A055243 -id:A055243 - cp's OEIS frontend

A001628 Convolved Fibonacci numbers.

1, 3, 9, 22, 51, 111, 233, 474, 942, 1836, 3522, 6666, 12473, 23109, 42447, 77378, 140109, 252177, 451441, 804228, 1426380, 2519640, 4434420, 7777860, 13599505, 23709783, 41225349, 71501422, 123723351, 213619683, 368080793, 633011454, 1086665562, 1862264196

Offset: 0

N. J. A. Sloane, Simon Plouffe

a(n-2) = (((-i)^(n-2))/2)*(d^2/dx^2)S(n,x)|A049310%20for%20the%20S-polynomials.%20-%20_Wolfdieter%20Lang">{x=i}, n>=2. Second derivative of Chebyshev S-polynomials evaluated at x=i (imaginary unit) multiplied by ((-i)^(n-2))/2. See A049310 for the S-polynomials. - _Wolfdieter Lang, Apr 04 2007
a(n) = number of weak compositions of n in which exactly 2 parts are 0 and all other parts are either 1 or 2. - Milan Janjic, Jun 28 2010
Number of 4-cycles in the Fibonacci cube Gamma[n+3] (see the Klavzar reference, p. 511). - Emeric Deutsch, Apr 17 2014

			G.f. = 1 + 3*x + 9*x^2 + 22*x^3 + 51*x^4 + 111*x^5 + 233*x^6 + 474*x^7 + ...

T. Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001, p. 375.
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).

G. C. Greubel, Table of n, a(n) for n = 0..4500 (terms 0..500 from T. D. Noe)
Daniel Birmajer, Juan Gil, and Michael D. Weiner, Linear recurrence sequences and their convolutions via Bell polynomials, arXiv:1405.7727 [math.CO], 2014.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Linear Recurrence Sequences and Their Convolutions via Bell Polynomials, Journal of Integer Sequences, 18 (2015), #15.1.2.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Verner E. Hoggatt, Jr., Letters to N. J. A. Sloane, 1974-1975
Verner E. Hoggatt, Jr. and Marjorie Bicknell-Johnson, Fibonacci convolution sequences, Fib. Quart., 15 (1977), 117-122.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Sandi Klavžar, Structure of Fibonacci cubes: a survey, preprint.
Sandi Klavžar, Structure of Fibonacci cubes: a survey, J. Comb. Optim., 25, 2013, 505-522
Toufik Mansour, Generalization of some identities involving the Fibonacci numbers, arXiv:math/0301157 [math.CO], 2003.
Pieter Moree, Convoluted convolved Fibonacci numbers, arXiv:math/0311205 [math.CO], 2003.
Pieter Moree, Convoluted Convolved Fibonacci Numbers, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 18.
Jeffrey B. Remmel and J. L. B. Tiefenbruck, Q-analogues of convolutions of Fibonacci numbers, Australasian Journal of Combinatorics, Volume 64(1) (2016), Pages 166-193.
John Riordan, Letter to N. J. A. Sloane, Oct. 1970
Eric Weisstein's World of Mathematics, Graph Cycle
Eric Weisstein's World of Mathematics, Fibonacci Cube Graph
Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).

a(n) = A037027(n+2,2) (Fibonacci convolution triangle).
Cf. A055243 (first differences).
Cf. A291915 (6-cycles).

[(&+[Binomial(k,n-k)*Binomial(k+2,2): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 10 2018

A001628:=-1/(z**2+z-1)**3; [Simon Plouffe in his 1992 dissertation.]
a:= n-> (Matrix(6, (i, j)-> `if` (i=j-1, 1, `if` (j=1, [3, 0, -5, 0, 3, 1][i], 0)))^n)[1,1]: seq (a(n), n=0..29); # Alois P. Heinz, Aug 01 2008

CoefficientList[Series[1/(-z^2 - z + 1)^3, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
(* Start Eric W. Weisstein, Sep 05 2017 *)
Table[Derivative[2][Fibonacci[n + 2, #] &][1]/2, {n, 20}]
Derivative[2][Fibonacci[Range[20] + 2, #] &][1]/2
LinearRecurrence[{3, 0, -5, 0, 3, 1}, {1, 3, 9, 22, 51, 111}, 20]
Table[-I^(n + 1) Derivative[2][ChebyshevU[n + 1, -#/2] &][I]/2, {n, 20}]
(* End *)

