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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000129 Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).

Original entry on oeis.org

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Sometimes also called lambda numbers.
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.
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
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.
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
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
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
Number of 132-avoiding two-stack sortable permutations.
From Herbert Kociemba, Jun 02 2004: (Start)
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.
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)
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
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson
Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004
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
a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004
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
(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
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
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
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
lower principal convergents: A002315/A001653
upper principal convergents: A001541/A001542
intermediate convergents: A052542/A001333
lower intermediate convergents: A005319/A001541
upper intermediate convergents: A075870/A002315
principal and intermediate convergents: A143607/A002965
lower principal and intermediate convergents: A143608/A079496
upper principal and intermediate convergents: A143609/A084068. (End)
Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008
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
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
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
Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 2008
From Charlie Marion, Jan 07 2009: (Start)
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:
let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
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)
and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
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
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
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);
for example, if k=1 and n=3, then a(1,n) = a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
Cf. A001333, A142238, A142239, A153313, A153314, A153315, A153316, A153317, A153318.
(End)
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
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
Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009
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
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
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
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
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
Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010
Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010
[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
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
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
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
A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012
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
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
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
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
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
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
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
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
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
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015
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
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
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
Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017
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
(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
Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
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
Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023
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

Examples

			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 + ...
From _Enrique Navarrete_, Dec 15 2023: (Start)
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.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6,           1,     8;
5+1,         2,    10;
4+2,         2,     6;
3+3,         1,     4;
4+1+1,       3,     9;
3+2+1,       6,    12;
2+2+2,       1,     1;
3+1+1+1,     4,     8;
2+2+1+1,     6,     6;
2+1+1+1+1,   5,     5;
1+1+1+1+1+1, 1,     1;
for a total of a(6)=70 compositions of n=6. (End).

References

  • J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 76.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
  • 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.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
  • Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Paulo Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 43.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 3.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 61.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
  • 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).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.

Links

Crossrefs

Partial sums of A001333.
2nd row of A172236.
a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
Cf. A002203, A096669, A096670, A097075, A097076, A051927, A005409.
Cf. A175181 (Pisano periods), A214028 (Entry points), A214027 (number of zeros in a fundamental period).
A077985 is a signed version.
INVERT transform of Fibonacci numbers (A000045).
Cf. A038207.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A034867, A131980, A133156, A143808, A135387, A153346, A001622, A006497, A014176 (growth power), A098316, A154325, A021083, A243399, A008555.
Cf. A048739.
Cf. A073133.
Cf. A041085.
Cf. A157103, A352361.
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).

Programs

  • 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
  • Haskell
    a000129 n = a000129_list !! n
a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
-- Reinhard Zumkeller, Jan 05 2012, Feb 05 2011
  • 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
  • Maple
    A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
seq(simplify(A000129(n)), n=0..31); # Peter Luschny, Dec 17 2015
  • Mathematica
    CoefficientList[Series[x/(1 - 2*x - x^2), {x, 0, 60}], x] (* Stefan Steinerberger, Apr 08 2006 *)
Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *)
a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *)
Table[Fibonacci[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, May 08 2016 *)
Fibonacci[Range[0, 20], 2] (* Eric W. Weisstein, Sep 30 2017 *)
a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)
  • Maxima
    a[0]:0$
a[1]:1$
a[n]:=2*a[n-1]+a[n-2]$
A000129(n):=a[n]$
makelist(A000129(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
  • Maxima
    makelist((%i)^(n-1)*ultraspherical(n-1,1,-%i),n,0,24),expand; /* Emanuele Munarini, Mar 07 2018 */
  • PARI
    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
  • PARI
    {a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */
  • PARI
    {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */
  • PARI
    a(n)=([2, 1; 1, 0]^n)[2,1] \\ Charles R Greathouse IV, Mar 04 2014
  • PARI
    {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */
  • Python
    from itertools import islice
def A000129_gen(): # generator of terms
    a, b = 0, 1
    yield from [a,b]
    while True:
        a, b = b, a+2*b
        yield b
A000129_list = list(islice(A000129_gen(),20)) # Chai Wah Wu, Jan 11 2022
  • Sage
    [lucas_number1(n, 2, -1) for n in range(30)]  # Zerinvary Lajos, Apr 22 2009

Formula

G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation.
a(2n+1)=A001653(n). a(2n)=A001542(n). - Ira Gessel, Sep 27 2002
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
a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
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
a(n) = integer nearest a(n-1)/(sqrt(2) - 1), where a(0) = 1. - Clark Kimberling
a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(2n) = 2*a(n)*A001333(n). - John McNamara, Oct 30 2002
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.
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
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003
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
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
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
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004
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
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
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
a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006
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
a(r+s) = a(r)*a(s+1) + a(r-1)*a(s). - Lekraj Beedassy, Sep 03 2006
a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A006645. - Sergio Falcon, Nov 22 2006
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers (A002203):
For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
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(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(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(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)
a(n+1)*a(n) = 2*Sum_{k=0..n} a(k)^2 (a similar relation holds for A001333). - Creighton Dement, Aug 28 2007
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
Equals row sums of unsigned triangle A133156. - Gary W. Adamson, Apr 21 2008
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
a(n) = A000045(n) + Sum_{k=1..n-1} A000045(k)*a(n-k). - Roger L. Bagula and Gary W. Adamson, Sep 07 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
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.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x).
a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024]
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
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
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010
From Charlie Marion, Apr 13 2011: (Start)
a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
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
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
Sum{n>=1} (-1)^(n-1)/(a(n)*a(n+1)) = sqrt(2) - 1. - Vladimir Shevelev, Feb 22 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(-n) = -(-1)^n * a(n). - Michael Somos, Jun 01 2013
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
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
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
a(n) = Sum_{k=1,3,5,...<=n} C(n,k)*2^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(2n) = 2*a(n)*(a(n-1) + a(n)). - John Blythe Dobson, Mar 08 2014
a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014
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
a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014
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
a(n) = round(sqrt(a(2n) + a(2n-1)))/2. - Richard R. Forberg, Jun 22 2014
a(n) = Product_{k divides n} A008555(k). - Tom Edgar, Jan 28 2015
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
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015
a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017
a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319);
(2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(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;
(4) a(n) - a(n-2*k+1) = (A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2);
(5) Sum_{k=n..n+9} a(k) = 41*A001333(n+5). (End)
From Kai Wang, Dec 30 2019: (Start)
a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
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.
A002203(m+r)*A002203(n+s) - A002203(m+s)*A002203(n+r) = (-1)^(n+s)*8*a(m-n)*a(r-s).
A002203(m+r)*A002203(n+s) - 8*a(m+s)*a(n+r) = (-1)^(n+s)*A002203(m-n)*A002203(r-s).
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)
From Kai Wang, Jan 12 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
a(m-n) = (-1)^n (a(m)*A002203(n) - A002203(m)*a(n))/2.
a(m+n) = (a(m)*A002203(n) + A002203(m)*a(n))/2.
A002203(n)^2 - A002203(n+r)*A002203(n-r) = (-1)^(n-r-1)*8*a(r)^2.
A002203(m)*A002203(n+1) - A002203(m+1)*A002203(n) = (-1)^(n-1)*8*a(m-n).
A002203(m-n) = (-1)^(n)*(A002203(m)*A002203(n) - 8*a(m)*a(n) )/2.
A002203(m+n) = (A002203(m)*A002203(n) + 8*a(m)*a(n) )/2. (End)
From Kai Wang, Mar 03 2020: (Start)
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
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
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
From Peter Bala, Aug 20 2022: (Start)
Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319.
Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
G.f. = 1/(1 - Sum_{k>=1} Fibonacci(k)*x^k). - Enrique Navarrete, Dec 17 2023
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024
a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024

