cp's OEIS Frontend

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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A143908 Eigentriangle by rows, n terms of (3 * A026150) followed by A026150(n).

Original entry on oeis.org

1, 3, 1, 3, 3, 4, 3, 3, 12, 10, 3, 3, 12, 30, 28, 3, 3, 12, 30, 84, 76, 3, 3, 12, 30, 84, 228, 208, 3, 3, 12, 30, 84, 228, 624, 568, 3, 3, 12, 30, 84, 228, 624, 1704, 1552
Offset: 0

Views

Author

Gary W. Adamson, Sep 06 2008

Keywords

Comments

Right border = A026150: (1, 1, 4, 10, 28, 76, 208, 568,...).
Row sums = (1, 4, 10, 28, 76,...) = INVERT transform of (1, 3, 3, 3,...).
Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
1;
3, 1;
3, 3, 4;
3, 3, 12, 10;
3, 3, 12, 30, 28;
3, 3, 12, 30, 84, 76;
3, 3, 12, 30, 84, 228, 208;
...
		

Crossrefs

Formula

Eigentriangle by rows, n terms of (3 * A026150) followed by A026150(n).

A322941 Coefficients of orthogonal polynomials p(n, x) where p(n, 0) is A026150 with 1 prepended. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 4, 7, 4, 1, 10, 22, 17, 6, 1, 28, 68, 64, 31, 8, 1, 76, 208, 230, 138, 49, 10, 1, 208, 628, 796, 568, 252, 71, 12, 1, 568, 1880, 2680, 2208, 1170, 414, 97, 14, 1, 1552, 5584, 8832, 8232, 5052, 2140, 632, 127, 16, 1, 4240, 16480, 28608, 29712, 20676, 10160, 3598, 914, 161, 18, 1
Offset: 0

Views

Author

Peter Luschny, Jan 06 2019

Keywords

Examples

			The first few polynomials are:
[0] p(0, x) = 1;
[1] p(1, x) = x + 1;
[2] p(2, x) = x^2 +  2*x + 1;
[3] p(3, x) = x^3 +  4*x^2 +  7*x + 4;
[4] p(4, x) = x^4 +  6*x^3 + 17*x^2 +  22*x + 10;
[5] p(5, x) = x^5 +  8*x^4 + 31*x^3 +  64*x^2 +  68*x + 28;
[6] p(6, x) = x^6 + 10*x^5 + 49*x^4 + 138*x^3 + 230*x^2 + 208*x + 76;
The triangle starts:
[0]    1;
[1]    1,    1;
[2]    1,    2,    1;
[3]    4,    7,    4,    1;
[4]   10,   22,   17,    6,    1;
[5]   28,   68,   64,   31,    8,    1;
[6]   76,  208,  230,  138,   49,   10,   1;
[7]  208,  628,  796,  568,  252,   71,  12,   1;
[8]  568, 1880, 2680, 2208, 1170,  414,  97,  14,  1;
[9] 1552, 5584, 8832, 8232, 5052, 2140, 632, 127, 16, 1;
		

Crossrefs

Row sums are A322940, alternating row sums are A000007.

Programs

  • Maple
    p := proc(n) option remember;
    `if`(n < 3, [1, x+1, x^2 + 2*x + 1][n+1], (x+2)*p(n-1) + 2*p(n-2));
    sort(expand(%)) end: seq(print(p(n)), n=0..11); # Computes the polynomials.
    seq(seq(coeff(p(n), x, k), k=0..n), n=0..10);

Formula

p(n, x) = (x+2)*p(n-1, x) + 2*p(n-2, x) for n >= 3.
T(n, k) = [x^k] p(n, x).

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

Views

Author

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.
(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.

Crossrefs

Partial sums of A001333.
2nd row of A172236.
a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
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. A048739.
Cf. A073133.
Cf. A041085.
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

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199
Offset: 0

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Comments

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
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) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
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 numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 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 b(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.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
(0 1 0 1)
(1 0 2 0)
(0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....2
.2..|..1.....2.....4
.3..|..1.....4.....4.....8
.4..|..1.....4....12.....8....16
.5..|..1.....6....12....32....16....32
.6..|..1.....6....24....32....80....32....64
.7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
_
|| _
|||_|||
|_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Examples

			Convergents are 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.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...
		

