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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A001998 Bending a piece of wire of length n+1; walks of length n+1 on a tetrahedron; also non-branched catafusenes with n+2 condensed hexagons.

Original entry on oeis.org

1, 2, 4, 10, 25, 70, 196, 574, 1681, 5002, 14884, 44530, 133225, 399310, 1196836, 3589414, 10764961, 32291602, 96864964, 290585050, 871725625, 2615147350, 7845353476, 23535971854, 70607649841, 211822683802, 635467254244, 1906400965570, 5719200505225, 17157599124190
Offset: 0

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Author

Keywords

Comments

The wire stays in the plane, there are n bends, each is R,L or O; turning the wire over does not count as a new figure.
Equivalently, walks of n+1 steps on a tetrahedron, visiting n+2 vertices, with n "corners"; the symmetry group is S4, reversing a walk does not count as different. Simply interpret R,L,O as instructions to turn R, turn L, or retrace the last step. Walks are not self-avoiding.
Also, it appears that a(n) gives the number of equivalence classes of n-tuples of 0, 1 and 2, where two n-tuples are equivalent if one can be obtained from the other by a sequence of operations R and C, where R denotes reversal and C denotes taking the 2's complement (C(x)=2-x). This has been verified up to a(19)=290585050. Example: for n=3 there are ten equivalence classes {000, 222}, {001, 100, 122, 221}, {002, 022, 200, 220}, {010, 212}, {011, 110, 112, 211}, {012, 210}, {020, 202}, {021, 102, 120, 201}, {101, 121}, {111}, so a(3)=10. - John W. Layman, Oct 13 2009
There exists a bijection between chains of n+2 hexagons and the above described equivalence classes of n-tuples of 0,1, and 2. Namely, for a given chain of n+2 hexagons we take the sequence of the numbers of vertices of degree 2 (0, 1, or 2) between the consecutive contact vertices on one side of the chain; switching to the other side we obtain the 2's complement of this sequence; reversing the order of the hexagons, we obtain the reverse sequence. The inverse mapping is straightforward. For example, to a linear chain of 7 hexagons there corresponds the 5-tuple 11111. - Emeric Deutsch, Apr 22 2013
If we treat two wire bends (or walks, or tuples) related by turning over (or reversing) as different in any of the above-given interpretations of this sequence, we get A007051 (or A124302). Also, a(n-1) is the sum of first 3 terms in n-th row of A284949, see crossrefs therein. - Andrey Zabolotskiy, Sep 29 2017
a(n-1) is the number of color patterns (set partitions) in an unoriented row of length n using 3 or fewer colors (subsets). - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 3-paths with n+6 vertices. (A 3-path with order n at least 5 can be constructed from a 4-clique by iteratively adding a new 3-leaf (vertex of degree 3) adjacent to an existing 3-clique containing an existing 3-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)
a(n) is also the number of distinct planar embeddings of the (n+1)-alkane graph (up to at least n=9, and likely for all n). - Eric W. Weisstein, May 21 2024

Examples

			There are 2 ways to bend a piece of wire of length 2 (bend it or not).
For n=4 and a(n-1)=10, the 6 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, and ABBC.  The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB. - _Robert A. Russell_, Oct 28 2018
		

References

  • A. T. Balaban, Enumeration of Cyclic Graphs, pp. 63-105 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976; see p. 75.
  • S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Enumeration of tree-like octagonal systems: catapolyoctagons, ACH Models in Chem. 134 (1997), 55-70.
  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]
  • R. C. Read, The Enumeration of Acyclic Chemical Compounds, pp. 25-61 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976. [I think this reference does not mention this sequence. - N. J. A. Sloane, Aug 10 2006]
  • 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).

Crossrefs

Column 3 of A320750, offset by one. Column k = 0 of A323942, offset by two.
Cf. A124302 (oriented), A107767 (chiral), A182522 (achiral), with varying offsets.
Column 3 of A320750.
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • GAP
    a:=[];; for n in [2..45] do if n mod 2 =0 then Add(a,((3^((n-2)/2)+1)/2)^2); else Add(a,  3^((n-3)/2)+(1/4)*(3^(n-2)+1)); fi; od; a; # Muniru A Asiru, Oct 28 2018
  • Maple
    A001998 := proc(n) if n = 0 then 1 elif n mod 2 = 1 then (1/4)*(3^n+4*3^((n-1)/2)+1) else (1/4)*(3^n+2*3^(n/2)+1); fi; end;
    A001998:=(-1+3*z+2*z**2-8*z**3+3*z**4)/(z-1)/(3*z-1)/(3*z**2-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence with an extra leading 1
  • Mathematica
    a[n_?OddQ] := (1/4)*(3^n + 4*3^((n - 1)/2) + 1); a[n_?EvenQ] := (1/4)*(3^n + 2*3^(n/2) + 1); Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Jan 25 2013, from formula *)
    LinearRecurrence[{4,0,-12,9},{1,2,4,10},30] (* Harvey P. Dale, Apr 10 2013 *)
    Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
    k=3; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,k}]/2,{n,40}] (* Robert A. Russell, Oct 28 2018 *)
  • PARI
    Vec((1-2*x-4*x^2+6*x^3)/((1-x)*(1-3*x)*(1-3*x^2)) + O(x^50)) \\ Colin Barker, May 15 2016
    

