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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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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Author

N. J. A. Sloane and R. K. Guy

Keywords

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.
Cf. A000129, A142238-A142239, A153313-A153318.
(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.

Links

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. A048211, A153588, A174283, A174284, A174285, A174286, A176499, A176500, A176501, A176502.
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

A055099 Expansion of g.f.: (1 + x)/(1 - 3*x - 2*x^2).

Original entry on oeis.org

1, 4, 14, 50, 178, 634, 2258, 8042, 28642, 102010, 363314, 1293962, 4608514, 16413466, 58457426, 208199210, 741512482, 2640935866, 9405832562, 33499369418, 119309773378, 424928058970, 1513403723666, 5390067288938, 19197009314146, 68371162520314, 243507506189234
Offset: 0

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Author

Wolfdieter Lang, Apr 26 2000

Keywords

Comments

Row sums of triangle A054458.
a(n) = term (1,1) in M^n, M = the 3 X 3 matrix [1,1,1; 1,1,1; 2,2,1]. - Gary W. Adamson, Mar 12 2009
Equals the INVERT transform of A001333: (1, 3, 7, 17, 41, 99, ...). - Gary W. Adamson, Aug 14 2010
a(n) is the number of one sided n-step walks taking steps from {(0,1), (-1,0), (1,0), (1,1)}. - Shanzhen Gao, May 13 2011
Number of quaternary words of length n on {0,1,2,3} containing no subwords 03 or 30. - Philippe Deléham, Apr 27 2012
Pisano period lengths: 1, 1, 4, 1, 24, 4, 48, 1, 12, 24, 30, 4, 12, 48, 24, 2, 272, 12, 18, 24, ... - R. J. Mathar, Aug 10 2012
a(n) = A007481(2*n+1) - A007481(2*n) = A007481(2*(n+1)) - A007481(2*n+1). - Reinhard Zumkeller, Oct 25 2015
Number of length-n words on a,b,c,d avoiding aa and ab. For n >= 1, the number of such words ending with a or the number of those ending with b is A007482(n-1), and the number of those ending with c or the number of those ending with d is a(n-1). - Jianing Song, Jun 01 2022

Examples

			a(3) = 50 because among the 4^3 = 64 quaternary words of length 3 only 14 namely 003, 030, 031, 032, 033, 103, 130, 203, 230, 300, 301, 302, 303, 330 contain the subwords 03 or 30. - _Philippe Deléham_, Apr 27 2012

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (Problem 2.4.6).

Links

Crossrefs

Column k=2 of A255256.
Cf. A001333, A002203, A007481, A007482, A054458, A104934, A202396.

Programs

  • Haskell
    a055099 n = a007481 (2 * n + 1) - a007481 (2 * n)
-- Reinhard Zumkeller, Oct 25 2015
  • Magma
    I:=[1,4]; [n le 2 select I[n] else 3*Self(n-1) + 2*Self(n-2): n in [1..41]]; // G. C. Greubel, Jun 27 2021
  • Maple
    a := proc(n) option remember; `if`(n < 2, [1, 4][n+1], (3*a(n-1) + 2*a(n-2))) end:
seq(a(n), n=0..23); # Peter Luschny, Jan 06 2019
  • Mathematica
    max = 24; cv = ContinuedFraction[ Sqrt[2], max] // Convergents // Numerator; Series[ 1/(1 - cv.x^Range[max]), {x, 0, max}] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Jun 21 2013, after Gary W. Adamson *)
LinearRecurrence[{3, 2}, {1, 4}, 24] (* Jean-François Alcover, Sep 23 2017 *)
  • Sage
    [(i*sqrt(2))^(n-1)*( i*sqrt(2)*chebyshev_U(n, -3*i/(2*sqrt(2))) + chebyshev_U(n-1, -3*i/(2*sqrt(2))) ) for n in (0..40)] # G. C. Greubel, Jun 27 2021