References

  • M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • 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, April 2004, see p. 16.
  • J. Devillet, J.-L. Marichal, and B. Teheux, Classifications of quasitrivial semigroups, Semigroup Forum, 100 (2020), 743-764.
  • Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111.
  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • 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.
  • A. F. Horadam, R. P. Loh, and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Volume 1 (1986), p. 203, Example 4.1.2.
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
  • R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.

Crossrefs

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

Programs

  • Haskell
    a001333 n = a001333_list !! n
    a001333_list = 1 : 1 : zipWith (+)
                           a001333_list (map (* 2) $ tail a001333_list)
    -- Reinhard Zumkeller, Jul 08 2012
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018
    
  • Maple
    A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
    Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
    with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
    a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
    seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
    A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
    A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022
  • Mathematica
    Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
    Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
    a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
    Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
    a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
    LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
    Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
    Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
    CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
    Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)
  • PARI
    {a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */
    
  • PARI
    {a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */
    
  • PARI
    a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021
    
  • PARI
    { for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009
    
  • Python
    from functools import cache
    @cache
    def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
    print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    from sage.combinat.sloane_functions import recur_gen2
    it = recur_gen2(1,1,2,1)
    [next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008
    
  • Sage
    [lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A002605 a(n) = 2*(a(n-1) + a(n-2)), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 6, 16, 44, 120, 328, 896, 2448, 6688, 18272, 49920, 136384, 372608, 1017984, 2781184, 7598336, 20759040, 56714752, 154947584, 423324672, 1156544512, 3159738368, 8632565760, 23584608256, 64434348032, 176037912576, 480944521216, 1313964867584
Offset: 0

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Author

Keywords

Comments

Individually, both this sequence and A028859 are convergents to 1 + sqrt(3). Mutually, both sequences are convergents to 2 + sqrt(3) and 1 + sqrt(3)/2. - Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Nov 04 2001
The number of (s(0), s(1), ..., s(n+1)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1, 2, ..., n + 1, s(0) = 2, s(n+1) = 3. - Herbert Kociemba, Jun 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(4). - Cino Hilliard, Sep 25 2005
The Hankel transform of this sequence is [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Nov 21 2007
[1, 3; 1, 1]^n *[1, 0] = [A026150(n), a(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3). - Gary W. Adamson, Mar 21 2008
a(n+1) is the number of ways to tile a board of length n using red and blue tiles of length one and two. - Geoffrey Critzer, Feb 07 2009
Starting with offset 1 = INVERT transform of the Jacobsthal sequence, A001045: (1, 1, 3, 5, 11, 21, ...). - Gary W. Adamson, May 12 2009
Starting with "1" = INVERTi transform of A007482: (1, 3, 11, 39, 139, ...). - Gary W. Adamson, Aug 06 2010
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A026150, without the first leading 1. - Johannes W. Meijer, Aug 15 2010
The sequence 0, 1, -2, 6, -16, 44, -120, 328, -896, ... (with alternating signs) is the Lucas U(-2,-2)-sequence. - R. J. Mathar, Jan 08 2013
a(n+1) counts n-walks (closed) on the graph G(1-vertex;1-loop,1-loop,2-loop,2-loop). - David Neil McGrath, Dec 11 2014
Number of binary strings of length 2*n - 2 in the regular language (00+11+0101+1010)*. - Jeffrey Shallit, Dec 14 2015
For n >= 1, a(n) equals the number of words of length n - 1 over {0, 1, 2, 3} in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Dec 17 2015
a(n+1) is the number of compositions of n into parts 1 and 2, both of two kinds. - Gregory L. Simay, Sep 20 2017
Number of associative, quasitrivial, and order-preserving binary operations on the n-element set {1, ..., n} that have neutral elements. - J. Devillet, Sep 28 2017
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
Starting with 1, 2, 6, 16, ..., number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the third and fourth elements. - Sergey Kitaev, Dec 09 2020
a(n) is the number of tilings of a 2 X n board missing one corner cell, with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares). Compare to A127864. - Greg Dresden and Yilin Zhu, Jul 17 2025

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, p. 16.

Crossrefs

First differences are given by A026150.
a(n) = A073387(n, 0), n>=0 (first column of triangle).
Equals (1/3) A083337. First differences of A077846. Pairwise sums of A028860 and abs(A077917).
a(n) = A028860(n)/2 apart from the initial terms.
Row sums of A081577 and row sums of triangle A156710.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A175289 (Pisano periods).
Cf. A002530.
Cf. A127864.

