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.

Showing 1-7 of 7 results.

A001629 Self-convolution of Fibonacci numbers.

Original entry on oeis.org

0, 0, 1, 2, 5, 10, 20, 38, 71, 130, 235, 420, 744, 1308, 2285, 3970, 6865, 11822, 20284, 34690, 59155, 100610, 170711, 289032, 488400, 823800, 1387225, 2332418, 3916061, 6566290, 10996580, 18394910, 30737759, 51310978, 85573315, 142587180, 237387960, 394905492
Offset: 0

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Keywords

Comments

Number of elements in all subsets of {1,2,...,n-1} with no consecutive integers. Example: a(5)=10 because the subsets of {1,2,3,4} that have no consecutive elements, i.e., {}, {1}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}, the total number of elements is 10. - Emeric Deutsch, Dec 10 2003
If g is either of the real solutions to x^2-x-1=0, g'=1-g is the other one and phi is any 2 X 2-matricial solution to the same equation, not of the form gI or g'I, then Sum'_{i+j=n-1} g^i phi^j = F_n + (A001629(n) - A001629(n-1)g')*(phi-g'I), where i,j >= 0, F_n is the n-th Fibonacci number and I is the 2 X 2 identity matrix... - Michele Dondi (blazar(AT)lcm.mi.infn.it), Apr 06 2004
Number of 3412-avoiding involutions containing exactly one subsequence of type 321.
Number of binary sequences of length n with exactly one pair of consecutive 1's. - George J. Schaeffer (gschaeff(AT)andrew.cmu.edu), Sep 02 2004
For this sequence the n-th term is given by (nF(n+1)-F(n)+nF(n-1))/5 where F(n) is the n-th Fibonacci number. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 20 2005
If an unbiased coin is tossed n times then there are 2^n possible strings of H and T. Out of these, number of strings with exactly one 'HH' is given by a(n) where a(n) denotes n-th term of this sequence. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), May 04 2005
a(n) is half the number of horizontal dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; a(n+1) = the number of vertical dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; thus 2*a(n)+a(n+1)=n*F(n+1) = the number of dominoes in all domino tilings of a 2 X n rectangle, where F=A000045, the Fibonacci sequence. - Roberto Tauraso, May 02 2005; Graeme McRae, Jun 02 2006
a(n+1) = (-i)^(n-1)*(d/dx)S(n,x)|A049310%20for%20the%20S-polynomials.%20-%20_Wolfdieter%20Lang">{x=i}, where i is the imaginary unit, n >= 1. First derivative of Chebyshev S-polynomials evaluated at x=i multiplied by (-i)^(n-1). See A049310 for the S-polynomials. - _Wolfdieter Lang, Apr 04 2007
For n >= 4, a(n) is the number of weak compositions of n-2 in which exactly one part is 0 and all other parts are either 1 or 2. - Milan Janjic, Jun 28 2010
For n greater than 1, a(n) equals the absolute value of (1 - (1/2 - i/2)*(1 + (-1)^(n + 1))) times the x-coefficient of the characteristic polynomial of the (n-1) X (n-1) tridiagonal matrix with i's along the main diagonal (i is the imaginary unit), 1's along the superdiagonal and the subdiagonal and 0's everywhere else (see Mathematica code below). - John M. Campbell, Jun 23 2011
For n > 0: a(n) = Sum_{k=1..n-1} (A039913(n-1,k)) / 2. - Reinhard Zumkeller, Oct 07 2012
The right-hand side of a binomial-coefficient identity [Gauthier]. - N. J. A. Sloane, Apr 09 2013
a(n) is the number of edges in the Fibonacci cube Gamma(n-1) (see the Klavzar 2005 reference, p. 149). Example: a(3)=2; indeed, the Fibonacci cube Gamma(2) is the path P(3) having 2 edges. - Emeric Deutsch, Aug 10 2014
a(n) is the number of c(i)'s, including repetitions, in p(n), where p(n)/q(n) is the n-th convergent p(n)/q(n) of the formal infinite continued fraction [c(0), c(1), ...]; e.g., the number of c(i)'s in p(3) = c(0)*c(1)*c(2)*c(3) + c(0)*c(1) + c(0)*c(3) + c(2)*c(3) + 1 is a(5) = 10. - Clark Kimberling, Dec 23 2015
Also the number of maximal and maximum cliques in the (n-1)-Fibonacci cube graph. - Eric W. Weisstein, Sep 07 2017
a(n+1) is the total number of fixed points in all permutations p on 1, 2, ..., n such that |k-p(k)| <= 1 for 1 <= k <= n. - Katharine Ahrens, Sep 03 2019
From Steven Finch, Mar 22 2020: (Start)
a(n+1) is the total binary weight (cf. A000120) of all A000045(n+2) binary sequences of length n not containing any adjacent 1's.
The only three 2-bitstrings without adjacent 1's are 00, 01 and 10. The bitsums of these are 0, 1 and 1. Adding these give a(3)=2.
The only five 3-bitstrings without adjacent 1's are 000, 001, 010, 100 and 101. The bitsums of these are 0, 1, 1, 1 and 2. Adding these give a(4)=5.
The only eight 4-bitstrings without adjacent 1's are 0000, 0001, 0010, 0100, 1000, 0101, 1010 and 1001. The bitsums of these are 0, 1, 1, 1, 1, 2, 2, and 2. Adding these give a(5)=10. (End)
Number of tilings of a 1 X n strip with monominoes (1 X 1 squares) and at least one domino (1 X 2 rectangles), where exactly one of the dominoes is colored gold. - Greg Dresden and Jiachen Weng, Jul 31 2025