A002315 NSW numbers: a(n) = 6*a(n-1) - a(n-2); also a(n)^2 - 2*b(n)^2 = -1 with b(n) = A001653(n+1).

Original entry on oeis.org

1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319, 54608393, 318281039, 1855077841, 10812186007, 63018038201, 367296043199, 2140758220993, 12477253282759, 72722761475561, 423859315570607, 2470433131948081, 14398739476117879, 83922003724759193
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Named after the Newman-Shanks-Williams reference.
Also numbers k such that A125650(3*k^2) is an odd perfect square. Such numbers 3*k^2 form a bisection of A125651. - Alexander Adamchuk, Nov 30 2006
For positive n, a(n) corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs). - Lekraj Beedassy, Feb 06 2007
Also numbers m such that m^2 is a centered 16-gonal number; or a number of the form 8k(k+1)+1, where k = A053141(m) = {0, 2, 14, 84, 492, 2870, ...}. - Alexander Adamchuk, Apr 21 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling, Aug 27 2008
The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators=A075870, denominators=A002315. - Clark Kimberling, Aug 27 2008
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->oo} a(n) = x*(k*x+1)^n, k = (a(1)-3), x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Numbers k such that (ceiling(sqrt(k*k/2)))^2 = (1+k*k)/2. - Ctibor O. Zizka, Nov 09 2009
A001109(n)/a(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009
The values 2(a(n)^2+1) are all perfect squares, whose square root is given by A075870. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
a(n) represents all positive integers K for which 2(K^2+1) is a perfect square. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(8)'s along the main diagonal, and i's along the superdiagonal and subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
Integers k such that A000217(k-2) + A000217(k-1) + A000217(k) + A000217(k+1) is a square (cf. A202391). - Max Alekseyev, Dec 19 2011
Integer square roots of floor(k^2/2 - 1) or A047838. - Richard R. Forberg, Aug 01 2013
Remark: x^2 - 2*y^2 = +2*k^2, with positive k, and X^2 - 2*Y^2 = +2 reduce to the present Pell equation a^2 - 2*b^2 = -1 with x = k*X = 2*k*b and y = k*Y = k*a. (After a proposed solution for k = 3 by Alexander Samokrutov.) - Wolfdieter Lang, Aug 21 2015
If p is an odd prime, a((p-1)/2) == 1 (mod p). - Altug Alkan, Mar 17 2016
a(n)^2 + 1 = 2*b(n)^2, with b(n) = A001653(n), is the necessary and sufficient condition for a(n) to be a number k for which the diagonal of a 1 X k rectangle is an integer multiple of the diagonal of a 1 X 1 square. If squares are laid out thus along one diagonal of a horizontal 1 X a(n) rectangle, from the lower left corner to the upper right, the number of squares is b(n), and there will always be a square whose top corner lies exactly within the top edge of the rectangle. Numbering the squares 1 to b(n) from left to right, the number of the one square that has a corner in the top edge of the rectangle is c(n) = (2*b(n) - a(n) + 1)/2, which is A055997(n). The horizontal component of the corner of the square in the edge of the rectangle is also an integer, namely d(n) = a(n) - b(n), which is A001542(n). - David Pasino, Jun 30 2016
(a(n)^2)-th triangular number is a square; a(n)^2 = A008843(n) is a subsequence of A001108. - Jaroslav Krizek, Aug 05 2016
a(n-1)/A001653(n) is the closest rational approximation of sqrt(2) with a numerator not larger than a(n-1). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. a(n-1)/A001653(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Consider the quadrant of a circle with center (0,0) bounded by the positive x and y axes. Now consider, as the start of a series, the circle contained within this quadrant which kisses both axes and the outer bounding circle. Consider further a succession of circles, each kissing the x-axis, the outer bounding circle, and the previous circle in the series. See Holmes link. The center of the n-th circle in this series is ((A001653(n)*sqrt(2)-1)/a(n-1), (A001653(n)*sqrt(2)-1)/a(n-1)^2), the y-coordinate also being its radius. It follows that a(n-1) is the cotangent of the angle subtended at point (0,0) by the center of the n-th circle in the series with respect to the x-axis. - Graham Holmes, Aug 31 2019
There is a link between the two sequences present at the numerator and at the denominator of the fractions that give the coordinates of the center of the kissing circles. A001653 is the sequence of numbers k such that 2*k^2 - 1 is a square, and here, we have 2*A001653(n)^2 - 1 = a(n-1)^2. - Bernard Schott, Sep 02 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i+4*n+2) - G(i))/(2*G(i+2*n+1)) as long as G(i+2*n+1) != 0. - Klaus Purath, Mar 25 2021
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022
3*a(n-1) is the n-th almost Lucas-cobalancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022
In Moret-Blanc (1881) on page 259 some solution of m^2 - 2n^2 = -1 are listed. The values of m give this sequence, and the values of n give A001653. - Michael Somos, Oct 25 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 8 = A028884(1). In general, the following applies to all sequences (t) satisfying t(i) = 6t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun and the Sep 28 2008 formula on A038723 by Michael Somos. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028884(t(1)-6) always applies.
If (t) is a sequence satisfying t(k) = 7t(k-1) - 7t(k-2) + t(k-3) or t(k) = 6t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) - t(k))/(t(k+n+1) - t(k+n)) always applies, as long as t(k+n+1) - t(k+n) != 0 for integer k and n >= 0. (End)