Programs

  • Haskell
    a002605 n = a002605_list !! n
    a002605_list =
       0 : 1 : map (* 2) (zipWith (+) a002605_list (tail a002605_list))
    -- Reinhard Zumkeller, Oct 15 2011
    
  • Magma
    [Floor(((1 + Sqrt(3))^n - (1 - Sqrt(3))^n)/(2*Sqrt(3))): n in [0..30]]; // Vincenzo Librandi, Aug 18 2011
    
  • Magma
    [n le 2 select n-1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+2*a[n-2]od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 15 2008
    a := n -> `if`(n<3, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -2));
    seq(simplify(a(n)), n=0..29); # Peter Luschny, Dec 16 2015
  • Mathematica
    Expand[Table[((1 + Sqrt[3])^n - (1 - Sqrt[3])^n)/(2Sqrt[3]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    a[n_]:=(MatrixPower[{{1,3},{1,1}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{2, 2}, {0, 1}, 30] (* Robert G. Wilson v, Apr 13 2013 *)
    Round@Table[Fibonacci[n, Sqrt[2]] 2^((n - 1)/2), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
    nxt[{a_,b_}]:={b,2(a+b)}; NestList[nxt,{0,1},30][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)
  • PARI
    Vec(x/(1-2*x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    A002605(n)=([2,2;1,0]^n)[2,1] \\ M. F. Hasler, Aug 06 2018
    
  • Sage
    [lucas_number1(n,2,-2) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009
    
  • Sage
    a = BinaryRecurrenceSequence(2,2)
    print([a(n) for n in (0..29)])  # Peter Luschny, Aug 29 2016
    

Formula

a(n) = (-I*sqrt(2))^(n-1)*U(n-1, I/sqrt(2)) where U(n, x) is the Chebyshev U-polynomial. - Wolfdieter Lang
G.f.: x/(1 - 2*x - 2*x^2).
From Paul Barry, Sep 17 2003: (Start)
E.g.f.: x*exp(x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = (1 + sqrt(3))^(n-1)*(1/2 + sqrt(3)/6) + (1 - sqrt(3))^(n-1)*(1/2 - sqrt(3)/6), for n>0.
Binomial transform of 1, 1, 3, 3, 9, 9, ... Binomial transform is A079935. (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(n - k, k)*2^(n - k). - Paul Barry, Jul 13 2004
a(n) = A080040(n) - A028860(n+1). - Creighton Dement, Jan 19 2005
a(n) = Sum_{k=0..n} A112899(n,k). - Philippe Deléham, Nov 21 2007
a(n) = Sum_{k=0..n} A063967(n,k). - Philippe Deléham, Nov 03 2006
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*sqrt(3)).
a(n) = Sum_{k=0..n} binomial(n, 2*k + 1) * 3^k.
Binomial transform of expansion of sinh(sqrt(3)x)/sqrt(3) (0, 1, 0, 3, 0, 9, ...). E.g.f.: exp(x)*sinh(sqrt(3)*x)/sqrt(3). - Paul Barry, May 09 2003
a(n) = (1/3)*Sum_{k=1..5} sin(Pi*k/2)*sin(2*Pi*k/3)*(1 + 2*cos(Pi*k/6))^n, n >= 1. - Herbert Kociemba, Jun 02 2004
a(n+1) = ((3 + sqrt(3))*(1 + sqrt(3))^n + (3 - sqrt(3))*(1 - sqrt(3))^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Jun 29 2009
Antidiagonals sums of A081577. - J. M. Bergot, Dec 15 2012
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k + 2 + 2*x)/(x*(4*k + 4 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = 2^(n - 1)*hypergeom([1 - n/2, (1 - n)/2], [1 - n], -2) for n >= 3. - Peter Luschny, Dec 16 2015
Sum_{k=0..n} a(k)*2^(n-k) = a(n+2)/2 - 2^n. - Greg Dresden, Feb 11 2022
a(n) = 2^floor(n/2) * A002530(n). - Gregory L. Simay, Sep 22 2022
From Peter Bala, May 08 2024: (Start)
G.f.: x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (k + 2*x + 1)/(1 + k*x) )
Also x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} (2*x)^n *( x*Product_{k = 1..n} (m*k + 2 - m + x)/(1 + 2*m*k*x) ) for arbitrary m (both series are telescoping). (End)
a(n) = A127864(n-1) + A127864(n-2). - Greg Dresden and Yilin Zhu, Jul 17 2025