Examples

			G.f. = x^2 + 2*x^3 + 5*x^4 + 10*x^5 + 20*x^6 + 38*x^7 + 71*x^8 + 130*x^9 + ... - _Michael Somos_, Jun 24 2018
		

References

  • Donald E. Knuth, Fundamental Algorithms, Addison-Wesley, 1968, p. 83, Eq. 1.2.8--(17). - Don Knuth, Feb 26 2019
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Chapter 15, page 187, "Hosoya's Triangle", and p. 375, eq. (32.13).
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 101.
  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 183, Nr.(98).

Crossrefs

Row sums of triangles A058071, A134510, A134836.
First differences of A006478.

Programs

  • GAP
    List([0..40],n->Sum([0..n],k->Fibonacci(k)*Fibonacci(n-k))); # Muniru A Asiru, Jun 24 2018
    
  • Haskell
    a001629 n = a001629_list !! (n-1)
    a001629_list = f [] $ tail a000045_list where
       f us (v:vs) = (sum $ zipWith (*) us a000045_list) : f (v:us) vs
    -- Reinhard Zumkeller, Jan 18 2014, Oct 16 2011
    
  • Magma
    I:=[0,0,1,2]; [n le 4 select I[n] else 2*Self(n-1)+Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 19 2014
    
  • Maple
    a:= n-> (<<2|1|0|0>, <1|0|1|0>, <-2|0|0|1>, <-1|0|0|0>>^n)[1,3]:
    seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
    # Alternative:
    A001629 := n -> `if`(n<2, 0, (n-1)*hypergeom([1-n/2, (3-n)/2], [1-n], -4)):
    seq(simplify(A001629(n)), n=0..37); # Peter Luschny, Apr 10 2018
  • Mathematica
    Table[Sum[Binomial[n-i, i] i, {i, 0, n}], {n, 0, 34}] (* Geoffrey Critzer, May 04 2009 *)
    Table[Abs[(1 -(1/2 -I/2)(1 - (-1)^n))*Coefficient[CharacteristicPolynomial[ Array[KroneckerDelta[#1, #2] I + KroneckerDelta[#1 + 1, #2] + KroneckerDelta[#1 -1, #2] &, {n-1, n-1}], x], x]], {n,2,50}] (* John M. Campbell, Jun 23 2011 *)
    LinearRecurrence[{2,1,-2,-1}, {0,0,1,2}, 40] (* Harvey P. Dale, Aug 26 2013 *)
    CoefficientList[Series[x^2/(1-x-x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 19 2014 *)
    Table[(2nFibonacci[n-1] + (n-1)Fibonacci[n])/5, {n, 0, 40}] (* Vladimir Reshetnikov, May 08 2016 *)
    Table[With[{fibs=Fibonacci[Range[n]]},ListConvolve[fibs,fibs]],{n,-1,40}]//Flatten (* Harvey P. Dale, Aug 19 2018 *)
  • PARI
    Vec(1/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n)[2,4] \\ Charles R Greathouse IV, Jul 20 2016
    