Examples

			G.f. = 1 + 7*x + 41*x^2 + 239*x^3 + 1393*x^4 + 8119*x^5 + 17321*x^6 + ... - _Michael Somos_, Jun 26 2022

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.
  • Paulo Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 247.
  • 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).
  • P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.
  • P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

Links

Crossrefs

Bisection of A001333. Cf. A001109, A001653. A065513(n)=a(n)-1.
First differences of A001108 and A055997. Bisection of A084068 and A088014. Cf. A077444.
Cf. A125650, A125651, A125652.
Row sums of unsigned triangle A127675.
Cf. A053141, A075870. Cf. A000045, A002878, A004146, A026003, A100047, A119915, A192425, A088165 (prime subsequence), A057084 (binomial transform), A108051 (inverse binomial transform).
See comments in A301383.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.
Cf. A000129, A001653, A003499, A358682, A002203.

Programs

  • Haskell
    a002315 n = a002315_list !! n
a002315_list = 1 : 7 : zipWith (-) (map (* 6) (tail a002315_list)) a002315_list
-- Reinhard Zumkeller, Jan 10 2012
  • Magma
    I:=[1,7]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Maple
    A002315 := proc(n)
    option remember;
    if n = 0 then
        1 ;
    elif n = 1 then
        7;
    else
        6*procname(n-1)-procname(n-2) ;
    end if;
end proc: # Zerinvary Lajos, Jul 26 2006, modified R. J. Mathar, Apr 30 2017
a:=n->abs(Im(simplify(ChebyshevT(2*n+1,I)))):seq(a(n),n=0..20); # Leonid Bedratyuk, Dec 17 2017
# third Maple program:
a:= n-> (<<0|1>, <-1|6>>^n. <<1, 7>>)[1, 1]:
seq(a(n), n=0..22);  # Alois P. Heinz, Aug 25 2024
  • Mathematica
    a[0] = 1; a[1] = 7; a[n_] := a[n] = 6a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 20}] (* Robert G. Wilson v, Jun 09 2004 *)
Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {1,7},20]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
Table[ If[n>0, a=b; b=c; c=6b-a, b=-1; c=1], {n, 0, 20}] (* Jean-François Alcover, Oct 19 2012 *)
LinearRecurrence[{6, -1}, {1, 7}, 20] (* Bruno Berselli, Apr 03 2018 *)
a[ n_] := -I*(-1)^n*ChebyshevT[2*n + 1, I]; (* Michael Somos, Jun 26 2022 *)
  • PARI
    {a(n) = subst(poltchebi(abs(n+1)) - poltchebi(abs(n)), x, 3)/2};
  • PARI
    {a(n) = if(n<0, -a(-1-n), polsym(x^2-2*x-1, 2*n+1)[2*n+2]/2)};
  • PARI
    {a(n) = my(w=3+quadgen(32)); imag((1+w)*w^n)};
  • PARI
    for (i=1,10000,if(Mod(sigma(i^2+1,2),2)==1,print1(i,",")))
  • PARI
    {a(n) = -I*(-1)^n*polchebyshev(2*n+1, 1, I)}; /* Michael Somos, Jun 26 2022 */

