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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A005251 a(0) = 0, a(1) = a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

0, 1, 1, 1, 2, 4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426, 525456, 922111, 1618192, 2839729, 4983377, 8745217, 15346786, 26931732, 47261895, 82938844, 145547525, 255418101, 448227521
Offset: 0

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Author

Keywords

Comments

a(n+3) is the number of n-bit sequences that avoid 010. Example: For n=4 the 12 sequences are all 4-bit sequences except 0100, 0101, 0010, 1010. - David Callan, Mar 25 2004
a(n+2) is the number of compositions (ordered partitions) of n where no two adjacent parts are != 1, see example. - Joerg Arndt, Jan 26 2013
a(n+1) is the number of compositions of n avoiding the part 2. - Joerg Arndt, Jul 13 2014
Number of different positive braids with n crossings of 3 strands.
This is a_2(n) in the Doroslovacki reference. Note that there is a typo in the paper in the formula for a_2(n): the upper bound in the inner sum should be "n-i" not "i-1". - Max Alekseyev, Jun 26 2007
a(n) is the number of peakless Motzkin paths of length n-1 with no UHH...HD's starting at level > 0 (here n > 0 and U=(1,1), H=(1,0), D=(1,-1)). Example: a(5)=7 because from all 8 peakless Motzkin paths of length 5 (see A004148) only UUHDD does not qualify. - Emeric Deutsch, Sep 13 2004
Equals the INVERT transform of (1, 0, 1, 1, 1, ...); equivalent to a(n) = a(n-1) + a(n-3) + a(n-4) + ... - Gary W. Adamson, Apr 27 2009
a(n) is the number of length n-1 words on {0,1} such that each string of 1's is followed by a string of at least two 0's. For example, a(5) = 4 because we have: 0000, 0100, 1000, and 1100. - Geoffrey Critzer, Aug 09 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 0; 0, 1, 1; 1, 0, 0] or [1, 0, 1; 1, 1, 0; 0, 1, 0] or [1, 1, 0; 0, 0, 1; 1, 0, 1] or [1, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
For n >= 2, a(n) is the number of (n-2)-length binary words with no isolated zeros. - Milan Janjic, Mar 07 2015
Apart from the first three terms, the total number of bargraphs of semiperimeter n of height at most two for n >= 2 starts 1, 2, 4, 7, 12, ... - Arnold Knopfmacher, Nov 02 2016
Number of DD-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are DD-equivalent iff the positions of pattern DD are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Nov 25 2019: (Start)
For n > 0, also the number of subsets of {1, ..., n - 3} such that if x and x + 2 are both in the subset, then so is x + 1. For example, the a(3) = 1 through a(7) = 12 subsets are:
{} {} {} {} {}
{1} {1} {1} {1}
{2} {2} {2}
{1,2} {3} {3}
{1,2} {4}
{2,3} {1,2}
{1,2,3} {1,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
(End)
The two-dimensional version, which counts sets of pairs where, if two pairs are separated by graph-distance 2, then the intermediate pair or pairs are also in the set, is A329871. - Gus Wiseman, Nov 30 2019
a(n+1) is the number of ways to tile a strip of length n with squares, dominoes, and tetrominoes, where the first tile cannot be a domino. - Greg Dresden and Myanna Nash, Aug 18 2020

Examples

			From _Joerg Arndt_, Jan 26 2013: (Start)
The a(5+2) = 12 compositions of 5 where no two adjacent parts are != 1 are
  [ 1]  [ 1 1 1 1 1 ]
  [ 2]  [ 1 1 1 2 ]
  [ 3]  [ 1 1 2 1 ]
  [ 4]  [ 1 1 3 ]
  [ 5]  [ 1 2 1 1 ]
  [ 6]  [ 1 3 1 ]
  [ 7]  [ 1 4 ]
  [ 8]  [ 2 1 1 1 ]
  [ 9]  [ 2 1 2 ]
  [10]  [ 3 1 1 ]
  [11]  [ 4 1 ]
  [12]  [ 5 ]
(End)
G.f. = x + x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 12*x^7 + 21*x^8 + 37*x^9 + ...
		