  • SageMath
    def A001629(n): return (1/5)*(n*lucas_number2(n, 1, -1) - fibonacci(n))
    [A001629(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

Formula

G.f.: x^2/(1 - x - x^2)^2. - Simon Plouffe in his 1992 dissertation
a(n) = A037027(n-1, 1), n >= 1 (Fibonacci convolution triangle).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), n > 3.
a(n) = Sum_{k=0..n} A000045(k)*A000045(n-k).
a(n+1) = Sum_{i=0..F(n)} A007895(i), where F = A000045, the Fibonacci sequence. - Claude Lenormand (claude.lenormand(AT)free.fr), Feb 04 2001
a(n) = Sum_{k=0..floor(n/2)-1} (k+1)*binomial(n-k-1, k+1). - Emeric Deutsch, Nov 15 2001
a(n) = floor( (1/5)*(n - 1/sqrt(5))*phi^n + 1/2 ) where phi=(1+sqrt(5))/2 is the golden ratio. - Benoit Cloitre, Jan 05 2003
a(n) = a(n-1) + A010049(n-1) for n > 0. - Emeric Deutsch, Dec 10 2003
a(n) = Sum_{k=0..floor((n-2)/2)} (n-k-1)*binomial(n-k-2, k). - Paul Barry, Jan 25 2005
a(n) = ((n-1)*F(n) + 2*n*F(n-1))/5, F(n)=A000045(n) (see Vajda and Koshy reference).
F'(n, 1), the first derivative of the n-th Fibonacci polynomial evaluated at 1. - T. D. Noe, Jan 18 2006
a(n) = a(n-1) + a(n-2) + F(n-1), where F=A000045, the Fibonacci sequence. - Graeme McRae, Jun 02 2006
a(n) = (1/5)*(n-1/sqrt(5))*((1+sqrt(5))/2)^n + (1/5)*(n+1/sqrt(5))*((1-sqrt(5))/2)^n. - Graeme McRae, Jun 02 2006
a(n) = A055244(n-1) - F(n-2). Example: a(6) = 20 = A055244(5) - F(3) = (23 - 3). - Gary W. Adamson, Jul 27 2007
a(n) = term (1,3) in the 4 X 4 matrix [2,1,0,0; 1,0,1,0; -2,0,0,1; -1,0,0,0]^n. - Alois P. Heinz, Aug 01 2008
a(n) = A214178(n,1) for n > 0. - Reinhard Zumkeller, Jul 08 2012
a(n) = ((n+1)*F(n-1) + (n-1)*F(n+1))/5. - Richard R. Forberg, Aug 04 2014
(n-2)*a(n) - (n-1)*a(n-1) - n*a(n-2) = 0, n > 1. - Michael D. Weiner, Nov 18 2014
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j), where F(n) = A000045 Fibonacci Numbers. - Carlos A. Rico A., Jul 14 2016
a(n) = (n*Lucas(n) - Fibonacci(n))/5, where Lucas = A000032, Fibonacci = A000045. - Vladimir Reshetnikov, Sep 27 2016
a(n) = (n-1)*hypergeom([1-n/2, (3-n)/2], [1-n], -4) for n >= 2. - Peter Luschny, Apr 10 2018
a(n) = -(-1)^n a(-n) for all n in Z. - Michael Somos, Jun 24 2018
E.g.f.: (1/50)*exp(-2*x/(1+sqrt(5)))*(2*sqrt(5)-5*(-1+sqrt(5))*x+exp(sqrt(5)*x)*(-2*sqrt(5)+5*(1+sqrt(5))*x)). - Stefano Spezia, Sep 03 2019
From Peter Bala, Jan 14 2025: (Start)
a(2*n+1) is even and a(2*n) has the same parity as Fibonacci(n).
For n >= 1, a(n) = (2/n)*Sum_{k = 0..n} k*Fibonacci(k)*Fibonacci(n-k). (End)