Formula

a(n) = (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)).
a(n) = A001109(n)+A001109(n+1).
a(n) = (1+sqrt(2))/2*(3+sqrt(8))^n+(1-sqrt(2))/2*(3-sqrt(8))^n. - Ralf Stephan, Feb 23 2003
a(n) = sqrt(2*(A001653(n+1))^2-1), n >= 0. [Pell equation a(n)^2 - 2*Pell(2*n+1)^2 = -1. - Wolfdieter Lang, Jul 11 2018]
G.f.: (1 + x)/(1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(n, 6)+S(n-1, 6) = S(2*n, sqrt(8)), S(n, x) = U(n, x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n, 6)= A001109(n+1).
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -8) = a(n). - Benoit Cloitre, Nov 10 2002
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)-b^((2n+1)/2))/2. a(n) = A077444(n)/2. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
a(n) = Sum_{k=0..n} 2^k*binomial(2*n+1, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 08 2003
Same as: i such that sigma(i^2+1, 2) mod 2 = 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
a(n) = L(n, -6)*(-1)^n, where L is defined as in A108299; see also A001653 for L(n, +6). - Reinhard Zumkeller, Jun 01 2005
a(n) = A001652(n)+A046090(n); e.g., 239=119+120. - Charlie Marion, Nov 20 2003
A001541(n)*a(n+k) = A001652(2n+k) + A001652(k)+1; e.g., 3*1393 = 4069 + 119 + 1; for k > 0, A001541(n+k)*a(n) = A001652(2n+k) - A001652(k-1); e.g., 99*7 = 696 - 3. - Charlie Marion, Mar 17 2003
a(n) = Jacobi_P(n,1/2,-1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
P_{2n}+P_{2n+1} where P_i are the Pell numbers (A000129). Also the square root of the partial sums of Pell numbers: P_{2n}+P_{2n+1} = sqrt(Sum_{i=0..4n+1} P_i) (Santana and Diaz-Barrero, 2006). - David Eppstein, Jan 28 2007
a(n) = 2*A001652(n) + 1 = 2*A046729(n) + (-1)^n. - Lekraj Beedassy, Feb 06 2007
a(n) = sqrt(A001108(2*n+1)). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
a(n) = sqrt(8*A053141(n)*(A053141(n) + 1) + 1). - Alexander Adamchuk, Apr 21 2007
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8), a(1)=1. - Richard Choulet, Sep 18 2007
a(n) = A001333(2*n+1). - Ctibor O. Zizka, Aug 13 2008
a(n) = third binomial transform of 1, 4, 8, 32, 64, 256, 512, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n) = (-1)^(n-1)*(1/sqrt(-1))*cos((2*n - 1)*arcsin(sqrt(2)). - Artur Jasinski, Feb 17 2010 *WRONG*
a(n+k) = A001541(k)*a(n) + 4*A001109(k)*A001653(n); e.g., 8119 = 17*239 + 4*6*169. - Charlie Marion, Feb 04 2011
In general, a(n+k) = A001541(k)*a(n)) + sqrt(A001108(2k)*(a(n)^2+1)). See Sep 18 2007 entry above. - Charlie Marion, Dec 07 2011
a(n) = floor((1+sqrt(2))^(2n+1))/2. - Thomas Ordowski, Jun 12 2012
(a(2n-1) + a(2n) + 8)/(8*a(n)) = A001653(n). - Ignacio Larrosa Cañestro, Jan 02 2015
(a(2n) + a(2n-1))/a(n) = 2*sqrt(2)*( (1 + sqrt(2))^(4*n) - (1 - sqrt(2))^(4*n))/((1 + sqrt(2))^(2*n+1) + (1 - sqrt(2))^(2*n+1)). [This was my solution to problem 5325, School Science and Mathematics 114 (No. 8, Dec 2014).] - Henry Ricardo, Feb 05 2015
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 7, 0, 41, 0, 239, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -4, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047.
b(n) = 1/2*((-1)^n - 1)*Pell(n) + 1/2*(1 + (-1)^(n+1))*Pell(n+1). The o.g.f. is x*(1 + x^2)/(1 - 6*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*A026003(n-1)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*Pell(n)*(-x)^n.
Exp( Sum_{n >= 1} 8*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*x^n.
Exp( Sum_{n >= 1} (-8)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 8*A119915(n)*(-x)^n. Cf. A002878, A004146, A113224, and A192425. (End)
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + cosh(2*sqrt(2)*x))*exp(3*x). - Ilya Gutkovskiy, Jun 30 2016
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^ceiling(k/2). - David Pasino, Jul 09 2016
a(n) = A001541(n) + 2*A001542(n). - A.H.M. Smeets, May 28 2017
a(n+1) = 3*a(n) + 4*b(n), b(n+1) = 2*a(n) + 3*b(n), with b(n)=A001653(n). - Zak Seidov, Jul 13 2017
a(n) = |Im(T(2n-1,i))|, i=sqrt(-1), T(n,x) is the Chebyshev polynomial of the first kind, Im is the imaginary part of a complex number, || is the absolute value. - Leonid Bedratyuk, Dec 17 2017
a(n) = sinh((2*n + 1)*arcsinh(1)). - Bruno Berselli, Apr 03 2018
a(n) = 5*a(n-1) + A003499(n-1), a(0) = 1. - Ivan N. Ianakiev, Aug 09 2019
From Klaus Purath, Mar 25 2021: (Start)
a(n) = A046090(2*n)/A001541(n).
a(n+1)*a(n+2) = a(n)*a(n+3) + 48.
a(n)^2 + a(n+1)^2 = 6*a(n)*a(n+1) + 8.
a(n+1)^2 = a(n)*a(n+2) + 8.
a(n+1) = a(n) + 2*A001541(n+1).
a(n) = 2*A046090(n) - 1. (End)
3*a(n-1) = sqrt(8*b(n)^2 + 8*b(n) - 7), where b(n) = A358682(n). - Stefano Spezia, Nov 26 2022
a(n) = -(-1)^n - 2 + Sum_{i=0..n} A002203(i)^2. - Adam Mohamed, Aug 22 2024
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 3), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
For arbitrary x, a(n+x)^2 - 6*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 8 with a(n) := (1/2)*((1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1)) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(2) * A001542(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A081554(n) + 1/A081554(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 2*(1 - 1/A055997(k+2))). (End)

A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).

Original entry on oeis.org

0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204, 5090723162050694, 29670908962269962, 172934730611569080
Offset: 0

Views

Author

Wolfdieter Lang

Keywords

Comments

Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).
The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007
Jason Holt observes that a pair drawn from a drawer with A053141(n)+1 red socks and A001652(n) - A053141(n) blue socks will as likely as not be matching reds: (A053141+1)*A053141/((A001652+1)*A001652) = 1/2, n>0. - Bill Gosper, Feb 07 2010
The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009
Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017

Links

Crossrefs

Cf. A000129, A001108, A001109, A001652, A001653.
Cf. A011900, A029549, A053142, A075528, A103200.
Partial sums of A001542.
Cf. A000194, A001541, A002024, A016278, A046090, A055997, A077259, A077288, A077398, A084068.

