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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A124959 Triangle read by rows: T(n,k) = a(k)*binomial(n,k) (0 <= k <= n), where a(0)=1, a(1)=2, a(k) = a(k-1) + 3*a(k-2) for k >= 2 (a(k) = A006138(k)).

Original entry on oeis.org

1, 1, 2, 1, 4, 5, 1, 6, 15, 11, 1, 8, 30, 44, 26, 1, 10, 50, 110, 130, 59, 1, 12, 75, 220, 390, 354, 137, 1, 14, 105, 385, 910, 1239, 959, 314, 1, 16, 140, 616, 1820, 3304, 3836, 2512, 725, 1, 18, 180, 924, 3276, 7434, 11508, 11304, 6525, 1667, 1, 20, 225, 1320, 5460, 14868, 28770, 37680, 32625, 16670, 3842
Offset: 0

Views

Author

Gary W. Adamson, Nov 13 2006

Keywords

Comments

Sum of entries in row n = A006190(n+1).

Examples

			First few rows of the triangle:
  1;
  1,   2;
  1,   4,   5;
  1,   6,  15,  11;
  1,   8,  30,  44,  26;
  1,  10,  50, 110, 130,  59;
  ...
		

Crossrefs

Programs

  • Magma
    function b(k)
      if k lt 2 then return k+1;
      else return b(k-1) + 3*b(k-2);
      end if;
      return b;
    end function;
    [Binomial(n,k)*b(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 19 2019
    
  • Maple
    a:=proc(n) if n=0 then 1 elif n=1 then 2 else a(n-1)+3*a(n-2) fi end: T:=(n,k)->a(k)*binomial(n,k): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    T[n_, k_]:= T[n, k]= Simplify[(I*Sqrt[3])^(k-1)*Binomial[n,k]*(I*Sqrt[3]* ChebyshevU[k, 1/(2*I*Sqrt[3])] + ChebyshevU[k-1, 1/(2*I*Sqrt[3])])];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 19 2019 *)
  • PARI
    b(k) = if(k<2, k+1, b(k-1) + 3*b(k-2));
    T(n,k) = binomial(n,k)*b(k);
    for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Nov 19 2019
    
  • Sage
    @CachedFunction
    def b(k):
        if (k<2): return k+1
        else: return b(k-1) + 3*b(k-2)
    [[binomial(n, k)*b(k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 19 2019

Extensions

Edited by N. J. A. Sloane, Dec 03 2006

A016116 a(n) = 2^floor(n/2).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Powers of 2 doubled up. The usual OEIS policy is to omit the duplicates in such cases (when this would become A000079). This is an exception.
Number of symmetric compositions of n: e.g., 5 = 2+1+2 = 1+3+1 = 1+1+1+1+1 so a(5) = 4; 6 = 3+3 = 2+2+2 = 1+4+1 = 2+1+1+2 = 1+2+2+1 = 1+1+2+1+1 = 1+1+1+1+1+1 so a(6) = 8. - Henry Bottomley, Dec 10 2001
This sequence is the number of digits of each term of A061519. - Dmitry Kamenetsky, Jan 17 2009
Starting with offset 1 = binomial transform of [1, 1, -1, 3, -7, 17, -41, ...]; where A001333 = (1, 1, 3, 7, 17, 41, ...). - Gary W. Adamson, Mar 25 2009
a(n+1) is the number of symmetric subsets of [n]={1,2,...,n}. A subset S of [n] is symmetric if k is an element of S implies (n-k+1) is an element of S. - Dennis P. Walsh, Oct 27 2009
INVERT and inverse INVERT transforms give A006138, A039834(n-1).
The Kn21 sums, see A180662, of triangle A065941 equal the terms of this sequence. - Johannes W. Meijer, Aug 15 2011
First differences of A027383. - Jason Kimberley, Nov 01 2011
Run lengths in A079944. - Jeremy Gardiner, Nov 21 2011
Number of binary palindromes (A006995) between 2^(n-1) and 2^n (for n>1). - Hieronymus Fischer, Feb 17 2012
Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012
Range of row n of the Circular Pascal array of order 4. - Shaun V. Ault, May 30 2014
a(n) is the number of permutations of length n avoiding both 213 and 312 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Also, the decimal representation of the diagonal from the origin to the corner (and from the corner to the origin except for the initial term) of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 190", based on the 5-celled von Neumann neighborhood when initialized with a single black (ON) cell at stage zero. - Robert Price, May 10 2017
a(n + 1) + n - 1, n > 0, is the number of maximal subsemigroups of the monoid of partial order-preserving or -reversing mappings on a set with n elements. See the East et al. link. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Number of symmetric stairs with n cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. See A005418. - Christian Barrientos, May 11 2018
For n >= 4, a(n) is the exponent of the group of the Gaussian integers in a reduced system modulo (1+i)^(n+2). See A302254. - Jianing Song, Jun 27 2018
a(n) is the number of length-(n+1) binary sequences, denoted , with s(1)=1 and with s(i+1)=s(i) for odd i. - Dennis P. Walsh, Sep 06 2018
a(n+1) is the number of subsets of {1,2,..,n} in which all differences between successive elements of subsets are even. For example, for n = 7, a(6) = 8 and the 8 subsets are {7}, {1,7}, {3,7}, {5,7}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. For odd differences between elements see Comment in A000045 (Fibonacci numbers). - Enrique Navarrete, Jul 01 2020
Also, the number of walks of length n on the graph x--y--z, starting at x. - Sean A. Irvine, May 30 2025