Formula

a(n) = if n mod 2 = 0 then ((3^((n-2)/2)+1)/2)^2 else 3^((n-3)/2)+(1/4)*(3^(n-2)+1).
G.f.: (1-2*x-4*x^2+6*x^3) / ((1-x)*(1-3*x)*(1-3*x^2)). - Corrected by Colin Barker, May 15 2016
a(n) = 4*a(n-1)-12*a(n-3)+9*a(n-4), with a(0)=1, a(1)=2, a(2)=4, a(3)=10. - Harvey P. Dale, Apr 10 2013
a(n) = (1+3^n+3^(1/2*(-1+n))*(2-2*(-1)^n+sqrt(3)+(-1)^n*sqrt(3)))/4. - Colin Barker, May 15 2016
E.g.f.: (2*sqrt(3)*sinh(sqrt(3)*x) + 3*exp(2*x)*cosh(x) + 3*cosh(sqrt(3)*x))/6. - Ilya Gutkovskiy, May 15 2016
From Robert A. Russell, Oct 28 2018: (Start)
a(n-1) = (A124302(n) + A182522(n)) / 2 = A124302(n) - A107767(n-1) = A107767(n-1) + A182522(n).
a(n-1) = Sum_{j=1..k} (S2(n,j) + Ach(n,j)) / 2, where k=3 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n-1) = A057427(n) + A056326(n) + A056327(n). (End)
a(2*n) = A007051(n)^2; a(2*n+1) = A007051(n)*A007051(n+1). - Todd Simpson, Mar 25 2024

Extensions

Offset and Maple code corrected by Colin Mallows, Nov 12 1999
Term added by Robert A. Russell, Oct 30 2018

A006452 a(n) = 6*a(n-2) - a(n-4).

Original entry on oeis.org

1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552, 12671, 26531, 73852, 154634, 430441, 901273, 2508794, 5253004, 14622323, 30616751, 85225144, 178447502, 496728541, 1040068261, 2895146102, 6061962064, 16874148071, 35331704123
Offset: 0

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Keywords

Comments

Solution to a Diophantine equation.
Integers k such that k^2-1 is a triangular number. - Benoit Cloitre, Apr 05 2002
For all elements "x" of the sequence, 8*x^2 - 7 is a square. - Gregory V. Richardson, Oct 07 2002
a(n) mod 10 is a sequence of period 12: repeat (1, 1, 2, 4, 1, 3, 4, 4, 3, 1, 4, 2). - Paul Curtz, Dec 07 2012
a(n)^2 - 1 = A006454(n - 1) is a Sophie Germain triangular number of the second kind as defined in A217278. - Raphie Frank, Feb 08 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6xy + y^2 + 7 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 34xy + y^2 + 252 = 0. - Colin Barker, Mar 04 2014
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n+1), for n >= 0, gives one half of all positive y solutions of the Pell equation x^2 - 2*y^2 = -7. The corresponding x-solutions are x(n) = A077446(n+1).
See a comment on A077446 for the first and second class solutions separately, and the connection to the Pell equation X^2 - 2*Y^2 = 14. (End)
For n > 0, a(n) is the n-th almost balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

Examples

			n = 3: 11^2 - 2*(2*4)^2 = -7 (see the Pell comment above);
(4*4)^2 - 2*11^2 = +14. - _Wolfdieter Lang_, Feb 26 2015
		

References

  • A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Jul/Aug 1973 pp. 73-74.
  • Jeffrey Shallit, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    I:=[1,1,2,4]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013
    