A046854 Triangle read by rows: T(n, k) = binomial(floor((n+k)/2), k), n >= k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 3, 4, 1, 1, 1, 3, 6, 4, 5, 1, 1, 1, 4, 6, 10, 5, 6, 1, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 1, 6, 21, 35, 70, 56, 84, 36, 45, 10, 11, 1, 1
Offset: 0

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Author

Keywords

Comments

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004
Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005
As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008
From Emeric Deutsch, Jun 18 2010: (Start)
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)
From L. Edson Jeffery, Mar 01 2011: (Start)
This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table
1
1
1 1
1 1
1 2 1
1 2 1
1 3 3 1
1 3 3 1
...,
then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with
1;
1, -1;
1, -1, -1;
1, -2, -1, 1;
1, -2, -3, 1, 1;
...
(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.
For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)
Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011
Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012
Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014
From Jeremy Dover, Jun 07 2016: (Start)
T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.
T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Examples

			Triangle begins:
  1;
  1 1;
  1 1 1;
  1 2 1 1;
  1 2 3 1 1;
  1 3 3 4 1 1;
  ...
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1978. [Emeric Deutsch, Jun 18 2010]

Crossrefs

Reflected version of A065941, which is considered the main entry. A deficient version is in A030111.
Cf. A055244. - Gary W. Adamson, Dec 23 2008

Programs

  • GAP
    Flat(List([0..16], n-> List([0..n], k-> Binomial(Int((n+k)/2), k) ))); # G. C. Greubel, Jul 13 2019
  • Haskell
    a046854 n k = a046854_tabl !! n !! k
    a046854_row n = a046854_tabl !! n
    a046854_tabl = [1] : f [1] [1,1] where
       f us vs = vs : f vs  (zipWith (+) (us ++ [0,0]) ([0] ++ vs))
    -- Reinhard Zumkeller, Apr 24 2013
    
  • Magma
    [Binomial(Floor((n+k)/2), k): k in [0..n], n in [0..16]]; // G. C. Greubel, Jul 13 2019
    
  • Maple
    A046854:= proc(n,k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n,k),k=0..n),n=0..16); # Nathaniel Johnston, Jun 30 2011
  • Mathematica
    Table[Binomial[Floor[(n+k)/2], k], {n,0,16}, {k,0,n}]//Flatten
    (* next program: Polynomials *)
    z = 12; f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1;
    t = Table[Factor[f[x, n]], {n, 1, z}]
    u = Flatten[CoefficientList[Numerator[t], x]] (* this sequence *)
    v = Flatten[CoefficientList[Denominator[t], x]]
    (* Clark Kimberling, Oct 13 2014 *)
  • PARI
    T(n,k) = binomial((n+k)\2, k); \\ G. C. Greubel, Jul 13 2019
    
  • Sage
    [[binomial(floor((n+k)/2), k) for k in (0..n)] for n in (0..16)] # G. C. Greubel, Jul 13 2019
    

Formula

T(n,k) = binomial(floor((n+k)/2), k).
G.f.: (1+x)/(1-x*y-x^2). - Ralf Stephan, Feb 13 2005
Triangle = A097806 * A168561, as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007
T(n,k) = A065941(n,n-k) = abs(A130777(n,k)) = abs(A066170(n,k)) = abs(A187660(n,k)). - Johannes W. Meijer, Aug 08 2011
For n > 1: T(n, k) = T(n-1, k-1) + T(n-2, k), 0 < k < n-1. - Reinhard Zumkeller, Apr 24 2013
T(n,k) = A168561(n,k) + A168561(n-1,k). - R. J. Mathar, Feb 10 2024

A194778 T(n,k)=Number of lower triangles of an (n+2k-2)X(n+2k-2) 0..k array with new values introduced in row major order 0..k and no element unequal to more than one horizontal or vertical neighbor.