Extensions

Edited by N. J. A. Sloane, Apr 15 2009

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647
Offset: 0

Views

Author

Keywords

Comments

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths: 1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
1, 1, 0, 0, 0, 0, ...
2, 0, 3, 0, 0, 0, ...
2, 0, 0, 3, 0, 0, ...
2, 0, 0, 0, 3, 0, ...
2, 0, 0, 0, 0, 3, ...
...
The triangle generated from M is:
1,
1, 1,
3, 1, 3,
11, 3, 3, 9,
41, 11, 9, 9, 27,
...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). - Gary W. Adamson, Jul 25 2016
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

Examples

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...
		

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
  • 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.

Crossrefs

Programs

  • Haskell
    a001075 n = a001075_list !! n
    a001075_list =
       1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
    -- Reinhard Zumkeller, Aug 11 2011
    
  • Magma
    I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
  • Maple
    A001075 := proc(n)
        orthopoly[T](n,2) ;
    end proc:
    seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018
  • Mathematica
    Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
    CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
    LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
    Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
    ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
    a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)
  • PARI
    {a(n) = subst(poltchebi(abs(n)), x, 2)};
    
  • PARI
    {a(n) = real((2 + quadgen(12))^abs(n))};
    
  • PARI
    {a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};
    
  • PARI
    a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016
    
  • PARI
    my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017
    
  • SageMath
    [lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009
    
  • SageMath
    def a(n):
        Q = QuadraticField(3, 't')
        u = Q.units()[0]
        return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014
    

Formula

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

Extensions

More terms from James Sellers, Jul 10 2000
Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082
Offset: 0

Views

Author

Keywords

Comments

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths: 1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.

Programs

  • Magma
    [Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011
    
  • Mathematica
    Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)
  • Maxima
    a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */
  • PARI
    a(n)=round(3^n/4)
    
  • Python
    for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018
    
  • Sage
    [round(3^n/4) for n in range(0,27)]
    

Formula

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction). (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) = x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

Extensions

More terms from Emeric Deutsch, Apr 01 2004
Edited by Ralf Stephan, Aug 30 2004

A098158 Triangle T(n,k) with diagonals T(n,n-k) = binomial(n, 2*k).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 0, 1, 6, 1, 0, 0, 0, 5, 10, 1, 0, 0, 0, 1, 15, 15, 1, 0, 0, 0, 0, 7, 35, 21, 1, 0, 0, 0, 0, 1, 28, 70, 28, 1, 0, 0, 0, 0, 0, 9, 84, 126, 36, 1, 0, 0, 0, 0, 0, 1, 45, 210, 210, 45, 1, 0, 0, 0, 0, 0, 0, 11, 165, 462, 330, 55, 1, 0, 0, 0, 0, 0, 0, 1, 66, 495, 924
Offset: 0

Views

Author

Paul Barry, Aug 29 2004

Keywords

Comments

Row sums are A011782. Inverse is A065547.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jul 29 2006
Sum of entries in column k is A001519(k+1) (the odd-indexed Fibonacci numbers). - Philippe Deléham, Dec 02 2008
Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k left-to-right minima. A left-to-right minimum in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j < i. - Tian Han, Nov 16 2023

Examples

			Rows begin
  1;
  0, 1;
  0, 1, 1;
  0, 0, 3, 1;
  0, 0, 1, 6, 1;
		

Crossrefs

Cf. A119900. - Philippe Deléham, Dec 02 2008

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n, 2*(n-k)) ))); # G. C. Greubel, Aug 01 2019
  • Magma
    [Binomial(n, 2*(n-k)): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
    
  • Mathematica
    Table[Binomial[n, 2*(n-k)], {n,0,12}, {k,0,n}]//Flatten (* Michael De Vlieger, Oct 12 2016 *)
  • PARI
    {T(n,k)=polcoeff(polcoeff((1-x*y)/((1-x*y)^2-x^2*y)+x*O(x^n), n, x) + y*O(y^k),k,y)} (Hanna)
    
  • PARI
    T(n,k) = binomial(n, 2*(n-k));
    for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019
    
  • Sage
    [[binomial(n, 2*(n-k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