Examples

			For n=5 the a(5)=4 symmetric subsets of [4] are {1,4}, {2,3}, {1,2,3,4} and the empty set. - _Dennis P. Walsh_, Oct 27 2009
For n=5 the a(5)=4 length-6 binary sequences are <1,1,0,0,0,0>, <1,1,0,0,1,1>, <1,1,1,1,0,0> and <1,1,1,1,1,1>. - _Dennis P. Walsh_, Sep 06 2018
		

Crossrefs

a(n) = A094718(3, n).
Cf. A001333.
See A052955 for partial sums (without the initial term).
A000079 gives the odd-indexed terms of a(n).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

Formula

a(n) = a(n-1)*a(n-2)/a(n-3) = 2*a(n-2) = 2^A004526(n).
G.f.: (1+x)/(1-2*x^2).
a(n) = (1/2 + sqrt(1/8))*sqrt(2)^n + (1/2 - sqrt(1/8))*(-sqrt(2))^n. - Ralf Stephan, Mar 11 2003
E.g.f.: cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, Jul 16 2003
The signed sequence (-1)^n*2^floor(n/2) has a(n) = (sqrt(2))^n(1/2 - sqrt(2)/4) + (-sqrt(2))^n(1/2 + sqrt(2)/4). It is the inverse binomial transform of A000129(n-1). - Paul Barry, Apr 21 2004
Diagonal sums of A046854. a(n) = Sum_{k=0..n} binomial(floor(n/2), k). - Paul Barry, Jul 07 2004
a(n) = a(n-2) + 2^floor((n-2)/2). - Paul Barry, Jul 14 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2), floor(k/2)). - Paul Barry, Jul 15 2004
E.g.f.: cosh(asinh(1) + sqrt(2)*x)/sqrt(2). - Michael Somos, Feb 28 2005
a(n) = Sum_{k=0..n} A103633(n,k). - Philippe Deléham, Dec 03 2006
a(n) = 2^(n/2)*((1 + (-1)^n)/2 + (1-(-1)^n)/(2*sqrt(2))). - Paul Barry, Nov 12 2009
a(n) = 2^((2*n - 1 + (-1)^n)/4). - Luce ETIENNE, Sep 20 2014

A006190 a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441, 8587845200883, 28363725910090
Offset: 0

Views

Author

Keywords

Comments

Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003
a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004
a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006
a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006
Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007
Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011
These numbers could also be called "bronze Fibonacci numbers". Indeed, 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; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013
Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013
The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014
a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERTi transform of A001076. - Gary W. Adamson, Jul 24 2015
Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018
Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Jan 25 2023: (Start)
Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023

Examples

			From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Pell number of parts, A000129(k), there are A000129(1)=1 type of 1, A000129(2)=2 types of 2, A000129(3)=5 types of 3, A000129(4)=12 types of 4, A000129(5)=29 types of 5 and A000129(6)=70 types of 6.
The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
  6,              1,     70;
  5+1,            2,     58;
  4+2,            2,     48;
  3+3,            1,     25;
  4+1+1,          3,     36;
  3+2+1,          6,     60;
  2+2+2,          1,      8;
  3+1+1+1,        4,     20;
  2+2+1+1,        6,     24;
  2+1+1+1+1,      5,     10;
  1+1+1+1+1+1,    1,      1;
for a total of a(6)=360 compositions of n=6. (End).
		

References

  • H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m).
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), this sequence (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).
Cf. A006497, A052906, A175182 (Pisano periods), A201001 (prime subsequence), A092936 (squares).
Cf. A243399.

Programs

  • GAP
    a:=[0,1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018
  • Haskell
    a006190 n = a006190_list !! n
    a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
    -- Reinhard Zumkeller, Feb 19 2011
    
  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n],n=0..30); # Emeric Deutsch, Sep 03 2007
    A006190:=-1/(-1+3*z+z**2); # Simon Plouffe in his 1992 dissertation, without the leading 0
    seq(combinat[fibonacci](n,3),n=0..30); # R. J. Mathar, Dec 07 2011
  • Mathematica
    a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *)
    LinearRecurrence[{3,1},{0,1},30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x,0,30}], x] (* Harvey P. Dale, Apr 20 2011 *)
    Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)
    Table[Fibonacci[n, 3], {n, 0, 30}] (* Vladimir Reshetnikov, May 08 2016 *)
  • PARI
    a(n)=if(n<1,0,contfracpnqn(vector(n,i,2+(i>1)))[2,1])
    