  • Maple
    A006452:=-(z-1)*(z**2+3*z+1)/(z**2+2*z-1)/(z**2-2*z-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's
  • Mathematica
    s=0;lst={1}; Do[s+=n;If[Sqrt[s+1]==Floor[Sqrt[s+1]],AppendTo[lst, Sqrt[s+1]]], {n,0,8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
    a[0]=a[1]= 1; a[2]=2; a[3]=4; a[n_]:= 6*a[n-2] -a[n-4]; Array[a, 30, 0] (* Robert G. Wilson v, Jun 11 2010 *)
    CoefficientList[Series[(1+x-4x^2-2x^3)/((1-2x-x^2)(1+2x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;1;2;4])[1,1] \\ Charles R Greathouse IV, May 10 2016
    
  • SageMath
    def A001333(n): return lucas_number2(n, 2, -1)/2
    def A006452(n): return (A001333(n+1) + (-1)^n *A001333(n-2))/4
    [A006452(n) for n in range(41)] # G. C. Greubel, Jan 22 2023

Formula

Bisection: a(2n) = A006452(n). a(2n+1) = A038723(n).
G.f.: ( 1+x-4*x^2-2*x^3 ) / ( (1-2*x-x^2)*(1+2*x-x^2) ).
From Gregory V. Richardson, Oct 07 2002: (Start)
For n (even), a(n) = ( ((3 + sqrt(8))^((n/2)+1) - (3 - sqrt(8))^((n/2)+1)) - 2*((3 + sqrt(8))^((n/2)-1) - (3 - sqrt(8))^((n/2)-1)) ) / (6*sqrt(8)).
For n (odd), a(n) = ( ((3 + sqrt(8))^((n+1)/2) - (3 - sqrt(8))^((n+1)/2)) - 2*((3 + sqrt(8))^((n-1)/2) - (3 - sqrt(8))^((n-1)/2)) ) / (2*sqrt(8)).
Limit_{n->oo} a(n)/a(n-2) = 3 + sqrt(8).
If n is odd, lim_{n->oo} a(n)/a(n-1) = (9 + 2*sqrt(8))/7.
If n is even, lim_{n->oo} a(n)/a(n-1) = (11 + 3*sqrt(8))/7. (End)
a(n+2) = (A001333(n+3) + (-1)^n *A001333(n))/4. - Paul Curtz, Dec 06 2012
a(n+2) = sqrt(17*a(n)^2 + 6*(sqrt(8*a(n)^2 - 7))*a(n)*sgn(2*n - 1) - 7) with a(0) = 1, a(1) = 1. - Raphie Frank, Feb 08 2013
a(n+2) = (A216134(n+2) - A216134(n))/2. - Raphie Frank, Feb 11 2013
E.g.f.: (2*cosh(sqrt(2)*x)*(2*cosh(x) - sinh(x)) + sqrt(2)*(3*cosh(x) - sinh(x))*sinh(sqrt(2)*x))/4. - Stefano Spezia, Nov 26 2022

Extensions

More terms from James Sellers, May 03 2000

A028297 Coefficients of Chebyshev polynomials of the first kind: triangle of coefficients in expansion of cos(n*x) in descending powers of cos(x).

Original entry on oeis.org

1, 1, 2, -1, 4, -3, 8, -8, 1, 16, -20, 5, 32, -48, 18, -1, 64, -112, 56, -7, 128, -256, 160, -32, 1, 256, -576, 432, -120, 9, 512, -1280, 1120, -400, 50, -1, 1024, -2816, 2816, -1232, 220, -11, 2048, -6144, 6912, -3584, 840, -72, 1, 4096, -13312, 16640, -9984
Offset: 0

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Keywords

Comments

Rows are of lengths 1, 1, 2, 2, 3, 3, ... (A008619).
This triangle is generated from A118800 by shifting down columns to allow for (1, 1, 2, 2, 3, 3, ...) terms in each row. - Gary W. Adamson, Dec 16 2007
Unsigned triangle = A034839 * A007318. - Gary W. Adamson, Nov 28 2008
Triangle, with zeros omitted, given by (1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, -1, 1, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 16 2011
From Wolfdieter Lang, Aug 02 2014: (Start)
This irregular triangle is the row reversed version of the Chebyshev T-triangle A053120 given by A039991 with vanishing odd-indexed columns removed.
If zeros are appended in each row n >= 1, in order to obtain a regular triangle (see the Philippe Deléham comment, g.f. and example) this becomes the Riordan triangle (1-x)/(1-2*x), -x^2/(1-2*x). See also the unsigned version A201701 of this regular triangle.
(End)
Apparently, unsigned diagonals of this array are rows of A200139. - Tom Copeland, Oct 11 2014
It appears that the coefficients are generated by the following: Let SM_k = Sum( d_(t_1, t_2)* eM_1^t_1 * eM_2^t_2) summed over all length 2 integer partitions of k, i.e., 1*t_1 + 2*t_2 = k, where SM_k are the averaged k-th power sum symmetric polynomials in 2 data (i.e., SM_k = S_k/2 where S_k are the k-th power sum symmetric polynomials, and where eM_k are the averaged k-th elementary symmetric polynomials, eM_k = e_k/binomial(2,k) with e_k being the k-th elementary symmetric polynomials. The data d_(t_1, t_2) form an irregular triangle, with one row for each k value starting with k=1. Thus this procedure and associated OEIS sequences A287768, A288199, A288207, A288211, A288245, A288188 are generalizations of Chebyshev polynomials of the first kind. - Gregory Gerard Wojnar, Jul 01 2017