Original entry on oeis.org

1, 7, 3, 43, 17, 6, 268, 105, 41, 12, 1740, 672, 265, 95, 23, 11862, 4490, 1736, 655, 219, 43, 85013, 31466, 11857, 4464, 1641, 493, 79, 639760, 231445, 85007, 31429, 11686, 4069, 1101, 143, 5045610, 1784788, 639753, 231395, 84727, 30608, 10121, 2427, 256
Offset: 1

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Author

R. H. Hardin Sep 02 2011

Keywords

Comments

Table starts
...1....7....43....268....1740....11862.....85013.....639760....5045610
...3...17...105....672....4490....31466....231445....1784788...14404218
...6...41...265...1736...11857....85007....639753....5045602...41615156
..12...95...655...4464...31429...231395...1784723...14404136..121420974
..23..219..1641..11686...84727...639325...5044981...41614291..358184223
..43..493..4069..30608..229869..1782111..14399939..121414559.1066897441
..79.1101.10121..80961..631373..5029741..41587186..358138793.3210593393
.143.2427.25025.214469.1745459.14321783.121261490.1066617315

Examples

			Some solutions for n=2 k=2
..0........0........0........0........0........0........0........0
..0.0......0.0......1.1......1.1......0.0......1.1......0.1......0.0
..1.1.1....0.0.1....1.1.2....1.1.1....0.0.0....1.1.1....0.1.1....0.0.0
..1.1.1.0..0.0.1.1..1.1.2.2..0.0.0.0..0.0.0.1..2.2.2.2..0.1.1.0..1.1.1.1
		

Crossrefs

Column 1 is A055244

A194540 T(n,k) is the number of lower triangles of an n X n 0..k array with new values introduced in row major order 0..k and no element equal to more than one horizontal or vertical neighbor.

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 4, 59, 12, 1, 4, 116, 1851, 23, 1, 4, 131, 16556, 119548, 43, 1, 4, 132, 43785, 7721920, 16039294, 79, 1, 4, 132, 62038, 79201795, 11525456507, 4460214471, 143, 1, 4, 132, 67377, 286773762, 627974590650, 54979732214116, 2572187445993
Offset: 1

Views

Author

R. H. Hardin, Aug 28 2011

Keywords

Examples

			Table starts:
...1.............1..................1......................1
...3.............4..................4......................4
...6............59................116....................131
..12..........1851..............16556..................43785
..23........119548............7721920...............79201795
..43......16039294........11525456507...........627974590650
..79....4460214471.....54979732214116......21399815343394998
.143.2572187445993.838387866908847478.3132305932642809585983
Some solutions for n=4 and k=3:
..0........0........0........0........0........0........0........0
..1.2......1.0......1.0......1.0......0.1......1.2......1.0......1.2
..3.1.1....0.1.0....2.0.3....2.3.1....2.3.0....1.3.3....2.3.3....0.3.1
..2.3.2.0..2.1.3.1..2.1.0.1..1.0.3.2..3.0.3.1..3.1.1.2..0.2.2.0..2.1.2.1
		

Crossrefs

Column 1 is A055244.

A131913 Product of the square matrix in A065941 and the column vector (1, 2, 3, ...)'.

Original entry on oeis.org

1, 3, 6, 13, 25, 48, 89, 163, 294, 525, 929, 1632, 2849, 4947, 8550, 14717, 25241, 43152, 73561, 125075, 212166, 359133, 606721, 1023168, 1722625, 2895843, 4861254, 8149933, 13646809, 22825200, 38136089, 63653827, 106146534, 176849517, 294401825, 489706272
Offset: 0

Views

Author

Gary W. Adamson, Jul 27 2007

Keywords

Examples

			a(4) = 25 = (1, 1, 3, 2, 1) dot (1, 2, 3, 4, 5) = (1 + 2 + 9 + 8 + 5), where (1, 1, 3, 2, 1) = row 4 of triangle A065941.
a(4) = 25 = A010049(4) + A001629(6) = 5 + 20.
a(5) = 48 = A055244(6) + A001629(4) = 43 + 5.
		

Crossrefs

Formula

a(n) = A010049(n) + A001629(n+2) = A055244(n+1) + A001629(n-1).
From Philippe Deléham, Dec 28 2013: (Start)
G.f.: (1+x-x^2)/(1-x-x^2)^2.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), a(0)=1, a(1)=3, a(2)=6, a(3)=13.
a(n) = a(n-1) + a(n-2) + 2*Fibonacci(n). (End)

A055245 Numerator sequence of mean length of certain stackings of n+1 squares on a double staircase.