  • PARI
    a(n)=([1,3;1,2]^n)[2,1] \\ Charles R Greathouse IV, Mar 06 2014
    
  • PARI
    concat([0],Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013
    
  • Sage
    [lucas_number1(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/(1 - 3*x - x^2).
From Benoit Cloitre, Jun 14 2003: (Start)
a(3*n) = 2*A041019(5*n-1), a(3*n+1) = A041019(5*n), a(3*n+2) = A041019(5*n+3).
a(2*n) = 3*A004190(n-1); a(3*n) = 10*A041613(n-1) for n >= 1. (End)
From Gary W. Adamson, Jun 15 2003: (Start)
a(n-1) + a(n+1) = A006497(n).
A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)
a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003
a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004
a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]{1,1}. - _L. Edson Jeffery, Aug 27 2013]
From Paul Barry, May 21 2006: (Start)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).
a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)
E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006
a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.
Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007
1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008
a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A006497(n) + a(n)*sqrt(13))/2.
Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End)
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a(n)^2 + 4*(-1)^n))/2;
(2) a^2(n+1) - 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(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(n) = sqrt(13*(A006497(n))^2 + (-1)^(n-1)*52)/13. - Vladimir Shevelev, Mar 13 2013
Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));
(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(8) a(n) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;
(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)
From Kai Wang, Feb 10 2020: (Start)
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.
arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).
arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).
The same formula holds for Fibonacci numbers and Pell numbers. (End)
a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022
G.f. = 1/(1-Sum_{k>=1} P(k)*x^k), P(k)=A000129(k) (with a(0)=1). - Enrique Navarrete, Dec 17 2023
G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
Sum_{n>=0} a(n)/phi^(3*n) = 1, where phi = A001622 is the golden ratio. - Diego Rattaggi, Apr 07 2025

Extensions

Second formula corrected by Johannes W. Meijer, Jun 02 2010

A026597 Expansion of (1+x)/(1-x-4*x^2).

Original entry on oeis.org

1, 2, 6, 14, 38, 94, 246, 622, 1606, 4094, 10518, 26894, 68966, 176542, 452406, 1158574, 2968198, 7602494, 19475286, 49885262, 127786406, 327327454, 838473078, 2147782894, 5501675206, 14092806782, 36099507606, 92470734734
Offset: 0

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Author

Keywords

Comments

This sequence can generated by the following formula: a(n) = a(n-1) + 4*a(n-2) when n > 2; a[1] = 1, a[2] = 2. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 21 2004
An elephant sequence, see A175654 and A175655. For the corner squares just one A[5] vector, with decimal value 325, leads to the sequence given above. For the central square this vector leads to a companion sequence that is 4 times this very same sequence with n >= -1. - Johannes W. Meijer, Aug 15 2010
Equals INVERTi transform of A180168. - Gary W. Adamson, Aug 14 2010
Start with a single cell at coordinates (0, 0), then iteratively subdivide the grid into 2 X 2 cells and remove the cells that have one '1' in their modulo 3 coordinates. a(n) is the number of cells after n iterations. Cell configuration converges to a fractal with approximate dimension 1.357. - Peter Karpov, Apr 20 2017
Also, the number of walks of length n starting at vertex 1 in the graph with 4 vertices and edges {{0,1}, {0,2}, {0,3}, {1,2}, {2,3}}. - Sean A. Irvine, Jun 02 2025

Crossrefs

Programs

  • Magma
    [n le 2 select n else Self(n-1) + 4*Self(n-2): n in [1..41]]; // G. C. Greubel, Dec 08 2021
  • Mathematica
    LinearRecurrence[{1,4},{1,2},40] (* Harvey P. Dale, Nov 28 2011 *)
  • Sage
    [(2*i)^n*( chebyshev_U(n, -i/4) - (i/2)*chebyshev_U(n-1, -i/4) ) for n in (0..40)] # G. C. Greubel, Dec 08 2021
    

Formula

G.f.: (1+x)/(1-x-4*x^2).
a(n) = T(n,0) + T(n,1) + ... + T(n,2*n), T given by A026584.
a(n) = Sum_{k=0..n} binomial(floor((2*n-k-1)/2), n-k)*2^k. - Paul Barry, Feb 11 2005
a(n) = A006131(n) + A006131(n-1), n >= 1. - R. J. Mathar, Oct 20 2006
a(n) = Sum_{k=0..n} binomial(floor((2*n-k)/2),n-k)*4^floor(k/2). - Paul Barry, Feb 02 2007
Inverse binomial transform of A007482: (1, 3, 11, 39, 139, 495, ...). - Gary W. Adamson, Dec 04 2007
a(n) = Sum_{k=0..n+1} A122950(n+1,k)*3^(n+1-k). - Philippe Deléham, Jan 04 2008
a(n) = (1/2 + 3*sqrt(17)/34)*(1/2 + sqrt(17)/2)^n + (1/2 - 3*sqrt(17)/34)*(1/2 - sqrt(17)/2)^n. - Antonio Alberto Olivares, Jun 07 2011
a(n) = (2*i)^n*( chebyshevU(n, -i/4) - (i/2)*chebyshevU(n-1, -i/4) ). - G. C. Greubel, Dec 08 2021
E.g.f.: exp(x/2)*(17*cosh(sqrt(17)*x/2) + 3*sqrt(17)*sinh(sqrt(17)*x/2))/17. - Stefano Spezia, Jan 31 2023

Extensions

Better name from Ralf Stephan, Jul 14 2013

A105476 Number of compositions of n when each even part can be of two kinds.