T(n,k) = binomial(n,2*(n-k)).
From Tom Copeland, Oct 10 2016: (Start)
E.g.f.: exp(t*x) * cosh(t*sqrt(x)).
O.g.f.: (1/2) * ( 1 / (1 - (1 + sqrt(1/x))*x*t) + 1 / (1 - (1 - sqrt(1/x))*x*t) ).
Row polynomial: x^n * ((1 + sqrt(1/x))^n + (1 - sqrt(1/x))^n) / 2. (End)
Column k is generated by the polynomial Sum_{j=0..floor(k/2)} C(k, 2j) * x^(k-j). - Paul Barry, Jan 22 2005
G.f.: (1-x*y)/((1-x*y)^2 - x^2*y). - Paul D. Hanna, Feb 25 2005
Sum_{k=0..n} x^k*T(n,k)= A009116(n), A000007(n), A011782(n), A006012(n), A083881(n), A081335(n), A090139(n), A145301(n), A145302(n), A145303(n), A143079(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Dec 04 2006, Oct 15 2008, Oct 19 2008
T(n,k) = T(n-1,k-1) + Sum_{i=0..k-1} T(n-2-i,k-1-i); T(0,0)=1; T(n,k)=0 if n < 0 or k < 0 or n < k. E.g.: T(8,5) = T(7,4) + T(6,4) + T(5,3) + T(4,2) + T(3,1) + T(2,0) = 7+15+5+1+0+0 = 28. - Philippe Deléham, Dec 04 2006
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A011782(n), A001333(n), A026150(n), A046717(n), A084057(n), A002533(n), A083098(n), A084058(n), A003665(n), A002535(n), A133294(n), A090042(n), A125816(n), A133343(n), A133345(n), A120612(n), A133356(n), A125818(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 respectively. - Philippe Deléham, Dec 24 2007
Sum_{k=0..n} T(n,k)*(-x)^(n-k) = A000012(n), A146559(n), A087455(n), A138230(n), A006495(n), A138229(n) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Nov 14 2008
T(n,k) = A085478(k,n-k). - Philippe Deléham, Dec 02 2008
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = T(1,1) = 1, T(1,0) = 0 and T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 15 2012

A001834 a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085
Offset: 0

Views

Author

Keywords

Comments

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p = q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p = {x + (-1)^n}/3, r = p-(-1)^n, s = y for n > 1. - Lekraj Beedassy, Jul 19 2002
a(n)= A002531(1+2*n). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
361 written in base A001835(n+1) - 1 is the square of a(n). E.g., a(12) = 2672279, A001835(13) - 1 = 1542840. We have 361_(1542840) = 3*1542840 + 6*1542840 + 1 = 2672279^2. - Richard Choulet, Oct 04 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators=A001834, denominators=A001835. - Clark Kimberling, Aug 27 2008
General recurrence is a(n) = (a(1) - 1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} 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, primes in it A121534. a(1) = 5 gives A001834, primes in it A086386. a(1) = 6 gives A030221, primes in it A299109. a(1) = 7 gives A002315, primes in it A088165. a(1) = 8 gives A033890, primes in it not in OEIS (do there exist any?). a(1) = 9 gives A057080, primes in {71, 34649, 16908641, ...}. a(1) = 10 gives A057081, primes in it {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030192. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2*n) X (2*n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
x-values in the solution to 3x^2 + 6 = y^2 (see A082841 for the y-values). - Sture Sjöstedt, Nov 25 2011
Pisano period lengths: 1, 1, 2, 4, 3, 2, 8, 4, 6, 3, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
The aerated sequence (b(n))A100047%20for%20a%20connection%20with%20Chebyshev%20polynomials.%20-%20_Peter%20Bala">{n>=1} = [1, 0, 5, 0, 19, 0, 71, 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 = -2, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - _Peter Bala, Mar 22 2015
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001835 and with any member of A001075. - René Gy, Feb 26 2018
From Wolfdieter Lang, Oct 15 2020: (Start)
((-1)^n)*a(n) = X(n) = (-1)^n*(S(n, 4) + S(n-1, 4)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 4*X*Y = +6, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form of discriminant 12, representing 6, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

Examples

			G.f. = 1 + 5*x + 19*x^2 + 71*x^3 + 265*x^4 + 989*x^5 + 3691*x^6 + ...
		

References

  • Bastida, Julio R. 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)
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • 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.