Programs

  • Haskell
    a053141 n = a053141_list !! n
a053141_list = 0 : 2 : map (+ 2)
   (zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)
-- Reinhard Zumkeller, Jan 10 2012
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!(2*x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018
  • Maple
    A053141 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,2]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016
  • Mathematica
    Join[{a=0,b=1}, Table[c=6*b-a+1; a=b; b=c, {n,60}]]*2 (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
a[n_] := Floor[1/8*(2+Sqrt[2])*(3+2*Sqrt[2])^n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 28 2013 *)
Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
  • PARI
    concat(0,Vec(2/(1-x)/(1-6*x+x^2)+O(x^30))) \\ Charles R Greathouse IV, May 14 2012
  • PARI
    {x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};
a(n) = round((P(2*n+1) - 1)/2);
for(n=0, 30, print1(a(n), ", ")) \\ G. C. Greubel, Jul 15 2018
  • Sage
    [(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # G. C. Greubel, Apr 27 2020

Formula

a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by Pontus von Brömssen, Sep 11 2024]
a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.
G.f.: 2*x/((1-x)*(1-6*x+x^2)).
Let c(n) = A001109(n). Then a(n+1) = a(n)+2*c(n+1), a(0)=0. This gives a generating function (same as existing g.f.) leading to a closed form: a(n) = (1/8)*(-4+(2+sqrt(2))*(3+2*sqrt(2))^n + (2-sqrt(2))*(3-2*sqrt(2))^n). - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003
For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - Charlie Marion, Jul 01 2003
For n and j >= 1, A001109(j+1)*A001652(n) - A001109(j)*A001652(n-1) + a(j) = A001652(n+j). - Charlie Marion, Jul 07 2003
From Antonio Alberto Olivares, Jan 13 2004: (Start)
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
a(n) = -(1/2) - (1-sqrt(2))/(4*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(4*sqrt(2))*(3+2*sqrt(2))^n. (End)
a(n) = sqrt(2)*cosh((2*n+1)*log(1+sqrt(2)))/4 - 1/2 = (sqrt(1+4*A029549)-1)/2. - Bill Gosper, Feb 07 2010 [typo corrected by Vaclav Kotesovec, Feb 05 2016]
a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
From Charlie Marion, Oct 18 2004: (Start)
For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.
For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)
a(n) = A084068(n)*A084068(n+1). - Kenneth J Ramsey, Aug 16 2007
Let G(n,m) = (2*m+1)*a(n)+ m and H(n,m) = (2*m+1)*b(n)+m where b(n) is from the sequence A001652 and let T(a) = a*(a+1)/2. Then T(G(n,m)) + T(m) = 2*T(H(n,m)). - Kenneth J Ramsey, Aug 16 2007
Let S(n) equal the average of two adjacent terms of G(n,m) as defined immediately above and B(n) be one half the difference of the same adjacent terms. Then for T(i) = triangular number i*(i+1)/2, T(S(n)) - T(m) = B(n)^2 (setting m = 0 gives the square triangular numbers). - Kenneth J Ramsey, Aug 16 2007
a(n) = A001108(n+1) - A001109(n+1). - Dylan Hamilton, Nov 25 2010
a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - G. C. Greubel, Apr 27 2020
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/4. - Stefano Spezia, Mar 16 2024
a(n) = A000194(A029549(n)) = A002024(A075528(n)). - Pontus von Brömssen, Sep 11 2024

Extensions

Name corrected by Zak Seidov, Apr 11 2011

A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

Original entry on oeis.org

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

Examples

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

Links

Crossrefs

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

Programs

  • Magma
    [(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017
  • Maple
    with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018
  • Mathematica
    a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)
  • PARI
    my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

Formula

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (End)

A005013 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=1, a(3)=4. Alternates Fibonacci (A000045) and Lucas (A000032) sequences for even and odd n.

Original entry on oeis.org

0, 1, 1, 4, 3, 11, 8, 29, 21, 76, 55, 199, 144, 521, 377, 1364, 987, 3571, 2584, 9349, 6765, 24476, 17711, 64079, 46368, 167761, 121393, 439204, 317811, 1149851, 832040, 3010349, 2178309, 7881196, 5702887, 20633239, 14930352, 54018521, 39088169, 141422324
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

S(n,sqrt(5)), with the Chebyshev polynomials A049310, is an integer sequence in the real quadratic number field Q(sqrt(5)) with basis numbers <1,phi>, phi:=(1+sqrt(5))/2. S(n,sqrt(5)) = A(n) + 2*B(n)*phi, with A(n)= a(n+1)*(-1)^n and B(n)= A147600(n-1), n>=0, with A147600(-1):=0.
a(n) = p(n+1) where p(x) is the unique degree-(n-1) polynomial such that p(k) = Fibonacci(k) for k = 1, ..., n. - Michael Somos, Jan 08 2012
Row sums of A227431. - Richard R. Forberg, Jul 29 2013
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 5 and Q = 1. It is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence satisfies a linear recurrence of order four. - Peter Bala, Apr 18 2014
The sequence of convergents of the 2-periodic continued fraction [0; 1, -5, 1, -5, ...] = 1/(1 - 1/(5 - 1/(1 - 1/(5 - ...)))) = (1/2)*(5 - sqrt(5)) begins [0/1, 1/1, 5/4, 4/3, 15/11, 11/8, 40/29, ...]; the denominators give the present sequence. The sequence of numerators [0, 1, 5, 4, 15, 11, 40, ...] is A203976. Cf. A108412 and A026741. - Peter Bala, May 19 2014
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have (1/2)*a(2*n + 1) = 1/2 o 1/2 o ... o 1/2 (2*n + 1 terms) and (1/2)*sqrt(5)* a(2*n) = 1/2 o 1/2 o ... o 1/2 (2*n terms). Cf. A084068 and A049629. - Peter Bala, Mar 23 2018

Examples

			G.f. = x + x^2 + 4*x^3 + 3*x^4 + 11*x^5 + 8*x^6 + 29*x^7 + 21*x^8 + 76*x^9 + ...
a(3) = 4 since p(x) = (x^2 - 3*x + 4) / 2 interpolates p(1) = 1, p(2) = 1, p(3) = 2, and p(4) = 4. - _Michael Somos_, Jan 08 2012

References

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

Links

Crossrefs

Cf. A000032, A000045, A001906, A002878, A005247, A096140, A026741, A108412, A084068, A049629.