Original entry on oeis.org

1, 1, 3, 6, 15, 33, 78, 177, 411, 942, 2175, 5001, 11526, 26529, 61107, 140694, 324015, 746097, 1718142, 3956433, 9110859, 20980158, 48312735, 111253209, 256191414, 589951041, 1358525283, 3128378406, 7203954255, 16589089473, 38200952238, 87968220657
Offset: 0

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Author

Emeric Deutsch, Apr 09 2005

Keywords

Comments

Row sums of A105475.
Starting (1, 3, 6, 15, ...) = sum of (n-1)-th row terms of triangle A140168. - Gary W. Adamson, May 10 2008
a(n) is also the number of compositions of n using 1's and 2's such that each run of like numbers can be grouped arbitrarily. For example, a(4) = 15 because 4 = (1)+(1)+(1)+(1) = (1+1)+(1)+(1) = (1)+(1+1)+(1) = (1)+(1)+(1+1) = (1+1)+(1+1) = (1+1+1)+(1) = (1)+(1+1+1) = (1+1+1+1) = (2)+(1)+(1) = (1)+(2)+(1) = (1)+(1)+(2) = (2)+(1+1) = (1+1)+(2) = (2)+(2) = (2+2). - Martin J. Erickson (erickson(AT)truman.edu), Dec 09 2008
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 69, 261, 321 and 324, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A006138. - Johannes W. Meijer, Aug 15 2010
Inverse INVERT transform of the left shifted sequence gives A000034.
Eigensequence of the triangle
1,
2, 1,
1, 2, 1,
2, 1, 2, 1,
1, 2, 1, 2, 1,
2, 1, 2, 1, 2, 1,
1, 2, 1, 2, 1, 2, 1,
2, 1, 2, 1, 2, 1, 2, 1 ... - Paul Barry, Feb 10 2011
Pisano period lengths: 1, 3, 1, 6, 24, 3, 24, 6, 1, 24, 120, 6, 156, 24, 24, 12, 16, 3, 90, 24, ... - R. J. Mathar, Aug 10 2012

Examples

			a(3)=6 because we have (3),(1,2),(1,2'),(2,1),(2',1) and (1,1,1).
		

Crossrefs

Programs

  • GAP
    a:=[1,3];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; Concatenation([1], a); # G. C. Greubel, Jan 15 2020
  • Magma
    I:=[1,1,3]; [n le 3 select I[n] else Self(n-1)+3*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Jul 21 2013
    
  • Magma
    R:=PowerSeriesRing(Integers(), 33); Coefficients(R!( 1/(1-(x/(1-x))-x^2/(1-x^2)))); // Marius A. Burtea, Jan 15 2020
    
  • Maple
    G:=(1-z^2)/(1-z-3*z^2): Gser:=series(G,z=0,35): 1,seq(coeff(Gser,z^n),n=1..33);
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x-3x^2), {x,0,35}], x] (* or *) Join[{1}, LinearRecurrence[{1, 3}, {1, 3}, 50]] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011; typo fixed by Vincenzo Librandi, Jul 21 2013 *)
    Table[Round[Sqrt[3]^(n-3)*(2*Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] +Fibonacci[n, 1/Sqrt[3]])], {n, 0, 40}] (* G. C. Greubel, Jan 15 2020 *)
  • PARI
    Vec((1-x^2)/(1-x-3*x^2)+O(x^40)) \\ Charles R Greathouse IV, Jun 13 2013
    
  • Sage
    def A105476_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1-x^2)/(1-x-3*x^2) ).list()
    A105476_list(40) # G. C. Greubel, Jan 15 2020
    

Formula

G.f.: (1-x^2) / (1-x-3*x^2).
a(n) = a(n-1) + 3*a(n-2) for n>=3.
a(n) = 3*A006138(n-2), n>=2.
a(n) = ((2+sqrt(13))*(1+sqrt(13))^n - (2-sqrt(13))*(1-sqrt(13))^n)/(3*2^n*sqrt(13)) for n>0. - Bruno Berselli, May 24 2011
G.f.: 1/(1 - Sum_{k>=1} x^k*(1+x^k) ). - Joerg Arndt, Mar 09 2014
G.f.: 1/(1 - (x/(1-x)) - x^2/(1-x^2)) = 1/(1 - (x+2*x^2+x^3+2*x^4+x^5+2*x^6+...) ); in general 1/(1 - Sum_{j>=1} m(j)*x^j ) is the g.f. for compositions with m(k) sorts of part k. - Joerg Arndt, Feb 16 2015
a(n) = 3^((n-1)/2)*( 2*sqrt(3)*Fibonacci(n, 1/sqrt(3)) + Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020
E.g.f.: 1/3 + (2/39)*exp(x/2)*(13*cosh((sqrt(13)*x)/2) + 2*sqrt(13)*sinh((sqrt(13)*x)/2)). - Stefano Spezia, Jan 15 2020

A175654 Eight bishops and one elephant on a 3 X 3 chessboard. G.f.: (1 - x - x^2)/(1 - 3*x - x^2 + 6*x^3).