Examples

			Letting c = cos x, we have: cos 0x = 1, cos 1x = 1c; cos 2x = 2c^2-1; cos 3x = 4c^3-3c, cos 4x = 8c^4-8c^2+1, etc.
T4 = 8x^4 - 8x^2 + 1 = 8, -8, +1 = 2^(3) - (4)(2) + [2^(-1)](4)/2.
From _Wolfdieter Lang_, Aug 02 2014: (Start)
The irregular triangle T(n,k) begins:
n\k     1      2     3      4     5     6   7   8 ....
0:      1
1:      1
2:      2     -1
3:      4     -3
4:      8     -8     1
5:     16    -20     5
6:     32    -48    18     -1
7:     64   -112    56     -7
8:    128   -256   160    -32     1
9:    256   -576   432   -120     9
10:   512  -1280  1120   -400    50    -1
11:  1024  -2816  2816  -1232   220   -11
12:  2048  -6144  6912  -3584   840   -72   1
13:  4096 -13312 16640  -9984  2912  -364  13
14:  8192 -28672 39424 -26880  9408 -1568  98  -1
15: 16384 -61440 92160 -70400 28800 -6048 560 -15
...
T(4,x) = 8*x^4 -8*x^2 + 1*x^0, T(5,x) = 16*x^5 - 20*x^3 + 5*x^1, with Chebyshev's T-polynomials (A053120). (End)
From _Philippe Deléham_, Dec 16 2011: (Start)
The triangle (1,1,0,0,0,0,...) DELTA (0,-1,1,0,0,0,0,...) includes zeros and begins:
   1;
   1,   0;
   2,  -1,  0;
   4,  -3,  0,  0;
   8,  -8,  1,  0, 0;
  16, -20,  5,  0, 0, 0;
  32, -48, 18, -1, 0, 0, 0; (End)
		

References

  • I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, 5th ed., Section 1.335, p. 35.
  • S. Selby, editor, CRC Basic Mathematical Tables, CRC Press, 1970, p. 106. [From Rick L. Shepherd, Jul 06 2010]

Crossrefs

Cf. A028298.
Reflection of A008310, the main entry. With zeros: A039991.
Cf. A053120 (row reversed table including zeros).
Cf. A001333 (row sums 1), A001333 (alternating row sums). - Wolfdieter Lang, Aug 02 2014

Programs

  • Maple
    b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
    seq(T(n), n=0..15);  # Alois P. Heinz, Sep 04 2019
  • Mathematica
    t[n_] := (Cos[n x] // TrigExpand) /. Sin[x]^m_ /; EvenQ[m] -> (1 - Cos[x]^2)^(m/2) // Expand; Flatten[Table[ r = Reverse @ CoefficientList[t[n], Cos[x]]; If[OddQ[Length[r]], AppendTo[r,0]]; Partition[r,2][[All, 1]],{n, 0, 13}] ][[1 ;; 53]] (* Jean-François Alcover, May 06 2011 *)
    Tpoly[n_] := HypergeometricPFQ[{(1 - n)/2, -n/2}, {1/2}, 1 - x];
    Table[CoefficientList[Tpoly[n], x], {n, 0, 12}] // Flatten (* Peter Luschny, Feb 03 2021 *)