Crossrefs

A bisection of sequence A002531.
Cf. A001352, A001835, A086386 (prime members).
Cf. A026150.
a(n)^2+1 = A094347(n+1).

Programs

  • Haskell
    a001834 n = a001834_list !! (n-1)
    a001834_list = 1 : 5 : zipWith (-) (map (* 4) $ tail a001834_list) a001834_list
    -- Reinhard Zumkeller, Jan 23 2012
    
  • Magma
    I:=[1,5]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Maple
    f:=n->((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1); # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    a[0] = 1; a[1] = 5; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 25}] (* Robert G. Wilson v, Apr 24 2004 *)
    Table[Expand[((1+Sqrt[3])^(2*n+1)+(1+Sqrt[3])^(2*n+1))/2^(n+1)],{n, 0, 20}] (* Anton Vrba, Feb 14 2007 *)
    LinearRecurrence[{4, -1}, {1, 5}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Numerator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
    ] (* Complement of A002531 *)
    a[3, 20] (* Gerry Martens, Jun 07 2015 *)
    Round@Table[LucasL[2n+1, Sqrt[2]]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 + quadgen(12)) )}; /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( polchebyshev(n-1, 2) + polchebyshev(n, 2), x, 2)}; /* Michael Somos, Sep 19 2008 */
    
  • SageMath
    [(lucas_number2(n,4,1)-lucas_number2(n-1,4,1))/2 for n in range(1, 27)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - N. J. A. Sloane, Nov 10 2009
a(n) = (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n). - Dean Hickerson, Dec 01 2002
From Mario Catalani, Apr 11 2003: (Start)
With a = 2 + sqrt(3), b = 2 - sqrt(3): a(n) = (1/sqrt(2))(a^(n + 1/2) - b^(n + 1/2)).
a(n) - a(n-1) = A003500(n).
a(n) = sqrt(1 + 12*A061278(n) + 12*A061278(n)^2). (End)
a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - Anton Vrba, Feb 14 2007
G.f.: (1 + x)/((1 - 4*x + x^2)). Simon Plouffe in his 1992 dissertation.
a(n) = S(2*n, sqrt(6)) = S(n, 4) + S(n-1, 4); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 4) = A001353(n).
For all members x of the sequence, 3*x^2 + 6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A001571(n) + 1. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n - i)*binomial(2*n - i, i); then (-1)^n*q(n, -6) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n + 1, 2*k)*3^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = floor(sqrt(3)*A001835(n+1)). - Philippe Deléham, Mar 03 2004
a(n+1) - 2*a(n) = 3*A001835(n+1). Using the known relation A001835(n+1) = sqrt((a(n)^2 + 2)/3) it follows that a(n+1) - 2*a(n) = sqrt(3*(a(n)^2 + 2)). Therefore a(n+1)^2 + a(n)^2 - 4*a(n+1)*a(n) - 6 = 0. - Creighton Dement, Apr 18 2005
a(n) = L(n,-4)*(-1)^n, where L is defined as in A108299; see also A001835 for L(n,+4). - Reinhard Zumkeller, Jun 01 2005
a(n) = Jacobi_P(n, 1/2, -1/2, 2)/Jacobi_P(n, -1/2, 1/2, 1). - Paul Barry, Feb 03 2006
Equals binomial transform of A026150 starting (1, 4, 10, 28, 76, ...) and double binomial transform of (1, 3, 3, 9, 9, 27, 27, 81, 81, ...). - Gary W. Adamson, Nov 30 2007
Sequence satisfies 6 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(-1-n) = -a(n). - Michael Somos, Sep 19 2008
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(5*A125905(n) + A125905(n+1)).
E.g.f.: exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). (End)
a(n) = A061278(n+1) - A061278(n-1) for n>=2. - John P. McSorley, Jun 20 2020
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 2), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n) - 2*a(n-1) = 3 * A001835(n) for n >= 1.
For arbitrary x, a(n+x)^2 - 4*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 6 with a(n) := (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(6) * A001353(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6*sqrt(6) + 12) * A001353(n+1).
a(n+3/4) - a(n+1/4) = sqrt(2*sqrt(6) - 4) * A001075(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001352(n) + 1/A001352(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3*(1 - 2/A102206(k))). (End)

A046717 a(n) = 2*a(n-1) + 3*a(n-2), a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165, 3812798742493, 11438396227481
Offset: 0