Original entry on oeis.org

1, 2, 6, 14, 36, 86, 210, 500, 1194, 2822, 6660, 15638, 36642, 85604, 199626, 464630, 1079892, 2506550, 5811762, 13462484, 31159914, 72071654, 166599972, 384912086, 888906306, 2052031172, 4735527306, 10925175254, 25198866036, 58108609526, 133973643090
Offset: 0

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Author

Johannes W. Meijer, Aug 06 2010; edited Jun 21 2013

Keywords

Comments

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 or 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a bishop on the eight side and corner squares but on the center square the bishop flies into a rage and turns into a raging elephant.
In chaturanga, the old Indian version of chess, one of the pieces was called gaja, elephant in Sanskrit. The Arabs called the game shatranj and the elephant became el fil in Arabic. In Spain chess became chess as we know it today but surprisingly in Spanish a bishop isn't a Christian bishop but a Moorish elephant and it still goes by its original name of el alfil.
On a 3 X 3 chessboard there are 2^9 = 512 ways for an elephant to fly into a rage on the central square (off the center the piece behaves like a normal bishop). The elephant is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the corner squares the 512 elephants lead to 46 different elephant sequences, see the overview of elephant sequences and the crossreferences.
The sequence above corresponds to 16 A[5] vectors with decimal values 71, 77, 101, 197, 263, 269, 293, 323, 326, 329, 332, 353, 356, 389, 449 and 452. These vectors lead for the side squares to A000079 and for the central square to A175655.

References

  • Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.
  • David Hooper and Kenneth Whyld, The Oxford Companion to Chess, pp. 74, 366, 1992.

Crossrefs

Cf. Elephant sequences corner squares [decimal value A[5]]: A040000 [0], A000027 [16], A000045 [1], A094373 [2], A000079 [3], A083329 [42], A027934 [11], A172481 [7], A006138 [69], A000325 [26], A045623 [19], A000129 [21], A095121 [170], A074878 [43], A059570 [15], A175654 [71, this sequence], A026597 [325], A097813 [58], A057711 [27], 2*A094723 [23; n>=-1], A002605 [85], A175660 [171], A123203 [186], A066373 [59], A015518 [341], A134401 [187], A093833 [343].

Programs

  • Magma
    [n le 3 select Factorial(n) else 3*Self(n-1) +Self(n-2) -6*Self(n-3): n in [1..41]]; // G. C. Greubel, Dec 08 2021
    
  • Maple
    nmax:=28; m:=1; A[1]:=[0,0,0,0,1,0,0,0,1]: A[2]:=[0,0,0,1,0,1,0,0,0]: A[3]:=[0,0,0,0,1,0,1,0,0]: A[4]:=[0,1,0,0,0,0,0,1,0]: A[5]:=[0,0,1,0,0,0,1,1,1]: A[6]:=[0,1,0,0,0,0,0,1,0]: A[7]:=[0,0,1,0,1,0,0,0,0]: A[8]:=[0,0,0,1,0,1,0,0,0]: A[9]:=[1,0,0,0,1,0,0,0,0]: A:=Matrix([A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);
  • Mathematica
    LinearRecurrence[{3,1,-6}, {1,2,6}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2012 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -6,1,3]^n*[1;2;6])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
    
  • Sage
    [( (1-x-x^2)/((1-2*x)*(1-x-3*x^2)) ).series(x,n+1).list()[n] for n in (0..40)] # G. C. Greubel, Dec 08 2021

Formula

G.f.: (1 - x - x^2)/(1 - 3*x - x^2 + 6*x^3).
a(n) = 3*a(n-1) + a(n-2) - 6*a(n-3) with a(0)=1, a(1)=2 and a(2)=6.
a(n) = ((6+10*A)*A^(-n-1) + (6+10*B)*B^(-n-1))/13 - 2^n with A = (-1+sqrt(13))/6 and B = (-1-sqrt(13))/6.
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n)*2*A000244(n)/(A075118(n) - A006130(n-1)*sqrt(13)).
a(n) = b(n) - b(n-1) - b(n-2), where b(n) = Sum_{k=1..n} Sum_{j=0..k} binomial(j,n-3*k+2*j)*(-6)^(k-j)*binomial(k,j)*3^(3*k-n-j), n>0, b(0)=1, with a(0) = b(0), a(1) = b(1) - b(0). - Vladimir Kruchinin, Aug 20 2010
a(n) = 2*A006138(n) - 2^n = 2*(A006130(n) + A006130(n-1)) - 2^n. - G. C. Greubel, Dec 08 2021
E.g.f.: 2*exp(x/2)*(13*cosh(sqrt(13)*x/2) + 3*sqrt(13)*sinh(sqrt(13)*x/2))/13 - cosh(2*x) - sinh(2*x). - Stefano Spezia, Feb 12 2023

A063782 a(0) = 1, a(1) = 3; for n > 1, a(n) = 2*a(n-1) + 4*a(n-2).