References

  • S. Burckel, Efficient methods for three strand braids (submitted). [Apparently unpublished]
  • P. Chinn and S. Heubach, "Compositions of n with no occurrence of k", Congressus Numeratium, 2002, v. 162, pp. 33-51.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 205.
  • R. K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisection of Padovan sequence A000931.
Partial sums of A005314 shifted 3 times to the right, if we assume A005314(0) = 1.
Compositions without adjacent equal parts are A003242.
Compositions without isolated parts are A114901.
Row sums of A097230(n-2) for n>1.

Programs

  • Haskell
    a005251 n = a005251_list !! n
    a005251_list = 0 : 1 : 1 : 1 : zipWith (+) a005251_list
       (drop 2 $ zipWith (+) a005251_list (tail a005251_list))
    -- Reinhard Zumkeller, Dec 28 2011
    
  • Magma
    I:=[0,1,1,1]; [n le 4 select I[n] else Self(n-1)+Self(n-2)+Self(n-4): n in [1..45]]; // Vincenzo Librandi, Nov 30 2018
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1-x)/(1-2*x + x^2 - x^3) )); // Marius A. Burtea, Oct 24 2019
    
  • Maple
    A005251 := proc(n) option remember; if n <= 2 then n elif n = 3 then 4 else 2*A005251(n - 1) - A005251(n - 2) + A005251(n - 3); fi; end;
    A005251:=(-1+z)/(-1+2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation
    a := n -> `if`(n<=1, n, hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4)):
    seq(simplify(a(n)), n=0..36); # Peter Luschny, Apr 08 2018
  • Mathematica
    LinearRecurrence[{2,-1,1},{0,1,1},40]  (* Harvey P. Dale, May 05 2011 *)
    a[ n_]:= If[n<0, SeriesCoefficient[ -x(1-x)/(1 -x + 2x^2 -x^3), {x, 0, -n}], SeriesCoefficient[ x(1-x)/(1 -2x +x^2 -x^3), {x, 0, n}]] (* Michael Somos, Dec 13 2013 *)
    a[0] = 1; a[1] = a[2] = 0; a[n_] := a[n] = a[n-2] + a[n-3]; Table[a[2 n-1], {n, 1, 20}] (* Rigoberto Florez, Oct 15 2019 *)
    Table[If[n==0,0,Length[DeleteCases[Subsets[Range[n-3]],{_,x_,y_,_}/;x+2==y]]],{n,0,10}] (* Gus Wiseman, Nov 25 2019 *)
  • PARI
    Vec((1-x)/(1-2*x+x^2-x^3)+O(x^99)) /* Charles R Greathouse IV, Nov 20 2012 */
    
  • PARI
    {a(n) = if( n<0, polcoeff( -x*(1-x)/(1 -x +2*x^2 -x^3) + x*O(x^-n), -n), polcoeff( x*(1-x)/(1 -2*x +x^2 -x^3) + x*O(x^n), n))} /* Michael Somos, Dec 13 2013 */
    
  • SageMath
    [sum( binomial(n-j-1, 2*j) for j in (0..floor((n-1)/3)) ) for n in (0..50)] # G. C. Greubel, Apr 13 2022

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-3).
G.f.: z*(1-z)/(1 - 2*z + z^2 - z^3). - Emeric Deutsch, Sep 13 2004
23*a_n = 3*P_{2n+1} + 7*P_{2n} - 2*P_{2n-1}, where P_n are the Perrin numbers, A001608. - Don Knuth, Dec 09 2008
a(n+1) = Sum_{k=0..n} binomial(n-k, 2k). - Richard L. Ollerton, May 12 2004
From Henry Bottomley, Feb 21 2001: (Start)
a(n) = (Sum_{j
a(n) = A005314(n) - A005314(n-1).
a(n) = A049853(n-1) - a(n-1).
a(n) = A005314(n) - a(n-2). (End)
Conjecture: a(n+1) + |A078065(n)| = 2*A005314(n+1). - Creighton Dement, Dec 21 2004
a(n+2) has g.f. (F_3(-x) + F_2(-x))/(F_4(-x) + F_3(-x)) = 1/(-x+1/(-x+1/(-x+1))) where F_n(x) is the n-th Fibonacci polynomial; see A011973. - Qiaochu Yuan (qchu(AT)mit.edu), Feb 19 2009
a(n) = A173022(2^(n-2) - 1) for n > 1. - Reinhard Zumkeller, Feb 07 2010
BINOMIAL transform of A176971 is a(n+1). - Michael Somos, Dec 13 2013
a(n) = hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4) for n > 1. - Peter Luschny, Apr 08 2018
G.f.: z/(1-z-z^3-z^4-z^5-...) for the compositions of n-1 avoiding 2. The g.f. for the number of compositions of n avoiding the part k is 1/(1-z-...-z^(k-1) - z^(k+1)-...). - Gregory L. Simay, Sep 09 2018
If p,q,r are the three solutions to x^3 = 2x^2 - x + 1, then a(n) = (p-1)*p^n/((p-q)*(p-r)) + (q-1)*q^n/((q-p)*(q-r)) + (r-1)*r^n/((r-p)*(r-q)). - Greg Dresden and AnXing Yang, Aug 12 2025