Views

Author

Gervais Deroo and M. Deroo

Keywords

Comments

Form the digraph with matrix A = [0,1,1,1; 1,0,1,1; 1,1,0,1; 1,0,1,1]. Then the sequence 0,1,1,5,... or (3^(n-1)-(-1)^n)/2+0^n/3 with g.f. x(1-x)/(1-2x-3x^2) corresponds to the (1,2) term of A^n. - Paul Barry, Oct 02 2004
3*a(n+1) + a(n) = 4*A060925(n); a(n+1) = A015518(n) + A060925(n); a(n+1) - 6*A015518(n) = (-1)^n. - Creighton Dement, Nov 15 2004
The sequence corresponds to the (1,1) term of the matrix [1,2;2,1]^n. - Simone Severini, Dec 04 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
a(n)^2 + (2*A015518(n))^2 = a(2n). E.g., a(3) = 13, 2*A015518(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
Equals INVERTi transform of A104934: (1, 2, 8, 28, 100, 356, 1268, ...). - Gary W. Adamson, Jul 21 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square just one A[5] vector, with decimal value 341, leads to this sequence (without the first leading 1). For the corner squares this vector leads to the companion sequence A015518 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 2, 1, 4, 2, 6, 4, 2, 4, 10, 2, 6, 6, 4, 8, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
a(n) is the number of words of length n over a ternary alphabet whose position in the lexicographic order is a multiple of two. - Alois P. Heinz, Apr 13 2022
a(n) is the sum, for k=0..3, of the number of walks of length n between two vertices at distance k of the cube graph. - Miquel A. Fiol, Mar 09 2024

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

The first difference sequence of A015518.
Row sums of triangle A080928.
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. A015518.
Cf. A104934. - Gary W. Adamson, Jul 21 2010

Programs

  • Magma
    [n le 2 select 1 else 2*Self(n-1)+3*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Jul 21 2013
    
  • Magma
    [(3^n + (-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+3*a[n-2] od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 14 2008
    seq(denom(((-2)^(2*n)+6^(2*n))/((-2)^n+6^n)),n=0..26)
  • Mathematica
    Table[(3^n + (-1)^n)/2, {n, 0, 30}] (* Artur Jasinski, Dec 10 2006 *)
    CoefficientList[ Series[(1 - x)/(1 - 2x - 3x^2), {x, 0, 30}], x]  (* Robert G. Wilson v, Apr 04 2011 *)
    Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, Apr 04 2011 *)
  • PARI
    {a(n) = (3^n+(-1)^n)/2};
    for(n=0,30, print1(a(n), ", ")) /* modified by G. C. Greubel, Jan 07 2018 */
    
  • PARI
    x='x+O('x^30); Vec((1-x)/((1+x)*(1-3*x))) \\ G. C. Greubel, Jan 07 2018
    
  • Sage
    [lucas_number2(n,2,-3)/2 for n in range(0, 27)] # Zerinvary Lajos, Apr 30 2009
    

Formula

G.f.: (1-x)/((1+x)*(1-3*x)).
a(n) = (3^n + (-1)^n)/2.
a(n) = Sum_{k=0..n} binomial(n, 2k)2^(2k). - Paul Barry, Feb 26 2003
Binomial transform of A000302 (powers of 4) with interpolated zeros. Inverse binomial transform of A081294. - Paul Barry, Mar 17 2003
E.g.f.: exp(x)cosh(2x). - Paul Barry, Mar 17 2003
a(n) = ceiling(3^n/4) + floor(3^n/4) = ceiling(3^n/4)^2 - floor(3^n/4)^2. - Paul Barry, Jan 17 2005
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,j)C(n-j,k)*(1+(-1)^(j-k))/2. - Paul Barry, May 21 2006
a(n) = Sum_{k=0..n} A098158(n,k)*4^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = (3^n + (-1)^n)/2. - M. F. Hasler, Mar 20 2008
a(n) = 2 A015518(n) + (-1)^n; for n > 0, a(n) = A080925(n). - M. F. Hasler, Mar 20 2008
((1 + sqrt4)^n + (1 - sqrt4)^n)/2. The offset is 0. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Nov 22 2008
If p[1]=1 and p[i]=4 (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: G(0)/2, where G(k) = 1 + (-1)^k/(3^k - 3*9^k*x/(3*3^k*x + (-1)^k/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013

Extensions

Description corrected by and more terms from Michael Somos
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