Original entry on oeis.org

1, 3, 10, 32, 104, 336, 1088, 3520, 11392, 36864, 119296, 386048, 1249280, 4042752, 13082624, 42336256, 137003008, 443351040, 1434714112, 4642832384, 15024521216, 48620371968, 157338828800, 509159145472, 1647673606144
Offset: 0

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Author

Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Aug 17 2001

Keywords

Comments

Ratio of successive terms approaches sqrt(5) + 1.
From Sean A. Irvine, Jun 06 2025: (Start)
Also, number of walks of length n starting at vertex 1 in the following graph:
1---2
|\ /|
| 0 |
|/ \|
4---3. (End)

Examples

			As the INVERT transform of A006138, (1, 2, 5, 11, 26, 59, ...); a(4) = 104 = (26, 11, 5, 2, 1) dot (1, 1, 3, 10, 32) = (26 + 11 + 15 + 20 + 32).
		

Crossrefs

Cf. A006138. Row sums of A215244.

Programs

  • Maple
    a := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(2) fi: 2*a(n-1) + 4*a(n-2); end: for n from 1 to 50 do printf(`%d,`,a(n)+a(n-1)) od:
    f:=n-> simplify(expand((1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1 -sqrt(5))^n )); # N. J. A. Sloane, Aug 10 2012
  • Mathematica
    a[n_]:=(MatrixPower[{{1,5},{1,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
    LinearRecurrence[{2, 4}, {1, 3}, 100] (* G. C. Greubel, Feb 18 2017 *)
  • PARI
    { for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); if (n, write("b063782.txt", n, " ", a + a2)) ) } \\ Harry J. Smith, Aug 31 2009

Formula

For n >= 1, a(n) = 2^(n-1)*Fibonacci(n+3). - Vladeta Jovovic, Oct 25 2003
G.f.: (1 + x)/(1 - 2*x - 4*x^2). - R. J. Mathar, Feb 06 2010
Equals INVERT transform of A006138 and INVERTi transform of A179606. - Gary W. Adamson, Aug 14 2010
a(n) = (1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1-sqrt(5))^n. - Alexander R. Povolotsky, Aug 15 2010
It follows that a(n) is the nearest integer to (and is increasingly close to) (1/2 + 1/sqrt(5))*(1+sqrt(5))^n. - N. J. A. Sloane, Aug 10 2012
a(n) = A063727(n) + A063727(n-1).
a(n) = M^n(1, 1), with the matrix M= [[3, 1], [1, -1]]. Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S = A049310 and F = A000045. Motivated by A319053. - Wolfdieter Lang, Oct 08 2018

Extensions

More terms from James Sellers, Sep 25 2001
Edited (new offset, new initial term, etc.) by N. J. A. Sloane, Aug 19 2010

A322942 Jacobsthal triangle, coefficients of orthogonal polynomials J(n, x) where J(n, 0) are the Jacobsthal numbers (A001045 with a(0) = 1). Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 5, 3, 1, 5, 12, 10, 4, 1, 11, 27, 28, 16, 5, 1, 21, 62, 75, 52, 23, 6, 1, 43, 137, 193, 159, 85, 31, 7, 1, 85, 304, 480, 456, 290, 128, 40, 8, 1, 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1, 341, 1442, 2793, 3336, 2750, 1652, 742, 248, 61, 10, 1
Offset: 0

Views

Author

Peter Luschny, Jan 03 2019

Keywords

Comments

The name 'Jacobsthal triangle' used here is not standard.

Examples

			The first few polynomials are:
  J(0, x) = 1;
  J(1, x) = x + 1;
  J(2, x) = x^2 + 2*x + 1;
  J(3, x) = x^3 + 3*x^2 +  5*x + 3;
  J(4, x) = x^4 + 4*x^3 + 10*x^2 + 12*x + 5;
  J(5, x) = x^5 + 5*x^4 + 16*x^3 + 28*x^2 + 27*x + 11;
  J(6, x) = x^6 + 6*x^5 + 23*x^4 + 52*x^3 + 75*x^2 + 62*x + 21;
The triangle starts:
  [0]   1;
  [1]   1,   1;
  [2]   1,   2,    1;
  [3]   3,   5,    3,    1;
  [4]   5,  12,   10,    4,   1;
  [5]  11,  27,   28,   16,   5,   1;
  [6]  21,  62,   75,   52,  23,   6,   1;
  [7]  43, 137,  193,  159,  85,  31,   7,  1;
  [8]  85, 304,  480,  456, 290, 128,  40,  8, 1;
  [9] 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1;
		