Programs

  • GAP
    a:=[0,1,1,4];; for n in [5..40] do a[n]:=3*a[n-2]-a[n-4]; od; a; # Muniru A Asiru, Oct 21 2018
  • Haskell
    a005013 n = a005013_list !! n
a005013_list = alt a000045_list a000032_list where
   alt (f::fs) (:l:ls) = f : l : alt fs ls
-- Reinhard Zumkeller, Jan 10 2012
  • Magma
    I:=[0,1,1,4]; [n le 4 select I[n]  else 3*Self(n-2) - Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 09 2016
  • Maple
    with(combinat): A005013 := n-> if n mod 2 = 0 then fibonacci(n) else fibonacci(n+1)+fibonacci(n-1); fi;
A005013:=z*(z**2+z+1)/((z**2+z-1)*(z**2-z-1)); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    CoefficientList[Series[(x + x^2 + x^3)/(1 - 3x^2 + x^4), {x, 0, 40}], x]
f[n_] = Product[(1 + 4*Sin[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; a = Table[f[n], {n, 0, 30}]; Round[a]; FullSimplify[ExpandAll[a]] (* Roger L. Bagula and Gary W. Adamson, Nov 26 2008 *)
LinearRecurrence[{0, 3, 0, -1}, {0, 1, 1, 4}, 100] (* G. C. Greubel, Feb 08 2016 *)
  • PARI
    {a(n) = if( n%2, fibonacci(n+1) + fibonacci(n-1), fibonacci(n))}; /* Michael Somos, Jan 08 2012 */
  • PARI
    {a(n) = if( n<0, -a(-n), subst( polinterpolate( vector( n, k, fibonacci(k))), x, n+1))}; /* Michael Somos, Jan 08 2012 */

Formula

a(1) = a(2) = 1, a(3) = 4, a(n) = (a(n-1) * a(n-2) - 1) / a(n-3), unless n=3. a(-n) = -a(n).
a(2n) = A001906(n), a(2n+1) = A002878(n). a(n)=F(n+1)+(-1)^(n+1)F(n-1). - Mario Catalani (mario.catalani(AT)unito.it), Sep 20 2002
G.f.: x*(1+x+x^2)/((1-x-x^2)*(1+x-x^2)).
a(n) = Product_{k=1..floor((n-1)/2)} (1 + 4*sin(k*Pi/n)^2). - Roger L. Bagula and Gary W. Adamson, Nov 26 2008
Binomial transform is A096140. - Michael Somos, Apr 13 2012
From Peter Bala, Apr 18 2014: (Start)
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = (1/2)*(sqrt(5) + 1) and beta = (1/2)*(sqrt(5) - 1). Equivalently, a(n) = U(n-1, sqrt(5)/2) for n odd and a(n) = (1/sqrt(5))*U(n-1, sqrt(5)/2) for n even, where U(n,x) is the Chebyshev polynomial of the second kind. (End)
E.g.f.: (Phi/sqrt(5))*exp(-Phi*x)*(exp(x)-1)*(exp(sqrt(5)*x) - 1/(Phi)^2), where Phi = (1+sqrt(5))/2. - G. C. Greubel, Feb 08 2016
a(n) = (5^floor((n-1)/2)/2^(n-1))*Sum_{k=0..n-1} binomial(n-1,k)/5^floor(k/2). - Tony Foster III, Oct 21 2018
a(n) = hypergeom([(1 - n)/2, (n + 1) mod 2 - n/2], [1 - n], -4) for n >= 2. - Peter Luschny, Sep 03 2019

Extensions

Additional comments from Michael Somos, Jun 01 2000

A143608 A005319 and A002315 interleaved.

Original entry on oeis.org

0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700
Offset: 0

Views

Author

Originally submitted by Clark Kimberling, Aug 27 2008. Merged with an essentially identical sequence submitted by Kenneth J Ramsey, Jun 01 2012, by N. J. A. Sloane, Aug 02 2012

Keywords

Comments

Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496.
Sequence a(n) such that a(2*n) = sqrt(2*A001108(2*n)) and a(2*n+1) = sqrt(A001108(2*n+1)).
For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer.
If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).
Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Links

Crossrefs

Cf. A002315, A005319, A182431, A001108, A049629, A084068.

Programs

  • Magma
    I:=[0,1,4,7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018
  • Maple
    A143608 := proc(n)
    option remember;
    if n <= 3 then
        op(n+1,[0,1,4,7]) ;
    else
        6*procname(n-2)-procname(n-4) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012
  • Mathematica
    a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2,1]]
CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
LinearRecurrence[{0, 6, 0, -1}, {0, 1, 4, 7}, 31] (* Jean-François Alcover, Sep 21 2017 *)
  • PARI
    a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,6,0]^n*[0;1;4;7])[1,1] \\ Charles R Greathouse IV, Jun 11 2015
  • PARI
    concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016

Formula

a(2*n) = (a(2*n - 1) + a(2*n + 1))/2.
a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.
a(2*n) = 4*A001109(n).
a(2*n + 1) = 4*A001109(n) + A001541(n).
From Colin Barker, Jun 29 2012: (Start)
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)
2*a(n) = A078057(n) - A123335(n-1). - R. J. Mathar, Jul 04 2012
a(2n) = A005319(n); a(2n+1) = A002315(n). - R. J. Mathar, Jul 17 2009
a(n)*a(n+1) + 1 = A001653(n+1). - Charlie Marion, Dec 11 2012
a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016
a(n) = A084068(n) - A079496(n). - César Aguilera, Feb 14 2023

A079496 a(0) = a(1) = 1; thereafter a(2*n+1) = 2*a(2*n) - a(2*n-1), a(2*n) = 4*a(2*n-1) - a(2*n-2).

Original entry on oeis.org

1, 1, 3, 5, 17, 29, 99, 169, 577, 985, 3363, 5741, 19601, 33461, 114243, 195025, 665857, 1136689, 3880899, 6625109, 22619537, 38613965, 131836323, 225058681, 768398401, 1311738121, 4478554083, 7645370045, 26102926097, 44560482149, 152139002499, 259717522849, 886731088897
Offset: 0

Views

Author

Benoit Cloitre, Jan 20 2003

Keywords

Comments

a(1)=1, a(n) is the smallest integer > a(n-1) such that sqrt(2)*a(n) is closer and > to an integer than sqrt(2)*a(n-1) (i.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(2)*a(n)) < frac(sqrt(2)*a(n-1))).
n such that floor(sqrt(2)*n^2) = n*floor(sqrt(2)*n).
The sequence 1,1,3,5,17,... has g.f. (1+x-3x^2-x^3)/(1-6x^2+x^4); a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k-floor((n+1)/2)); a(n) = -(sqrt(2)-1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4) - (sqrt(2)+1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4); a(2n) = A001541(n) = A001333(2n); a(2n+1) = A001653(n) = A000129(2n+1). - Paul Barry, Jan 22 2005
The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496. - Clark Kimberling, Aug 27 2008
From Richard Choulet, May 09 2010: (Start)
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.
The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).
The general form of a(n) is given by:
a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and
a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)). (End)
The integer square roots of floor(n^2/2 + 1) or (A007590 + 1). - Richard R. Forberg, Aug 01 2013

Examples

			1 + x + 3*x^2 + 5*x^3 + 17*x^4 + 29*x^5 + 99*x^6 + 169*x^7 + 577*x^8 + ...