Formula

cos(n*x) = 2 * cos((n-1)*x) * cos(x) - cos((n-2)*x) (from CRC's Multiple-angle relations). - Rick L. Shepherd, Jul 06 2010
G.f.: (1-x) / (1-2x+y*x^2). - Philippe Deléham, Dec 16 2011
Sum_{k=0..n} T(n,k)*x^k = A011782(n), A000012(n), A146559(n), A087455(n), A138230(n), A006495(n), A138229(n) for x = 0, 1, 2, 3, 4, 5, 6, respectively. - Philippe Deléham, Dec 16 2011
T(n,k) = [x^k] hypergeom([1/2 - n/2, -n/2], [1/2], 1 - x). - Peter Luschny, Feb 03 2021
T(n,k) = (-1)^k * 2^(n-1-2*k) * A034807(n,k). - Hoang Xuan Thanh, Jun 21 2025

Extensions

More terms from David W. Wilson
Row length sequence and link to Abramowitz-Stegun added by Wolfdieter Lang, Aug 02 2014

A005409 Number of polynomials of height n: a(1)=1, a(2)=1, a(3)=4, a(n) = 2*a(n-1) + a(n-2) + 2 for n >= 4.

Original entry on oeis.org

1, 1, 4, 11, 28, 69, 168, 407, 984, 2377, 5740, 13859, 33460, 80781, 195024, 470831, 1136688, 2744209, 6625108, 15994427, 38613964, 93222357, 225058680, 543339719, 1311738120, 3166815961, 7645370044, 18457556051, 44560482148, 107578520349, 259717522848
Offset: 1

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Author

N. J. A. Sloane, S. M. Diano

Keywords

Comments

Starting with n=1, the sum of the antidiagonals of the array in a comment from Cloitre regarding A002002. - Gerald McGarvey, Aug 12 2004
Cumulative sum of A001333. - Sture Sjöstedt, Nov 15 2011
a(n) is the number of self-avoiding walks on a 3 rows X n columns grid of squares, starting top-left, ending bottom-left, using moves of R(ight), L(eft), U(p), D(own). E.g., for 3 X 1 there is just the path (D,D), and a(1) = 1. For 3 X 2, there are 4 paths (D,D) (D,R,D,L) (R,D,D,L) and (R,D,L,D) and a(2) = 4. - Toby Gottfried, Mar 04 2013
Define a triangle to have T(n,1) = n*(n-1)+1 and T(n,n) = n; the other terms T(r,c) = T(r-1,c) + T(r-1,c-1) + T(r-2,c-1). The sum of the terms in row(n+1) minus those in row(n) = a(n+2). - J. M. Bergot, Apr 30 2013
Since the terms of the sequence are all finite, it can be used in enumerating all polynomials with integer coefficients. Since each polynomial has only a finite number of roots, this enumeration can be used in turn to enumerate the algebraic numbers. Cantor uses this to derive the existence of transcendental numbers as a corollary of his stronger result that no enumerable sequence of real numbers can include all of them. - Morgan L. Owens, May 15 2022
For n > 1, also the rank of the (n-1)-Pell graph. - Eric W. Weisstein, Aug 01 2023

References

  • R. Courant and H. Robbins, What is Mathematics?, Oxford Univ. Press, 1941, p. 103.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A214931 (walks on grids with 4 rows), A006189 (grids with 3 columns).
Cf. A216211 (grids with 4 columns).

Programs

  • Haskell
    a005409 n = a005409_list !! (n-1)
    a005409_list = 1 : scanl1 (+) (tail a001333_list)
    -- Reinhard Zumkeller, Jul 08 2012
    
  • Magma
    [1] cat [n le 2 select n^2 else 2*Self(n-1) +Self(n-2) +2: n in [1..30]]; // G. C. Greubel, Apr 22 2021
    
  • Mathematica
    Join[{1}, RecurrenceTable[{a[1] == 1, a[2] == 4, a[n] == 2 a[n - 1] + a[n - 2] + 2}, a[n], {n, 30}]] (* Harvey P. Dale, Jul 27 2011 *)
    Join[{1}, CoefficientList[Series[(x + 1)/((x - 1) (x^2 + 2 x - 1)), {x, 0, 40}], x]] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    Join[{1}, Fibonacci[Range[2, 35], 2] -1] (* G. C. Greubel, Apr 22 2021 *)
    Join[{1}, LinearRecurrence[{3, -1, -1}, {1, 4, 11}, 20]] (* Eric W. Weisstein, Aug 01 2023 *)
  • PARI
    a(n)=polcoeff(1+x*(1+x)/(1-3*x+x^2+x^3)+x*O(x^n),n) \\ Paul D. Hanna
    
  • Sage
    [1]+[lucas_number1(n,2,-1) -1 for n in (2..35)] # G. C. Greubel, Apr 22 2021