Crossrefs

Row sums are A152035, alternating row sums are A000007, values at x=1/2 are A323232, values at x=0 (first column) are A152046.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
    g:= func< n,x | (&+[Binomial(n-k,k)*2^k*(x+1)^(n-2*k): k in [0..Floor(n/2)]]) >;
    f:= func< n,x | n le 1 select (x+1)^n else g(n,x) - 2*g(n-2,x) >;
    A322942:= func< n,k | Coefficient(R!( f(n,x) ), k) >;
    [A322942(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 20 2023
  • Maple
    J := proc(n) option remember;
    `if`(n < 3, [1, x+1, x^2 + 2*x + 1][n+1], (x+1)*J(n-1) + 2*J(n-2));
    sort(expand(%)) end: seq(print(J(n)), n=0..11); # Computes the polynomials.
    seq(seq(coeff(J(n), x, k), k=0..n), n=0..11);
  • Mathematica
    J[n_, x_]:= J[n, x]= If[n<3, (x+1)^n, (x+1)*J[n-1, x] + 2*J[n-2, x]];
    T[n_, k_]:= Coefficient[J[n, x], x, k];
    Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 17 2019 *)
    (* Second program *)
    A322942[n_, k_]:= Coefficient[Series[Boole[n==0] + (I*Sqrt[2])^n*(ChebyshevU[n, (x+1)/(2*Sqrt[2]*I)] + ChebyshevU[n-2, (x+ 1)/(2*Sqrt[2]*I)]), {x,0,50}], x, k];
    Table[A322942[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 20 2023 *)
  • Sage
    # use[riordan_square from A321620]
    riordan_square((2*x^2 - 1)/((x + 1)*(2*x - 1)), 9)
    

Formula

J(n, x) = (x+1)*J(n-1, x) + 2*J(n-2, x) for n >= 3.
T(n, k) = [x^k] J(n, x).
Equals the Riordan square (cf. A321620) generated by (2*x^2-1)/((x + 1)*(2*x - 1)).
Sum_{k=0..n} T(n, k) = A152035(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
From G. C. Greubel, Sep 20 2023: (Start)
T(n, k) = [x^k]( [n=0] + (i*sqrt(2))^n*(ChebyshevU(n, (x+1)/(2*sqrt(2)*i)) + ChebyshevU(n-2, (x+1)/(2*sqrt(2)*i))) ).
G.f.: (1 - 2*t^2)/(1 - (x+1)*t - 2*t^2).
Sum_{k=0..floor(n/2)} T(n-k, k) = (2/3)*[n=0] + A006138(n-1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = 2*[n=0] + Fibonacci(n-2). (End)

A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 7, 15, 9, 1, 10, 36, 54, 27, 1, 13, 66, 162, 189, 81, 1, 16, 105, 360, 675, 648, 243, 1, 19, 153, 675, 1755, 2673, 2187, 729, 1, 22, 210, 1134, 3780, 7938, 10206, 7290, 2187, 1, 25, 276, 1764, 7182, 19278, 34020, 37908, 24057, 6561, 1, 28, 351, 2592, 12474, 40824, 91854, 139968, 137781, 78732, 19683
Offset: 0

Views

Author

N. J. A. Sloane, May 03 2000

Keywords

Comments

Triangle T(n,k), 0<=k<=n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 10 2005
Triangle read by rows, n-th row = X^(n-1) * [1, 1, 0, 0, 0, ...] where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (3,3,3,...) in the subdiagonal; given row 0 = 1. - Gary W. Adamson, Jul 19 2008
Fusion of polynomial sequences P and Q given by p(n,x)=(x+2)^n and q(n,x)=(2x+1)^n; see A193722 for the definition of fusion of two sequences of polynomials or triangular arrays. - Clark Kimberling, Aug 04 2011

Examples

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  7,  15,   9;
  1, 10,  36,  54,   27;
  1, 13,  66, 162,  189,   81;
  1, 16, 105, 360,  675,  648,  243;
  1, 19, 153, 675, 1755, 2673, 2187, 729;
		

Crossrefs

Programs

  • Magma
    A038763:= func< n,k | n eq 0 select 1 else 3^(k-1)*(3*n-2*k)*Binomial(n,k)/n >;
    [A038763(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 27 2023
    
  • Mathematica
    A038763[n_,k_]:= If[n==0, 1, 3^(k-1)*(3*n-2*k)*Binomial[n,k]/n];
    Table[A038763[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2023 *)
  • PARI
    T(n,k) = if ((n<0) || (k<0), return(0)); if ((n==0) && (k==0), return(1)); if (n==1, if (k<=1, return(1))); T(n-1,k) + 3*T(n-1,k-1);
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", "))); \\ Michel Marcus, Jul 25 2023
    
  • SageMath
    def A038763(n,k): return 1 if (n==0) else 3^(k-1)*(3*n-2*k)*binomial(n,k)/n
    flatten([[A038763(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 27 2023