Formula

a(n) = a*c^n - b*d^n, a := (5 + sqrt(17))/(2*sqrt(17)), b := (5 - sqrt(17))/(2*sqrt(17)), c := (3 + sqrt(17))/2, d := (3 - sqrt(17))/2.
a(n) = Sum_{m=0..n} A054458(n, m).
a(n) = F32(n) + F32(n-1) with F32(n) = A007482(n), n >= 1, a(0) = 1.
a(n) = A007482(n) + A007482(n-1) = 2*A007482(n) - A104934(n). - R. J. Mathar, Jul 23 2010
a(n) = 3*a(n-1) + 2*a(n-2) with a(0) = 1, a(1) = 4. - Vincenzo Librandi, Dec 08 2010
a(n) = (Sum_{k = 0..n} A202396(n,k)*3^k)/2^n. - Philippe Deléham, Feb 05 2012
a(n) = (i*sqrt(2))^(n-1)*( i*sqrt(2)*ChebyshevU(n, -3*i/(2*sqrt(2))) + ChebyshevU(n-1, -3*i/(2*sqrt(2))) ). - G. C. Greubel, Jun 27 2021
a(n) = 2*a(n-1) + 2*A007482(n-1), n >= 1. - Jianing Song, Jun 01 2022
E.g.f.: exp(3*x/2)*(17*cosh(sqrt(17)*x/2) + 5*sqrt(17)*sinh(sqrt(17)*x/2))/17. - Stefano Spezia, May 24 2024

Extensions

Edited by N. J. A. Sloane, Jun 08 2010

A126501 Number of n-tuples of numbers [0..5] (leading zeros allowed) in which adjacent digits differ by 4 or less.

Original entry on oeis.org

1, 6, 34, 194, 1106, 6306, 35954, 204994, 1168786, 6663906, 37994674, 216628994, 1235123666, 7042134306, 40151166194, 228924368194, 1305226505746, 7441830001506, 42430056030514, 241917600158594, 1379308224915026
Offset: 0

Views

Author

R. H. Hardin, Dec 28 2006

Keywords

Comments

For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,5} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015
See PARI script for proof of g.f. - Andrew Howroyd, Apr 15 2017

Links

Crossrefs

Cf. Base 6 differing by three or less A126474, two or less A126393, one or less A126360.

Programs

  • Mathematica
    LinearRecurrence[{5, 4}, {1, 6}, 21] (* Jean-François Alcover, Oct 07 2017 *)
  • PARI
    \\ Proof of generating function
TransferGf(m,u,t,v,z)=vector(m,i,u(i))*matsolve(matid(m)-z*matrix(m,m,i,j,t(i,j)),vectorv(m,i,v(i)));
RowGf(d,m,z)=1+z*TransferGf(m, i->1, (i,j)->abs(i-j)<=d, j->1, z);
print(RowGf(4,6,x)); \\ Andrew Howroyd, Apr 15 2017

Formula

[Empirical] a(base,n) = a(base-1,n)+9^(n-1) for base>=4n-3; a(base,n) = a(base-1,n)+9^(n-1)-2 when base=4n-4.
From Philippe Deléham, Mar 24 2012: (Start)
G.f.: (1+x)/(1-5*x-4*x^2).
a(n) = 5*a(n-1) + 4*a(n-2), a(0) = 1, a(1) = 6.
a(n) = Sum_{k, 0<=k<=n} A054458(n,k)*3^k. (End)
Conjecture: a(n) = (2^(-1-n)*((5-sqrt(41))^n*(-7+sqrt(41)) + (5+sqrt(41))^n*(7+sqrt(41)))) / sqrt(41). - Colin Barker, Jan 20 2017

A126528 Number of base 7 n-digit numbers with adjacent digits differing by five or less.

Original entry on oeis.org

1, 7, 47, 317, 2137, 14407, 97127, 654797, 4414417, 29760487, 200635007, 1352612477, 9118849897, 61476161767, 414451220087, 2794088129357, 18836784876577, 126991149906247, 856130823820367, 5771740692453437, 38911098273822457, 262325293105201927
Offset: 0

Views

Author

R. H. Hardin, Dec 28 2006

Keywords

Comments

[Empirical] a(base,n)=a(base-1,n)+11^(n-1) for base>=5n-4; a(base,n)=a(base-1,n)+11^(n-1)-2 when base=5n-5.
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,6} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Links

Crossrefs

Cf. Base 7 differing by four or less A126502, three or less A126475, two or less A126394, one or less A126361.