Formula

a(n) = A000129(n) - 1, n > 1.
a(n) = ((1+sqrt(2))^n - (1-sqrt(2))^n)/(2*sqrt(2))-1 for n > 1, a(1)=1.
G.f.: 1 + x*(1+x)/( (1-x)*(1-2*x-x^2) ). - Simon Plouffe in his 1992 dissertation.
a(n) = 3*a(n-1) - a(n-2) - a(n-3). - Toby Gottfried, Mar 08 2013
(1, 4, 11, 28, ...) = (1, 2, 2, 2, ...) * the Pell sequence starting (1, 2, 5, 12, 29, ...); such that, for example: a(5) = (2, 2, 2, 1) dot (1, 2, 5, 12) = (2 + 4 + 10 + 12) = 48. - Gary W. Adamson May 21 2013
E.g.f.: 1 + exp(x)*(2*(cosh(sqrt(2)*x) - 1) + sqrt(2)*sinh(sqrt(2)*x))/2. - Stefano Spezia, Jun 26 2022

Extensions

Additional comments from Barry E. Williams

A048655 Generalized Pellian with second term equal to 5.

Original entry on oeis.org

1, 5, 11, 27, 65, 157, 379, 915, 2209, 5333, 12875, 31083, 75041, 181165, 437371, 1055907, 2549185, 6154277, 14857739, 35869755, 86597249, 209064253, 504725755, 1218515763, 2941757281, 7102030325, 17145817931, 41393666187, 99933150305, 241259966797, 582453083899
Offset: 0

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Author

Keywords

Comments

Equals binomial transform of A143095: (1, 4, 2, 8, 4, 16, 8, 32, ...). - Gary W. Adamson, Jul 23 2008

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x)/(1-2*x-x^2))); // G. C. Greubel, Jul 26 2018
  • Maple
    with(combinat): a:=n->3*fibonacci(n, 2)+fibonacci(n+1, 2): seq(a(n), n=0..26); # Zerinvary Lajos, Apr 04 2008
  • Mathematica
    a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{4},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
    LinearRecurrence[{2,1},{1,5},30] (* Harvey P. Dale, Nov 05 2011 *)
  • Maxima
    a[0]:1$
    a[1]:5$
    a[n]:=2*a[n-1]+a[n-2]$
    A048655(n):=a[n]$
    makelist(A048655(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    a(n)=([0,1; 1,2]^n*[1;5])[1,1] \\ Charles R Greathouse IV, Feb 09 2017
    

Formula

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=5.
a(n) = ((4+sqrt(2))(1+sqrt(2))^n - (4-sqrt(2))(1-sqrt(2))^n)/2*sqrt(2).
a(n) = P(n) - 3*P(n+1) + 2*P(n+2). - Creighton Dement, Jan 18 2005
G.f.: (1+3*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008
E.g.f.: exp(x)*(cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x)). - Vaclav Kotesovec, Feb 16 2015
a(n) = 3*Pell(n) + Pell(n+1), where Pell = A000129. - Vladimir Reshetnikov, Sep 27 2016

A083098 a(n) = 2*a(n-1) + 6*a(n-2).

Original entry on oeis.org

1, 1, 8, 22, 92, 316, 1184, 4264, 15632, 56848, 207488, 756064, 2757056, 10050496, 36643328, 133589632, 487039232, 1775616256, 6473467904, 23600633344, 86042074112, 313687948288, 1143628341248, 4169384372224, 15200538791936
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003

Keywords

Comments

a(n+1) = a(n) + 7*A083099(n-1); a(n+1)/A083099(n) converges to sqrt(7).
Binomial transform of expansion of cosh(sqrt(7)x) (A000420 with interpolated zeros: 1, 0, 7, 0, 49, 0, 343, 0, ...).
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 7 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(7). - Cino Hilliard, Sep 25 2005
a(n) is the number of compositions of n when there are 1 type of 1 and 7 types of other natural numbers. - Milan Janjic, Aug 13 2010

References

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

Crossrefs

The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Mathematica
    CoefficientList[Series[(1+6x)/(1-2x-6x^2), {x, 0, 25}], x]
    LinearRecurrence[{2, 6}, {1, 1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)
    a[n_] := Simplify[((1 + Sqrt[7])^n + (1 - Sqrt[7])^n)/2]; Array[a, 25, 0] (* Robert G. Wilson v, Sep 18 2013 *)
  • PARI
    x='x+O('x^30); Vec((1-x)/(1-2*x-6*x^2)) \\ G. C. Greubel, Jan 08 2018
    
  • Sage
    [lucas_number2(n,2,-6)/2 for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009
    

Formula

G.f.: (1-x)/(1-2*x-6*x^2).
a(n) = (1+sqrt(7))^n/2 + (1-sqrt(7))^n/2.
E.g.f.: exp(x)*cosh(sqrt(7)x).
a(n) = Sum_{k=0..n} A098158(n,k)*7^(n-k). - Philippe Deléham, Dec 26 2007
If p[1]=1, and p[i]=7, (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*(7*k-1)/(x*(7*k+6) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

A159684 Sturmian word: limit S(infinity) where S(0) = 0, S(1) = 0,1 and for n>=1, S(n+1) = S(n)S(n)S(n-1).