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
Offset: 0

Keywords

Comments

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

Examples

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 117.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182-183.

Crossrefs

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.

Programs

  • Haskell
    a011973 n k = a011973_tabf !! n !! k
    a011973_row n = a011973_tabf !! n
    a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
    -- Reinhard Zumkeller, Jul 14 2015
  • Maple
    a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
    T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013
  • Mathematica
    (* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
    (* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
    (* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
    Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
    CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
    CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};
    
  • Sage
    # Prints the table; cf. A145574.
    for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012
    

Formula

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A350842 Number of integer partitions of n with no difference -2.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 9, 12, 16, 24, 30, 40, 54, 69, 89, 118, 146, 187, 239, 297, 372, 468, 575, 711, 880, 1075, 1314, 1610, 1947, 2359, 2864, 3438, 4135, 4973, 5936, 7090, 8466, 10044, 11922, 14144, 16698, 19704, 23249, 27306, 32071, 37639, 44019, 51457, 60113
Offset: 0

Author

Gus Wiseman, Jan 20 2022

Keywords

Examples

			The a(1) = 1 through a(7) = 12 partitions:
  (1)  (2)   (3)    (4)     (5)      (6)       (7)
       (11)  (21)   (22)    (32)     (33)      (43)
             (111)  (211)   (41)     (51)      (52)
                    (1111)  (221)    (222)     (61)
                            (2111)   (321)     (322)
                            (11111)  (411)     (511)
                                     (2211)    (2221)
                                     (21111)   (3211)
                                     (111111)  (4111)
                                               (22111)
                                               (211111)
                                               (1111111)
		

Crossrefs

Heinz number rankings are in parentheses below.
The version for no difference 0 is A000009.
The version for subsets of prescribed maximum is A005314.
The version for all differences < -2 is A025157, non-strict A116932.
The version for all differences > -2 is A034296, strict A001227.
The opposite version is A072670.
The version for no difference -1 is A116931 (A319630), strict A003114.
The multiplicative version is A350837 (A350838), strict A350840.
The strict case is A350844.
The complement for quotients is counted by A350846 (A350845).
A000041 = integer partitions.
A027187 = partitions of even length.
A027193 = partitions of odd length (A026424).
A323092 = double-free partitions (A320340), strict A120641.
A325534 = separable partitions (A335433).
A325535 = inseparable partitions (A335448).
A350839 = partitions with a gap and conjugate gap (A350841).

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],FreeQ[Differences[#],-2]&]],{n,0,30}]

A350839 Number of integer partitions of n with a difference < -1 and a conjugate difference < -1.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 7, 11, 17, 26, 39, 54, 81, 108, 148, 201, 269, 353, 467, 601, 779, 995, 1272, 1605, 2029, 2538, 3171, 3941, 4881, 6012, 7405, 9058, 11077, 13478, 16373, 19817, 23953, 28850, 34692, 41599, 49802, 59461, 70905, 84321, 100155, 118694
Offset: 0

Author

Gus Wiseman, Jan 24 2022

Keywords

Comments

We define a difference of a partition to be a difference of two adjacent parts.