Programs

  • Mathematica
    LinearRecurrence[{6, 5}, {1, 7}, 25] (* Paolo Xausa, Aug 08 2024 *)
  • PARI
    Vec((1+x)/(1-6*x-5*x^2) + O(x^30)) \\ Colin Barker, Sep 08 2016

Formula

From Philippe Deléham, Mar 24 2012: (Start)
G.f.: (1+x)/(1-6*x-5*x^2).
a(n) = 6*a(n-1) + 5*a(n-2), a(0) = 1, a(1) = 7 .
a(n) = Sum_{k=0..=n} A054458(n,k)*4^k.
(End)
a(n) = A091928(n+1)/5. - Philippe Deléham, Mar 27 2012
a(n) = (((3-sqrt(14))^n * (-4+sqrt(14)) + (3+sqrt(14))^n * (4+sqrt(14)))) / (2*sqrt(14)). - Colin Barker, Sep 08 2016

A054459 A001333(n), n >= 1, convolved with itself.

Original entry on oeis.org

1, 6, 23, 76, 233, 682, 1935, 5368, 14641, 39406, 104935, 276996, 725849, 1890258, 4896415, 12624752, 32419297, 82951766, 211573047, 538086716, 1364974409, 3454480250, 8724052271, 21989264232, 55326056977, 138975010110
Offset: 0

Views

Author

Wolfdieter Lang, Apr 26 2000

Keywords

Comments

a(n) = A054458(n+1,1) (second column of convolution triangle).

References

  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149. (The sequence t_n, also the sequence lev_{n-1}.)

Links

Crossrefs

Cf. A054458, A001333, A002203.

Programs

  • Mathematica
    LinearRecurrence[{4, -2, -4, -1}, {1, 6, 23, 76}, 30] (* Paolo Xausa, Feb 06 2024 *)

Formula

a(n) = ((4*n+3)*LP(n)+(2*n+1)*LP(n-1))/4, n >= 1, with LP(n) = A001333(n+1), a(0) = 1.
G.f.: ((1+x)/(1-2*x-x^2))^2.
a(n) = 2*A006645(n+2) - A000129(n+1). - R. J. Mathar, Feb 05 2024

A209696 Triangle of coefficients of polynomials v(n,x) jointly generated with A209695; see the Formula section.

Original entry on oeis.org

1, 1, 3, 1, 6, 7, 1, 9, 23, 17, 1, 12, 48, 76, 41, 1, 15, 82, 204, 233, 99, 1, 18, 125, 428, 765, 682, 239, 1, 21, 177, 775, 1907, 2649, 1935, 577, 1, 24, 238, 1272, 4010, 7656, 8680, 5368, 1393, 1, 27, 308, 1946, 7506, 18358, 28548, 27312, 14641, 3363
Offset: 1

Views

Author

Clark Kimberling, Mar 13 2012

Keywords

Comments

Alternating row sums: 1,-2,2,-2,2,-2,2,-2,...
For a discussion and guide to related arrays, see A208510.
Subtriangle of the triangle given by (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 3, -2/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 24 2012
Mirror image of triangle in A054458. - Philippe Deléham, Mar 24 2012

Examples

			First five rows:
  1;
  1,  3;
  1,  6,  7;
  1,  9, 23, 17;
  1, 12, 48, 76, 41;
First three polynomials v(n,x):
  1
  1 + 3x
  1 + 6x + 7x^2.
From _Philippe Deléham_, Mar 24 2012: (Start)
(1, 0, 0, 0, 0, ...) DELTA (0, 3, -2/3, -1/3, 0, 0, ...) begins:
  1;
  1,  0;
  1,  3,  0;
  1,  6,  7,  0;
  1,  9, 23, 17,  0;
  1, 12, 48, 76, 41,  0; (End)

Crossrefs

Cf. A054458, A209695, A208510.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209695 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209696 *)

Formula

u(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x),
v(n,x) = 2x*u(n-1,x) + (x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Mar 24 2012: (Start)
As DELTA-triangle T(n,k) with 0 <= k <= n:
G.f.: (1-2*y*x-y^2*x^2)/(1-x-2*y*x-y*x^2-y^2*x^2).
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = 1, T(1,1) = T(2,2) = 0, T(2,1) = 3 and T(n,k) = 0 if k < 0 or if k > n. (End)

A054460 A001333(n), n >= 1, convolved twice with itself.