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Links

Crossrefs

Cf. A010914, A058580.
Cf. A000129, A084068.

Programs

  • Maple
    H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -1):
a := n -> `if`(n < 3, [1, 1, 3][n+1], 2^(n - 1)*H(n, irem(n, 2), 1/2)):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Sep 03 2019
  • Mathematica
    a[1] = 1; a[2] = 3; a[3] = 5; a[n_] := a[n] = (a[n-1]*a[n-2] + 2) / a[n-3]; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Jul 17 2013, after Paul D. Hanna *)
  • PARI
    {a(n) = n = abs(n); 2^((4-n)\2) * real( (10 + 7 * quadgen(8)) / 2 * (2 + quadgen(8))^(n-3) ) }  /* Michael Somos, Sep 03 2013 */
  • PARI
    {a(n) = polcoeff( (1 + x - 3*x^2 - x^3) / (1 - 6*x^2 + x^4) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Sep 03 2013 */

Formula

a(2n+1) - a(2n) = a(2n) - a(2n-1) = A001542(n).
a(2n+1) = ceiling((2+sqrt(2))/4*(3+2*sqrt(2))^n), a(2n) = ceiling(1/2*(3+2*sqrt(2))^n).
G.f.: (1 + x - 3*x^2 - x^3)/(1 - 6*x^2 + x^4).
a(n)*a(n+3) - a(n+1)*a(n+2) = 2. - Paul D. Hanna, Feb 22 2003
a(n) = 6*a(n-2) - a(n-4). - R. J. Mathar, Apr 04 2008
a(-n) = a(n) = A010914(n-3)*2^floor((4 - n)/2). - Michael Somos, Sep 03 2013
a(n) = (sqrt(2)*sqrt(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n))/(2+sqrt(2)+(-1)^n*(-2+sqrt(2))). - Gerry Martens, Jun 06 2015
a(n) = 2^(n - 1)*H(n, n mod 2, 1/2) for n >= 3 where H(n, a, b) = hypergeom([a - n/2, b - n/2], [1 - n], -1). - Peter Luschny, Sep 03 2019
a(n) == Pell(n)^(-1) (mod Pell(n+1)) where Pell(n) = A000129(n), use the identity a(n)*Pell(n) - A084068(n-1)*Pell(n+1) = 1, taken modulo Pell(n+1). - Gary W. Adamson, Nov 21 2023
E.g.f.: cosh(x)*(cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2)). - Stefano Spezia, Apr 21 2025

Extensions

a(0)=1 added by Michael Somos, Sep 03 2013

A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

Original entry on oeis.org

0, 6, 228, 8658, 328776, 12484830, 474094764, 18003116202, 683644320912, 25960481078454, 985814636660340, 37434995712014466, 1421544022419889368, 53981237856243781518, 2049865494514843808316, 77840907553707820934490, 2955904621546382351702304
Offset: 0

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 10*y^2 = 1. The corresponding x values are in A078986. - Vincenzo Librandi, Aug 08 2010 [edited by Jon E. Schoenfield, May 04 2014]

Examples

			G.f. = 6*x + 228*x^2 + 8658*x^3 + 328776*x^4 + ... - _Michael Somos_, Feb 24 2023

Links

Crossrefs

Cf. A001078, A001109, A001353, A001653, A005668, A060645, A084068, A084069, A221874.
Cf. A078986. - Vincenzo Librandi, Apr 14 2010

Programs

  • GAP
    a:=[0,6];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2020
  • Magma
    I:=[0,6]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 12 2020
  • Maple
    seq( simplify(6*ChebyshevU(n-1, 19)), n=0..20); # G. C. Greubel, Jan 12 2020
  • Mathematica
    LinearRecurrence[{38,-1},{0,6},30] (* Harvey P. Dale, Nov 01 2011 *)
6*ChebyshevU[Range[20]-2, 19] (* G. C. Greubel, Jan 12 2020 *)
  • PARI
    u=0; v=6; for(n=2,20, w=38*v-u; u=v; v=w; print1(w,","))
  • PARI
    vector(21, n, 6*polchebyshev(n-2, 2, 19) ) \\ G. C. Greubel, Jan 12 2020
  • Sage
    [6*chebyshev_U(n-1, 19) for n in (0..20)] # G. C. Greubel, Jan 12 2020

Formula

Numbers k such that 10*k^2 = floor(k*sqrt(10)*ceiling(k*sqrt(10))).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 37*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 39*(a(n-1) - a(n-2)) + a(n-3). (End)
From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: 6*x/(1 - 38*x + x^2).
a(n) = 6*A078987(n-1). (End)
a(n) = 6*ChebyshevU(n-1, 19). - G. C. Greubel, Jan 12 2020
a(n) = A005668(2*n). - Michael Somos, Feb 24 2023

A143609 Numerators of the upper principal and intermediate convergents to 2^(1/2).

Original entry on oeis.org

2, 3, 10, 17, 58, 99, 338, 577, 1970, 3363, 11482, 19601, 66922, 114243, 390050, 665857, 2273378, 3880899, 13250218, 22619537, 77227930, 131836323, 450117362, 768398401, 2623476242, 4478554083, 15290740090, 26102926097, 89120964298, 152139002499
Offset: 1

Views

Author

Clark Kimberling, Aug 27 2008

Keywords

Comments

The upper principal and intermediate convergents to 2^(1/2), beginning with
2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence;
essentially, numerators=A143609 and denominators=A084068.