Examples

			The a(5) = 1 through a(10) = 17 partitions:
  (311)  (411)   (511)    (422)     (522)      (622)
         (3111)  (4111)   (611)     (711)      (811)
                 (31111)  (3311)    (4221)     (4222)
                          (4211)    (4311)     (4411)
                          (5111)    (5211)     (5221)
                          (41111)   (6111)     (5311)
                          (311111)  (33111)    (6211)
                                    (42111)    (7111)
                                    (51111)    (42211)
                                    (411111)   (43111)
                                    (3111111)  (52111)
                                               (61111)
                                               (331111)
                                               (421111)
                                               (511111)
                                               (4111111)
                                               (31111111)
		

Crossrefs

Allowing -1 gives A144300 = non-constant partitions.
Taking one of the two conditions gives A239955, ranked by A073492, A065201.
These partitions are ranked by A350841.
A000041 = integer partitions, strict A000009.
A034296 = flat (contiguous) partitions, strict A001227.
A073491 = numbers whose prime indices have no gaps, strict A137793.
A090858 = partitions with a single hole, ranked by A325284.
A116931 = partitions with differences != -1, strict A003114.
A116932 = partitions with differences != -1 or -2, strict A025157.
A277103 = partitions with the same number of odd parts as their conjugate.
A350837 = partitions with no adjacent doublings, strict A350840.
A350842 = partitions with differences != -2, strict A350844, sets A005314.

Programs

  • Mathematica
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Table[Length[Select[IntegerPartitions[n],(Min@@Differences[#]<-1)&&(Min@@Differences[conj[#]]<-1)&]],{n,0,30}]

A350844 Number of strict integer partitions of n with no difference -2.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 3, 4, 4, 7, 7, 8, 11, 12, 15, 18, 21, 23, 31, 32, 40, 45, 54, 59, 73, 78, 94, 106, 122, 136, 161, 177, 203, 231, 259, 293, 334, 372, 417, 476, 525, 592, 663, 742, 821, 931, 1020, 1147, 1271, 1416, 1558, 1752, 1916, 2137, 2357, 2613, 2867
Offset: 0

Author

Gus Wiseman, Jan 21 2022

Keywords

Examples

			The a(1) = 1 through a(12) = 11 partitions (A..C = 10..12):
  1   2   3    4   5    6     7    8     9     A      B     C
          21       32   51    43   62    54    73     65    84
                   41   321   52   71    63    82     74    93
                              61   521   72    91     83    A2
                                         81    541    92    B1
                                         432   721    A1    543
                                         621   4321   632   651
                                                      821   732
                                                            741
                                                            921
                                                            6321
		

Crossrefs

The version for no difference 0 is A000009.
The version for no difference > -2 is A001227, non-strict A034296.
The version for no difference -1 is A003114 (A325160).
The version for subsets of prescribed maximum is A005314.
The version for all differences < -2 is A025157, non-strict A116932.
The opposite version is A072670.
The multiplicative version is A350840, non-strict A350837 (A350838).
The non-strict version is A350842.
A000041 counts integer partitions.
A027187 counts partitions of even length.
A027193 counts partitions of odd length (A026424).
A116931 counts partitions with no difference -1 (A319630).
A323092 counts double-free integer partitions (A320340) strict A120641.
A325534 counts separable partitions (A335433).
A325535 counts inseparable partitions (A335448).

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],FreeQ[Differences[#],0|-2]&]],{n,0,30}]

A060945 Number of compositions (ordered partitions) of n into 1's, 2's and 4's.

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 18, 31, 55, 96, 169, 296, 520, 912, 1601, 2809, 4930, 8651, 15182, 26642, 46754, 82047, 143983, 252672, 443409, 778128, 1365520, 2396320, 4205249, 7379697, 12950466, 22726483, 39882198, 69988378, 122821042, 215535903, 378239143, 663763424, 1164823609
Offset: 0