Formula

T(n, 0)=1; T(1, 1)=1; T(n, k)=0 for k>n; T(n, k) = T(n-1, k-1)*3 + T(n-1, k) for n >= 2.
Sum_{k=0..n} T(n,k) = A081294(n). - Philippe Deléham, Sep 22 2006
T(n, k) = A136158(n, n-k). - Philippe Deléham, Dec 17 2007
G.f.: (1-2*x*y)/(1-(3*y+1)*x). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Dec 27 2023: (Start)
T(n, 0) = A000012(n).
T(n, 1) = A016777(n-1).
T(n, 2) = A062741(n-1).
T(n, 3) = 9*A002411(n-2).
T(n, 4) = 27*A001296(n-3).
T(n, 5) = 81*A051836(n-4).
T(n, n) = A133494(n).
T(n, n-1) = A006234(n+2).
T(n, n-2) = A080420(n-2).
T(n, n-3) = A080421(n-3).
T(n, n-4) = A080422(n-4).
T(n, n-5) = A080423(n-5).
T(2*n, n) = 4*A098399(n-1) + (2/3)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A006138(n-1) + (2/3)*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A110523(n-1) + (4/3)*[n=0]. (End)

Extensions

More terms from Michel Marcus, Jul 25 2023

A274977 a(n) = a(n-1) + 3*a(n-2) with n>1, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 9, 27, 54, 135, 297, 702, 1593, 3699, 8478, 19575, 45009, 103734, 238761, 549963, 1266246, 2916135, 6714873, 15463278, 35607897, 81997731, 188821422, 434814615, 1001278881, 2305722726, 5309559369, 12226727547, 28155405654, 64835588295, 149301805257, 343808570142
Offset: 0

Views

Author

Bruno Berselli, Sep 13 2016

Keywords

Comments

a(n)/a(n+1) converges to 1/A209927 as n approaches infinity.

Examples

			Table of similar sequences (not extendable on the left side) where this recurrence can be applied to the first two terms:
----------------------------------------------------------------------
(*)      -  -  1, -1,  2, -1,  5,   2,  17,  23,   74,  143,  365, ...
A052533: -  -  1,  0,  3,  3, 12,  21,  57, 120,  291,  651, 1524, ...
(^)      -  0, 1,  1,  4,  7, 19,  40,  97, 217,  508, 1159, 2683, ...
A006138: -  -  1,  2,  5, 11, 26,  59, 137, 314,  725, 1667, 3842, ...
A105476: -  -  1,  3,  6, 15, 33,  78, 177, 411,  942, 2175, 5001, ...
(^)      0, 1, 1,  4,  7, 19, 40,  97, 217, 508, 1159, 2683, 6160, ...
A105963: -  -  1,  5,  8, 23, 47, 116, 257, 605, 1376, 3191, 7319, ...
A274977: -  -  1,  6,  9, 27, 54, 135, 297, 702, 1593, 3699, 8478, ...
A075118: -  2, 1,  7, 10, 31, 61, 154, 337, 799, 1810, 4207, 9637, ...
----------------------------------------------------------------------
(*) see version A140165.
(^) see A006130 and the signed versions A140167, A182228.
		

Crossrefs

Programs

  • GAP
    a:=[1,6];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020
  • Magma
    [n le 2 select 5*n-4 else Self(n-1)+3*Self(n-2): n in [1..40]];
    
  • Magma
    R:=PowerSeriesRing(Integers(), 32); Coefficients(R!((1 + 5*x)/(1- x-3*x^2))); // Marius A. Burtea, Jan 15 2020
    
  • Maple
    seq(coeff(series((1+5*x)/(1-x-3*x^2), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 15 2020
  • Mathematica
    RecurrenceTable[{a[n]==a[n-1] +3a[n-2], a[0]==1, a[1]==6}, a, {n,0,40}]
    Table[Round[Sqrt[3]^(n-1)*(Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] + 5*Fibonacci[n, 1/Sqrt[3]])], {n,0,40}] (* G. C. Greubel, Jan 15 2020 *)
    LinearRecurrence[{1,3},{1,6},40] (* Harvey P. Dale, Jul 11 2023 *)
  • PARI
    v=vector(40); v[1]=1; v[2]=6; for(n=3, #v, v[n]=v[n-1]+3*v[n-2]); v
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2
    a = recur_gen2(1, 6, 1, 3)
    [next(a) for n in range(40)]
    

Formula

G.f.: (1 + 5*x)/(1 - x - 3*x^2).
a(n) = ((13 + 11*sqrt(13))*(1 + sqrt(13))^n + (13 - 11*sqrt(13))*(1 - sqrt(13))^n)/(26*2^n).
3*a(n) + a(n+1) = 9*A105476(n+1).
3*a(n) - a(n+1) = 27*A006130(n-3) with n>1, A006130(-1) = 0.
a(n+1) - a(n) = 27*A105476(n-3) with n>2.
a(n) = 3^((n-1)/2)*( sqrt(3)*Fibonacci(n+1, 1/sqrt(3)) + 5*Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020
E.g.f.: (1/13)*exp(x/2)*(13*cosh((sqrt(13)*x)/2) + 11*sqrt(13)*sinh((sqrt(13)*x)/2)). - Stefano Spezia, Jan 15 2020
Showing 1-10 of 14 results. Next