Examples

			2*x + 3*x^2 + 10*x^3 + 17*x^4 + 58*x^5 + 99*x^6 + 338*x^7 + 577*x^8 + ...

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Links

Crossrefs

Cf. A001541, A001653, A010914, A084068.

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x (2 + 3 x - 2 x^2 - x^3)/(1 - 6 x^2 + x^4), {x, 0, 30}], x] (* Michael De Vlieger, Mar 27 2016 *)
  • PARI
    {a(n) = if( n<1, 0, polcoeff( x * (2 + 3*x - 2*x^2 - x^3) / (1 - 6*x^2 + x^4) + x * O(x^n), n))} /* Michael Somos, Sep 03 2013 */
  • PARI
    x='x+O('x^99); Vec(x*(2+3*x-2*x^2-x^3)/(1-6*x^2+x^4)) \\ Altug Alkan, Mar 27 2016

Formula

a(n) = 6 * a(n-2) - a(n-4). a(2*n) = A001541(n) if n>0. a(2*n + 1) = 2 * A001653(n + 1).- Michael Somos, Sep 03 2013
G.f.: x * (2 + 3*x - 2*x^2 - x^3) / (1 - 6*x^2 + x^4). - Michael Somos, Sep 03 2013
a(n) = (2+sqrt(2)+(-1)^n*(-2+sqrt(2)))*((-1+sqrt(2))^n+(1+sqrt(2))^n)/(4*sqrt(2)). - Colin Barker, Mar 27 2016

A108412 Expansion of (1 + x + x^2)/(1 - 4x^2 + x^4).

Original entry on oeis.org

1, 1, 5, 4, 19, 15, 71, 56, 265, 209, 989, 780, 3691, 2911, 13775, 10864, 51409, 40545, 191861, 151316, 716035, 564719, 2672279, 2107560, 9973081, 7865521, 37220045, 29354524, 138907099, 109552575, 518408351, 408855776, 1934726305
Offset: 0

Views

Author

Ralf Stephan, Jun 05 2005

Keywords

Comments

This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 6 and Q = 1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence satisfies a linear recurrence of order four. - Peter Bala, Apr 18 2014
The sequence of convergents of the 2-periodic continued fraction [0; 1, -6, 1, -6, ...] = 1/(1 - 1/(6 - 1/(1 - 1/(6 - ...)))) = 3 - sqrt(3) begins [0/1, 1/1, 6/5, 5/4, 24/19, 19/15, 90/71,...]. The present sequence is the sequence of denominators; the sequence of numerators of the continued fraction convergents [1, 6, 5, 24, 19, 90,...] is also a strong divisibility sequence. Cf. A005013 and A203976. - Peter Bala, May 19 2014
From Peter Bala, Mar 25 2018: (Start)
The following remarks assume an offset of 1.
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + (1/2)*y^2) + y*sqrt(1 + (1/2)*x^2). The operation o is commutative and associative with identity 0. We have a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and sqrt(6)*a(2*n) = (1 o 1 o ... o 1) (2*n terms). Cf. A005013 and A084068. For example, 1 o 1 = sqrt(6) and 1 o 1 o 1 = sqrt(6) o 1 = 5 = a(3).
From the obvious identity ( 1 o 1 o ... o 1 (2*n terms) ) o ( 1 o 1 o ... o 1 (2*m terms) ) = 1 o 1 o ... o 1 (2*n + 2*m terms) we find the relation a(2*n+2*m) = a(2*n)*sqrt(1 + 3*a(2*m)^2) + a(2*m)*sqrt(1 + 3*a(2*n)^2).
Similarly, from a(2*n+1) o a(2*m+1) = sqrt(6)*a(2*n+2*m+2) we find sqrt(6)*a(2*n+2*m+2) = a(2*n+1)*sqrt(1 + (1/2)*a(2*m+1)^2) + a(2*m+1)*sqrt(1 + (1/2)*a(2*n+1)^2). (End)

Examples

			G.f. = 1 + x + 5*x^2 + 4*x^3 + 19*x^4 + 15*x^5 + 71*x^6 + 56*x^7 + ...

Links

Crossrefs

Cf. A001353, A001834.
Cf. A026741, A005013, A084068.

Programs

  • Maple
    a := proc (n) if `mod`(n, 2) = 1 then 1/sqrt(2)*( ((sqrt(6) + sqrt(2))/2 )^n - ( (sqrt(6) - sqrt(2))/2 )^n) else 1/sqrt(12)*( ((sqrt(6) + sqrt(2))/2 )^n - ( (sqrt(6) - sqrt(2))/2 )^n) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018
  • Mathematica
    CoefficientList[Series[(1+x+x^2)/(1-4x^2+x^4),{x,0,40}],x] (* or *) LinearRecurrence[{0,4,0,-1},{1,1,5,4},40] (* Harvey P. Dale, Nov 15 2012 *)
  • PARI
    {a(n) = my( w = quadgen(24)); simplify( polchebyshev( n, 2, w/2) / if( n%2, w, 1))}; /* Michael Somos, Feb 10 2015 */

Formula

a(0)=a(1)=1, a(2)=5, a(n)a(n+3) - a(n+1)a(n+2) = -1.
a(0)=1, a(1)=1, a(2)=5, a(3)=4, a(n) = 4*a(n-2)-a(n-4). - Harvey P. Dale, Nov 15 2012
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = (1/2)*(sqrt(6) + sqrt(2)) (A188887) and beta = (1/2)*(sqrt(6) - sqrt(2)) (A101263). Equivalently, a(n) = U(n-1,sqrt(6)/2) for n odd and a(n) = (1/sqrt(6))*U(n-1,sqrt(6)/2) for n even, where U(n,x) is the Chebyshev polynomial of the second kind. - Peter Bala, Apr 18 2014
a(2*n) = A001834(n). a(2*n + 1) = A001353(n+1). - Michael Somos, Feb 10 2015
a(n) = -a(-2-n) for all n in Z. - Michael Somos, Feb 10 2015
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