Author

Len Smiley, May 07 2001

Keywords

Comments

Diagonal sums of A038137. - Paul Barry, Oct 24 2005
From Gary W. Adamson, Oct 28 2010: (Start)
INVERT transform of the aerated Fibonacci sequence (1, 0, 1, 0, 2, 0, 3, 0, 5, ...).
a(n) = term (4,4) in the n-th power of the matrix [0,1,0,0; 0,0,1,0; 0,0,0,1; 1,0,1,1]. (End)
Number of permutations satisfying -k <= p(i)-i <= r and p(i)-i not in I, i=1..n, with k=1, r=3, I={2}. - Vladimir Baltic, Mar 07 2012
Number of compositions of n if the summand 2 is frozen in place or equivalently, if the ordering of the summand 2 does not count. - Gregory L. Simay, Jul 18 2016
a(n) - a(n-2) = number of compositions of n with no 2's = A005251(n+1). - Gregory L. Simay, Jul 18 2016
In general, the number of compositions of n with summand k frozen in place is equal to the number of compositions of n with only summands 1,...,k,2k. - Gregory L. Simay, May 10 2017
In the same way that the sum of any two alternating terms of A006498 produces a term from A000045 (the Fibonacci sequence), so it could be thought of as a "meta-Fibonacci," and the sum of any two alternating terms of A013979 produces a term from A000930 (Narayana’s cows), so it could analogously be called "meta-Narayana’s cows," this sequence embeds (can generate) A000931 (the Padovan sequence), as the odd terms of A000931 are generated by the sum of successive elements (e.g. 1+2=3, 2+3=5, 3+6=9, 6+10=16) and its even terms are generated by the difference of "supersuccessive" (second-order successive or "alternating," separated by a single other term) terms (e.g. 10-3=7, 18-6=12, 31-10=21, 55-18=37) — or, equivalently, adding "supersupersuccessive" terms (separated by 2 other terms, e.g. 1+6=7, 2+10=12, 3+18=21, 6+31=37) — so it could be dubbed the "metaPadovan." - Michael Cohen and Yasuyuki Kachi, Jun 13 2024

Examples

			There are 18=a(6) compositions of 6 with the summand 2 frozen in place: (6), (51), (15), (4[2]), (33), (411), (141), (114), (3[2]1), (1[2]3), ([222]), (3111), (1311), (1131), (1113), ([22]11), ([2]1111), (111111). Equivalently, the position of the summand 2 does not affect the composition count. For example, (321)=(231)=(312) and (123)=(213)=(132).
		

Crossrefs

Cf. A000045 (1's and 2's only), A023359 (all powers of 2)
Same as unsigned version of A077930.
All of A060945, A077930, A181532 are variations of the same sequence. - N. J. A. Sloane, Mar 04 2012

Programs

  • Haskell
    a060945 n = a060945_list !! (n-1)
    a060945_list = 1 : 1 : 2 : 3 : 6 : zipWith (+) a060945_list
       (zipWith (+) (drop 2 a060945_list) (drop 3 a060945_list))
    -- Reinhard Zumkeller, Mar 23 2012
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40);
    Coefficients(R!( 1/(1-x-x^2-x^4) )); // G. C. Greubel, Apr 09 2021
    
  • Maple
    m:= 40; S:= series( 1/(1-x-x^2-x^4), x, m+1);
    seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Apr 09 2021
  • Mathematica
    LinearRecurrence[{1,1,0,1}, {1,1,2,3}, 39] (* or *)
    CoefficientList[Series[1/(1-x-x^2-x^4), {x, 0, 38}], x] (* Michael De Vlieger, May 10 2017 *)
  • PARI
    N=66; my(x='x+O('x^N));
    Vec(1/(1-x-x^2-x^4))
    /* Joerg Arndt, Oct 21 2012 */
    
  • Sage
    def A060945_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( 1/(1-x-x^2-x^4) ).list()
    A060945_list(40) # G. C. Greubel, Apr 09 2021

Formula

a(n) = a(n-1) + a(n-2) + a(n-4).
G.f.: 1 / (1 - x - x^2 - x^4).
a(n) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} C(i, n-k-i)*C(2*i-n+k, 3*k-2*n+2*i). - Paul Barry, Oct 24 2005
a(2n) = A238236(n), a(2n+1) = A097472(n). - Philippe Deléham, Feb 20 2014
a(n) + a(n+1) = A005314(n+2). - R. J. Mathar, Jun 17 2020

Extensions

a(0) = 1 prepended by Joerg Arndt, Oct 21 2012

A350841 Heinz numbers of integer partitions with a difference < -1 and a conjugate difference < -1.