Original entry on oeis.org

1, 9, 48, 204, 765, 2649, 8680, 27312, 83313, 247985, 723624, 2077164, 5880797, 16454865, 45577200, 125130432, 340882113, 922265721, 2479938368, 6631802220, 17646603933, 46744464745, 123314065944, 324085913136, 848801213425
Offset: 0

Views

Author

Wolfdieter Lang, Apr 26 2000

Keywords

Comments

a(n) = A054458(n+2,2) (third column of convolution triangle).

Links

Crossrefs

Cf. A054458, A054459, A001333, A002203.

Programs

  • Mathematica
    LinearRecurrence[{6, -9, -4, 9, 6, 1}, {1, 9, 48, 204, 765, 2649}, 30] (* Paolo Xausa, Feb 06 2024 *)

Formula

a(n) = (n+1)*(LP2(n+1)+2*LP2(n)+LP2(n-1))/8, n >= 1, LP2(n) = A054459(n), a(0) = 1.
a(n) = (n+1)*((10*n+11)*LP(n)+(4*n+5)*LP(n-1))/16, n >= 1, LP(n) = A001333(n+1).
G.f.: ((1+x)/(1-2*x-x^2))^3.
a(n) = +6*a(n-1) -9*a(n-2) -4*a(n-3) +9*a(n-4) +6*a(n-5) +a(n-6). - R. J. Mathar, Feb 05 2024

A062134 Triangle of coefficients of polynomials (rising powers) useful for convolutions of A001333(n+1), n >= 0 (associated Pell numbers).

Original entry on oeis.org

1, 2, 0, 8, 24, 16, 336, 832, 576, 128, 12480, 28480, 23680, 8960, 1280, 481920, 1208832, 1167360, 552960, 130560, 12288, 22786560, 61834752, 65709056, 35911680, 10895360, 1763328, 118784, 1280885760, 3645444096
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials pPL1(n,x) := Sum_{m=0..n} A062133(n,m)*x^m and pPL2(n,x) := Sum_{m=0..n} a(n,m)*x^m appear in the k-fold convolution of the associated Pell numbers PL(n) := A001333(n+1), n >= 0, as follows: PL(k; n) := A054458(n+k,k) = (2*pPL1(k,n)*PL(n+1)+pPL2(k,n)*PL(n))/(k!*8^k), k >= 0.

Examples

			Triangle begins:
  {1};
  {2,0};
  {8,24,16};
  {336,832,576,128};
  ...
pPL1(1,n) = 1+2*n.
pPL2(1,n) = 2.
PL(1; n) = A054459(n) = ((1+2*n)*PL(n+1)+PL(n))/4.

Crossrefs

Cf. A062133(n, m) (companion triangle), A054458(n, m) (convolution triangle).

A062133 Triangle of coefficients of polynomials (rising powers) useful for convolutions of A001333(n+1), n >= 0 (associated Pell numbers).

Original entry on oeis.org

0, 1, 2, 20, 36, 16, 456, 944, 672, 160, 14304, 33760, 28800, 10880, 1536, 575040, 1466752, 1413120, 666880, 157440, 14848, 27659520, 74774784, 79278080, 43330560, 13153280, 2128896, 143360, 1548126720
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials pPL1(n,x) := Sum_{m=0..n} a(n,m)*x^m, and pPL2(n,x) := Sum_{m=0..n} A062134(n,m)*x^m appear in the k-fold convolution of the associated Pell numbers PL(n) := A001333(n+1), n >= 0, as follows: PL(k; n) := A054458(n+k,k) = (2*pPL1(k,n)*PL(n+1)+pPL2(k,n)*PL(n))/(k!*8^k), k >= 0.

Examples

			Triangle begins:
  {0};
  {1,2};
  {20,36,16};
  {456,944,672,160};
  ...
pPL1(2,n) = 4*(5+9*n+4*n^2) = 4*(1+n)*(5+4*n).
pPL2(2,n) = 8*(1+3*n+2*n^2) = 8*(1+n)*(1+2*n).
PL(2; n) = A054460(n) = (1+n)*((5+4*n)*PL(n+1)+(1+2*n)*PL(n))/16.

Crossrefs

Cf. A062134(n, m) (companion triangle), A054458(n, m) (convolution triangle).
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