6;
%e 3+3, 1, 4;
%e 4+1+1, 3, 9;
%e 3+2+1, 6, 12;
%e 2+2+2, 1, 1;
%e 3+1+1+1, 4, 8;
%e 2+2+1+1, 6, 6;
%e 2+1+1+1+1, 5, 5;
%e 1+1+1+1+1+1, 1, 1;
%e for a total of a(6)=70 compositions of n=6. (End).
%p A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
%p a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # _Alois P. Heinz_, Aug 01 2008
%p A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
%p seq(simplify(A000129(n)), n=0..31); # _Peter Luschny_, Dec 17 2015
%t CoefficientList[Series[x/(1 - 2*x - x^2), {x, 0, 60}], x] (* _Stefan Steinerberger_, Apr 08 2006 *)
%t Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* _Artur Jasinski_, Dec 10 2006 *)
%t LinearRecurrence[{2, 1}, {0, 1}, 60] (* _Harvey P. Dale_, Jan 04 2012 *)
%t a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* _Michael Somos_, Jun 01 2013 *)
%t Table[Fibonacci[n, 2], {n, 0, 20}] (* _Vladimir Reshetnikov_, May 08 2016 *)
%t Fibonacci[Range[0, 20], 2] (* _Eric W. Weisstein_, Sep 30 2017 *)
%t a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* _Michael Somos_, Oct 30 2021 *)
%o (PARI) default(realprecision, 2000); for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ _Harry J. Smith_, Jun 12 2009
%o (PARI) {a(n) = imag( (1 + quadgen( 8))^n )}; /* _Michael Somos_, Jun 01 2013 */
%o (PARI) {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* _Michael Somos_, Jun 01 2013 */
%o (PARI) a(n)=([2, 1; 1, 0]^n)[2,1] \\ _Charles R Greathouse IV_, Mar 04 2014
%o (PARI) {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* _Michael Somos_, Oct 30 2021 */
%o (Sage) [lucas_number1(n, 2, -1) for n in range(30)] # _Zerinvary Lajos_, Apr 22 2009
%o (Haskell)
%o a000129 n = a000129_list !! n
%o a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
%o -- _Reinhard Zumkeller_, Jan 05 2012, Feb 05 2011
%o (Maxima)
%o a[0]:0$
%o a[1]:1$
%o a[n]:=2*a[n-1]+a[n-2]$
%o A000129(n):=a[n]$
%o makelist(A000129(n),n,0,30); /* _Martin Ettl_, Nov 03 2012 */
%o (Maxima) makelist((%i)^(n-1)*ultraspherical(n-1,1,-%i),n,0,24),expand; /* _Emanuele Munarini_, Mar 07 2018 */
%o (Magma) [0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // _Vincenzo Librandi_, Aug 08 2015
%o (GAP) a := [0,1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # _Muniru A Asiru_, Oct 16 2017
%o (Python)
%o from itertools import islice
%o def A000129_gen(): # generator of terms
%o a, b = 0, 1
%o yield from [a,b]
%o while True:
%o a, b = b, a+2*b
%o yield b
%o A000129_list = list(islice(A000129_gen(),20)) # _Chai Wah Wu_, Jan 11 2022
%Y Partial sums of A001333.
%Y 2nd row of A172236.
%Y a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
%Y Cf. A002203, A096669, A096670, A097075, A097076, A051927, A005409.
%Y Cf. A175181 (Pisano periods), A214028 (Entry points), A214027 (number of zeros in a fundamental period).
%Y A077985 is a signed version.
%Y INVERT transform of Fibonacci numbers (A000045).
%Y Cf. A038207.
%Y The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
%Y Cf. A034867, A131980, A133156, A143808, A135387, A153346, A001622, A006497, A014176 (growth power), A098316, A154325, A021083, A243399, A008555.
%Y Cf. A048739.
%Y Cf. A073133.
%Y Cf. A041085.
%Y Cf. A157103, A352361.
%Y Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
%K nonn,easy,core,cofr,nice,frac
%O 0,3
%A _N. J. A. Sloane_
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