Original entry on oeis.org

20, 28, 40, 44, 52, 56, 63, 68, 76, 80, 84, 88, 92, 99, 100, 104, 112, 116, 117, 124, 126, 132, 136, 140, 148, 152, 153, 156, 160, 164, 168, 171, 172, 176, 184, 188, 189, 196, 198, 200, 204, 207, 208, 212, 220, 224, 228, 232, 234, 236, 244, 248, 252, 260, 261
Offset: 1

Author

Gus Wiseman, Jan 26 2022

Keywords

Comments

We define a difference of a partition to be a difference of two adjacent parts.

Examples

			The terms together with their prime indices begin:
   20: (3,1,1)
   28: (4,1,1)
   40: (3,1,1,1)
   44: (5,1,1)
   52: (6,1,1)
   56: (4,1,1,1)
   63: (4,2,2)
   68: (7,1,1)
   76: (8,1,1)
   80: (3,1,1,1,1)
   84: (4,2,1,1)
   88: (5,1,1,1)
   92: (9,1,1)
   99: (5,2,2)
		

Crossrefs

Heinz number rankings are in parentheses below.
Taking just one condition gives (A073492) and (A065201), counted by A239955.
These partitions are counted by A350839.
A000041 = integer partitions, strict A000009.
A034296 = partitions with no gaps (A073491), strict A001227 (A073485).
A090858 = partitions with a single gap of size 1 (A325284).
A116931 = partitions with no successions (A319630), strict A003114.
A116932 = partitions with no successions or gaps of size 1, strict A025157.
A350842 = partitions with no gaps of size 1, strict A350844, sets A005314.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Select[Range[100],(Min@@Differences[Reverse[primeMS[#]]]<-1)&&(Min@@Differences[conj[primeMS[#]]]<-1)&]

A118430 Number of binary sequences of length n containing exactly one subsequence 010.

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 22, 47, 98, 199, 396, 777, 1508, 2900, 5534, 10492, 19782, 37119, 69358, 129118, 239578, 443229, 817822, 1505389, 2764986, 5068435, 9273928, 16940488, 30897020, 56271128, 102347564, 185922589, 337353688, 611462514
Offset: 0

Author

Emeric Deutsch, Apr 27 2006

Keywords

Comments

With only two 0's at the beginning, the convolution of A005314 with itself. Column 1 of A118429.

Examples

			a(4) = 4 because we have 0100, 0101, 0010 and 1010.
		

Crossrefs

Programs

  • Maple
    g:=z^3/(1-2*z+z^2-z^3)^2: gser:=series(g,z=0,40): seq(coeff(gser,z,n),n=0..38);
  • Mathematica
    LinearRecurrence[{4, -6, 6, -5, 2, -1}, {0, 0, 0, 1, 4, 10}, 40] (* Jean-François Alcover, May 11 2019 *)

Formula

G.f.: z^3/(1-2*z+z^2-z^3)^2.

A077855 Expansion of 1/((1-2*x+x^2-x^3)*(1-x)).

Original entry on oeis.org

1, 3, 6, 11, 20, 36, 64, 113, 199, 350, 615, 1080, 1896, 3328, 5841, 10251, 17990, 31571, 55404, 97228, 170624, 299425, 525455, 922110, 1618191, 2839728, 4983376, 8745216, 15346785, 26931731, 47261894, 82938843, 145547524, 255418100, 448227520, 786584465
Offset: 0

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

a(n) is the number of binary words of length n+2 such that there is at least one run of 0's and every run of 0's is of length >=2. a(1)=3 because we have: {0,0,0}, {0,0,1}, {1,0,0}. - Geoffrey Critzer, Jan 12 2013
INVERT transform of A099254: (1, 2, 1, -2, -4, -2, 3, 6, 3, ...). - Gary W. Adamson, Jan 11 2017
a(n) is the number of nonempty subsets A of {1, 2, ..., n+1} with the property that every element in A has at least one consecutive neighbor also in A. That is, for every element x in A, either x-1 is in A or x+1 is in A. - MingKun Yue, Mar 07 2025

Crossrefs

Cf. A018918, A099254, A005314 (first differences).