Original entry on oeis.org

0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0
Offset: 0

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Author

Philippe Deléham, Apr 19 2009

Keywords

Comments

Fixed point of morphism 0 -> 0,1; 1 -> 0,1,0.
This sequence corresponds to the case k = 1 of the Sturmian word S_k(infinity) as defined in A080764. See A171588 for the case k = 2. - Peter Bala, Nov 22 2013
This sequence is the {1->01}-transform of the Sturmian word A080764. - Clark Kimberling, May 17 2017
Also, sequence (1 if x/sqrt(2) is integer, 0 else) as x runs over the elements of N U N*sqrt(2) in increasing order, N = {0, 1, 2, ...}. See A144612 for the sqrt(3) analog. - M. F. Hasler, Feb 06 2025

Examples

			0 -> 0,1 -> 0,1,0,1,0 -> 0,1,0,1,0,0,1,0,1,0,0,1 ->...
		

Crossrefs

See A188037 for another version of this sequence. - N. J. A. Sloane, Mar 22 2011
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021

Programs

  • Haskell
    a159684 n = a159684_list !! n
    a159684_list = 0 : concat (iterate (concatMap s) [1])
       where s 0 = [0,1]; s 1 = [0,1,0]
    -- Reinhard Zumkeller, Oct 26 2013
    
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0, 1, 0}}] &, {1}, 6] (* Robert G. Wilson v, May 02 2009 *)
    SubstitutionSystem[{0->{0,1},1->{0,1,0}},{1},{6}][[1]] (* Harvey P. Dale, Dec 25 2021 *)
  • Python
    def aupto(nn):
        Snm1, Sn = [0], [0, 1]
        while len(Sn) < nn+1: Snm1, Sn = Sn, Sn + Sn + Snm1
        return Sn[:nn+1]
    print(aupto(104)) # Michael S. Branicky, Jul 23 2022
    
  • Python
    from math import isqrt
    def A159684(n): return -isqrt(m:=(n+1)**2<<1)+isqrt(m+(n<<2)+6)-1 # Chai Wah Wu, Aug 03 2022

Formula

From Peter Bala, Nov 22 2013: (Start)
a(n) = floor((n + 2)*(sqrt(2) - 1)) - floor((n + 1)*(sqrt(2) - 1)).
If we read the sequence as the decimal constant C = 0.01010 01010 01010 10010 10010 ... then C = sum {n >= 1} 1/10^floor(n*(1 + sqrt(2))).
The real number 9*C has the simple continued fraction expansion [0; 11, 1010, 10000100, 100000000000100000, 100000000000000000000000000001000000000000, ...], the partial quotients having the form 10^Pell(n)*(1 + 10^Pell(n+1)) = 10^A001333(n+1) + 10^A000129(n) (see Adams and Davison).
A rapidly converging series for C is C = 9*sum {n >= 1} 10^Pell(2*n-1)*(1 + 10^Pell(2*n))/( (10^Pell(2*n-1) - 1)*(10^Pell(2*n+1) - 1) ): for example, the first 10 terms of the series give a rational approximation to C accurate to more than 130 million decimal places. Compare with the Fibonacci words A005614 and A221150. (End)

Extensions

More terms from Robert G. Wilson v, May 02 2009

A001601 a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 17, 577, 665857, 886731088897, 1572584048032918633353217, 4946041176255201878775086487573351061418968498177
Offset: 0

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Keywords

Comments

Reduced numerators of Newton's iteration for sqrt(2). - Eric W. Weisstein
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Evaluation of the 2^n - 1 degree interpolating polynomial of 1/x at Chebyshev nodes in the interval (1,2): v = 1.0; for(i = 1, n, v *= 4*(a(i) - x*v)); v *= 2/a(n+1). - Jose Hortal, Apr 07 2012
Smallest positive integer x satisfying the Pell equation x^2 - 2^(2*n+1) * y^2 = 1, for n > 0. - A.H.M. Smeets, Sep 29 2017

References

  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 376.
  • 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).