Original entry on oeis.org

1, 1, 5, 12, 28, 61, 127, 257, 507, 982, 1872, 3523, 6557, 12089, 22105, 40128, 72380, 129809, 231611, 411337, 727455, 1281578, 2249856, 3936935, 6868537, 11950033, 20737613, 35901300, 62014396, 106897669, 183905143, 315806321, 541372131
Offset: 0

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Author

Wolfdieter Lang, May 10 2000

Keywords

Comments

Denominator sequence is A055244(n).

References

  • L. Turban, Lattice animals on a staircase and Fibonacci numbers, J.Phys. A 33 (2000) 2587-2595.

Crossrefs

Programs

  • Maple
    a:= n-> (Matrix([[1,-1,0,2,-9,25]]). Matrix(6, (i,j)-> if (i=j-1) then 1 elif j=1 then [3,0,-5,0,3,1][i] else 0 fi)^(n))[1,1]: seq(a(n), n=0..32); # Alois P. Heinz, Aug 05 2008
  • Mathematica
    CoefficientList[Series[(1-2x+2x^2+2x^3-3x^4-x^5)/(1-x-x^2)^3,{x,0,50}],x] (* or *) LinearRecurrence[{3,0,-5,0,3,1},{1,1,5,12,28,61},50] (* Harvey P. Dale, Aug 24 2014 *)

Formula

G.f.: (1-2*x+2*x^2+2*x^3-3*x^4-x^5)/(1-x-x^2)^3. (from Turban reference eq.(3.11)).
a(n) = ((5*n^2+3*n-27)*F(n)+(19*n+25)*F(n+1))/25 with F(n)=A000045(n) (Fibonacci numbers) (from Turban reference eq.(3.12)).
a(0)=1, a(1)=1, a(2)=5, a(3)=12, a(4)=28, a(5)=61, a(n)=3*a(n-1)- 5*a(n-3)+ 3*a(n-5)+a(n-6). - Harvey P. Dale, Aug 24 2014

A129718 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs of 1's (n >= 0, 0 <= k <= floor((n+1)/2)). A Fibonacci binary word is a binary word having no 00 subword. A run of 1's is a maximal subword of the form 11..1.

Original entry on oeis.org

1, 1, 1, 0, 3, 0, 4, 1, 0, 4, 4, 0, 4, 8, 1, 0, 4, 12, 5, 0, 4, 16, 13, 1, 0, 4, 20, 25, 6, 0, 4, 24, 41, 19, 1, 0, 4, 28, 61, 44, 7, 0, 4, 32, 85, 85, 26, 1, 0, 4, 36, 113, 146, 70, 8, 0, 4, 40, 145, 231, 155, 34, 1, 0, 4, 44, 181, 344, 301, 104, 9, 0, 4, 48, 221, 489, 532, 259, 43, 1
Offset: 0

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Author

Emeric Deutsch, May 12 2007

Keywords

Comments

Row n has 1+floor((n+1)/2) terms.
Row sums are the Fibonacci numbers (A000045).
T(n,k) = A129717(n,k-1) (since in each word the number of runs of 1's = 1 + the number of 101's).
Sum_{k=0..floor((n+1)/2)} k*T(n,k) = A055244(n) (n >= 1).

Examples

			T(6,3)=5 because we have 110101, 101101, 101010, 101011 and 010101.
Triangle starts:
  1;
  1,  1;
  0,  3;
  0,  4,  1;
  0,  4,  4;
  0,  4,  8,  1;
  0,  4, 12,  5;
		

Crossrefs

Programs

  • Maple
    G:=(1+z)*(1-z+t*z)/(1-z-t*z^2): Gser:=simplify(series(G,z=0,21)): for n from 0 to 18 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 17 do seq(coeff(P[n],t,j),j=0..ceil(n/2)) od; # yields sequence in triangular form

Formula

G.f. = G(t,z) = (1+z)(1-z+tz)/(1-z-tz^2).
T(n,k) = binomial(n-k,k-1) + 2*binomial(n-k-1,k-1) + binomial(n-k-2,k-1) for n >= 4 and 0 <= k < floor((n+1)/2).
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