Programs

  • Mathematica
    nn=40; a=x^2/(1-x); Drop[CoefficientList[Series[(a+1)/(1-x a/(1-x))/(1-x)-1/(1-x), {x,0,nn}], x], 2] (* Geoffrey Critzer, Jan 12 2013 *)
    LinearRecurrence[{3, -3, 2, -1}, {1, 3, 6, 11}, 36] (* or *)
    CoefficientList[ Series[1/(x^4 - 2 x^3 + 3 x^2 - 3 x + 1), {x, 0, 35}], x] (* Robert G. Wilson v, Nov 25 2016 *)
  • PARI
    Vec((1-x)^(-1)/(1-2*x+x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

Formula

G.f.: 1/((1-2*x+x^2-x^3)*(1-x)).
a(n) = A005251(n+4) - 1. a(n+1) - a(n) = A005314(n+2). - R. J. Mathar, Sep 19 2008
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4). - Seiichi Manyama, Nov 25 2016
a(n) = Sum_{i=1..(n+3)} binomial((n+3)-i, (n+3)-3*i). - Wesley Ivan Hurt, Jul 07 2020
a(n) ~ (48 - 11*r + 29*r^2) / (23 * r^n), where r = 0.569840290998... is the root of the equation r*(2 - r + r^2) = 1. - Vaclav Kotesovec, Apr 15 2024
From MingKun Yue, Mar 07 2025: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-3) + 1.
a(n) = a(n-1) + Sum_{i=1..(n-3)} a(i) + n. (End)

A193147 Expansion of 1/(1 - x - 2*x^3 - x^5).

Original entry on oeis.org

1, 1, 1, 3, 5, 8, 15, 26, 45, 80, 140, 245, 431, 756, 1326, 2328, 4085, 7168, 12580, 22076, 38740, 67985, 119305, 209365, 367411, 644761, 1131476, 1985603, 3484490, 6114853, 10730820, 18831276, 33046585, 57992715, 101770120, 178594110, 313410816, 549997641
Offset: 0

Author

Johannes W. Meijer, Jul 20 2011

Keywords

Comments

The Ze3 sums, see A180662 for the definition of these sums, of the "Races with Ties" triangle A035317 equal this sequence.
Number of tilings of a 5 X 2n rectangle with 5 X 1 pentominoes. - M. Poyraz Torcuk, Dec 18 2021

Crossrefs

Bisection of A003520.

Programs

  • Maple
    A193147 := proc(n) option remember: if n>=-4 and n<=-1 then 0 elif n=0 then 1 else procname(n-1) + 2*procname(n-3) + procname(n-5) fi: end: seq(A193147(n), n=0..32);
  • Mathematica
    Series[1/(1 - x - 2*x^3 - x^5), {x, 0, 32}] // CoefficientList[#, x]& (* Jean-François Alcover, Apr 02 2015 *)
  • Maxima
    a(n):=sum(sum(binomial(j,3*n-5*m+2*j)*binomial(2*m-n,j)*2^(3*n-5*m+2*j), j,0,2*m-n),m,floor((n+1)/2),n); /* Vladimir Kruchinin, Mar 10 2013 */

Formula

G.f.: 1/(1-x-2*x^3-x^5) = -1 / ( (1+x+x^2)*(x^3-x^2+2*x-1) ).
a(n) = a(n-1) + 2*a(n-3) + a(n-5) with a(n) = 0 for n= -4, -3, -2, -1 and a(0) = 1.
a(n) = (5*b(n+1) - 4*b(n) + 3*b(n-1) + 2*c(n) + 3*c(n-1))/7 with b(n) = A005314(n) and c(n) = A049347(n).
G.f.: 1 + x/(U(0)-x) where G(k)= 1 - x^2*(k+1)/(1 - 1/(1 + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2012
a(n) = Sum_{m=floor((n+1)/2)..n} Sum_{j=0..2*m-n} C(j,3*n-5*m+2*j) * C(2*m-n,j) * 2^(3*n-5*m+2*j). - Vladimir Kruchinin, Mar 10 2013
With offset 1, the INVERT transform of (1 + 2x^2 + x^4). - Gary W. Adamson, Mar 30 2017
a(n) = Sum_{k=0..floor(2*n/5)} binomial(2*n-4*k,k). - Seiichi Manyama, Jun 14 2024
Showing 1-10 of 32 results. Next