Crossrefs

a(n) = A001333(2^n).

Programs

  • Mathematica
    Table[Simplify[Expand[(1/2) ((1 + Sqrt[2])^(2^n) + (1 - Sqrt[2])^(2^n))]], {n, 0, 7}]  (* Artur Jasinski, Oct 10 2008 *)
    Join[{1},NestList[2#^2-1&,3,7]]  (* Harvey P. Dale, Mar 24 2011 *)
  • PARI
    a(n)=if(n<1,n==0,2*a(n-1)^2-1)

Formula

From Mario Catalani (mario.catalani(AT)unito.it), May 27 2003, May 30 2003: (Start)
a(n) = a(n-1)^2 + 2*A051009(n)^2 for n > 0.
a(n)^2 = 2*A051009(n+1)^2 + 1.
a(n) = Sum_{r=0..2^(n-1)} binomial(2^n, 2*r)*2^r. (End)
Expansion of 1/sqrt(2) as an infinite product: 1/sqrt(2) = Product_{k>=1} (1 - 1/(a(n)+1)). a(1)=3; a(n) = floor(1/(1-1/(sqrt(2)*Product_{k=1..n-1} 1-1/(a(k)+1)))). - Thomas Baruchel, Nov 06 2003
2*a(n+1) = A003423(n).
a(n) = (1/2)*((1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n > 1: a(n) - 1 = 4^n * Product_{i=1..n-2} a(i)^2. - Jose Hortal, Apr 13 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n>=1} (1 - 1/(2*a(n))).
sqrt(2) = Product_{n>=1} (1 + 1/a(n)).
a(n) = (1/2)*A003423(n-1). (End)
a(n) = cos(2^(n-1) * arccos(3)) = cosh(2^(n-1) * log(3 + 2*sqrt(2))) for n >= 1. - Daniel Suteu, Jul 28 2016
a(n+1) = T(2^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = A001541(2^(n-1)). - A.H.M. Smeets, May 28 2017

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025
Offset: 0

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Author

Keywords

Comments

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

Examples

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Crossrefs

Programs

  • GAP
    a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018
    
  • Magma
    I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018
  • Maple
    CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007
  • Mathematica
    LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
    CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)
  • PARI
    a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))
    
  • PARI
    vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015
    

Formula

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Extensions

Entry edited by N. J. A. Sloane, Sep 14 2007

A002534 a(n) = 2*a(n-1) + 9*a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 13, 44, 205, 806, 3457, 14168, 59449, 246410, 1027861, 4273412, 17797573, 74055854, 308289865, 1283082416, 5340773617, 22229288978, 92525540509, 385114681820, 1602959228221, 6671950592822, 27770534239633, 115588623814664
Offset: 0

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Author

Keywords

Comments

For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 2's along the main diagonal, and 3's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 19 2011

References

  • 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).
  • 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.

Crossrefs

Programs

  • Magma
    [Ceiling(((1+Sqrt(10))^n-(1-Sqrt(10))^n)/(2*Sqrt(10))): n in [0..30]]; // Vincenzo Librandi, Aug 15 2011
    
  • Maple
    A002534:=-z/(-1+2*z+9*z**2); # [Simon Plouffe in his 1992 dissertation.]
  • Mathematica
    LinearRecurrence[{2, 9}, {0, 1}, 30] (* T. D. Noe, Aug 18 2011 *)
  • PARI
    first(n) = Vec(x/(1 - 2*x - 9*x^2) + O(x^n), -n) \\ Iain Fox, Jan 17 2018
  • Sage
    [lucas_number1(n,2,-9) for n in range(0, 20)] # Zerinvary Lajos, Apr 22 2009
    

Formula

From Paul Barry, Sep 29 2004: (Start)
E.g.f.: exp(x)*sinh(sqrt(10)*x)/sqrt(10).
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*10^k. (End)
a(n) = ((1+sqrt(10))^n - (1-sqrt(10))^n)/(2*sqrt(10)). - Artur Jasinski, Dec 10 2006
G.f.: x/(1 - 2*x - 9*x^2) - Iain Fox, Jan 17 2018
From G. C. Greubel, Jan 03 2024: (Start)
a(n) = (3*i)^(n-1)*ChebyshevU(n-1, -i/3).
a(n) = 3^(n-1)*Fibonacci(n, 2/3), where Fibonacci(n, x) is the Fibonacci polynomial. (End)

Extensions

More terms from Johannes W. Meijer, Aug 18 2011
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