Vec((1 - x - x^2 )^-3+O(x^99)) \\ Charles R Greathouse IV, Jul 01 2011

G.f.: 1 / (1 - x - x^2)^3.
a(n) = A037027(n+2,2) (Fibonacci convolution triangle).
a(n) = ((5*n+16)*(n+1)*F(n+2)+(5*n+17)*(n+2)*F(n+1))/50, F=A000045. - Wolfdieter Lang, Apr 12 2000 (This formula coincides with eq. (32.14) of the Koshy reference, p. 375, if there n -> n+3. - Wolfdieter Lang, Aug 03 2012)
For n>2, a(n-2) = sum(i+j+k=n, F(i)*F(j)*F(k)). - Benoit Cloitre, Nov 01 2002
a(n) = F''(n+2, 1)/2, i.e. 1/2 times the 2nd derivative of the (n+2)th Fibonacci polynomial evaluated at 1. - T. D. Noe, Jan 18 2006
a(n) = Sum_{k=0..n} C(k,n-k)*C(k+2,2). - Paul Barry, Apr 13 2008
0 = n*a(n) - (n+2)*a(n-1) - (n+4)*a(n-2), n>1. - Michael D. Weiner, Nov 18 2014
a(n) = 3*a(n-1) - 5*a(n-3) + 3*a(n-5) + a(n-6). - Eric W. Weisstein, Sep 05 2017

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Nov 29 2009

Row sums: A000045(n+1), Fibonacci numbers.
A168561*A007318 = A037027, as lower triangular matrices. Diagonal sums : A077957. - Philippe Deléham, Dec 02 2009
T(n,k) is the number of compositions of n+1 into k+1 odd parts. Example: T(4,2)=3 because we have 5 = 1+1+3 = 1+3+1 = 3+1+1.
Coefficients of monic Fibonacci polynomials (rising powers of x). Ftilde(n, x) = x*Ftilde(n-1, x) + Ftilde(n-2, x), n >=0, Ftilde(-1,x) = 0, Ftilde(0, x) = 1. G.f.: 1/(1 - x*z - z^2). Compare with Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jul 29 2014

			The triangle T(n,k) begins:
n\k 0  1   2   3   4    5    6    7    8    9  10  11  12  13 14 15 ...
0:  1
1:  0  1
2:  1  0   1
3:  0  2   0   1
4:  1  0   3   0   1
5:  0  3   0   4   0    1
6:  1  0   6   0   5    0    1
7:  0  4   0  10   0    6    0    1
8:  1  0  10   0  15    0    7    0    1
9:  0  5   0  20   0   21    0    8    0    1
10: 1  0  15   0  35    0   28    0    9    0   1
11: 0  6   0  35   0   56    0   36    0   10   0   1
12: 1  0  21   0  70    0   84    0   45    0  11   0   1
13: 0  7   0  56   0  126    0  120    0   55   0  12   0   1
14: 1  0  28   0 126    0  210    0  165    0  66   0  13   0  1
15: 0  8   0  84   0  252    0  330    0  220   0  78   0  14  0  1
... reformatted by _Wolfdieter Lang_, Jul 29 2014.
------------------------------------------------------------------------

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
Tom Copeland, Addendum to Elliptic Lie Triad
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A162515 (rows reversed), A112552, A102426 (deflated).

A168561:=proc(n,k) if n-k mod 2 = 0 then binomial((n+k)/2,k) else 0 fi end proc:
seq(seq(A168561(n,k),k=0..n),n=0..12) ; # yields sequence in triangular form

Table[If[EvenQ[n + k], Binomial[(n + k)/2, k], 0], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 16 2017 *)

T(n,k) = if ((n+k) % 2, 0, binomial((n+k)/2,k));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print();); \\ Michel Marcus, Oct 09 2016

Sum_{k=0..n} T(n,k)*x^k = A059841(n), A000045(n+1), A000129(n+1), A006190(n+1), A001076(n+1), A052918(n), A005668(n+1), A054413(n), A041025(n), A099371(n+1), A041041(n), A049666(n+1), A041061(n), A140455(n+1), A041085(n), A154597(n+1), A041113(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 respectively. - Philippe Deléham, Dec 02 2009
T(2n,2k) = A085478(n,k). T(2n+1,2k+1) = A078812(n,k). Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000045(n+1), A006131(n), A015445(n), A168579(n), A122999(n) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Dec 02 2009
T(n,k) = binomial((n+k)/2,k) if (n+k) is even; otherwise T(n,k)=0.
G.f.: (1-z^2)/(1-t*z-z^2) if offset is 1.
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = 1, T(0,1) = 0. - Philippe Deléham, Feb 09 2012
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 09 2012
From R. J. Mathar, Feb 04 2022: (Start)
Sum_{k=0..n} T(n,k)*k = A001629(n+1).
Sum_{k=0..n} T(n,k)*k^2 = 0,1,4,11,... = 2*A055243(n)-A099920(n+1).
Sum_{k=0..n} T(n,k)*k^3 = 0,1,8,29,88,236,... = 12*A055243(n) -6*A001629(n+2) +A001629(n+1)-6*(A001872(n)-2*A001872(n-1)). (End)

Typo in name corrected (1(1-x^2) changed to 1/(1-x^2)) by Wolfdieter Lang, Nov 20 2010

A281056 T(n,k)=Number of nXk 0..1 arrays with no element equal to more than one of its horizontal and antidiagonal neighbors, with the exception of exactly two elements, and with new values introduced in order 0 sequentially upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 6, 6, 0, 2, 34, 68, 29, 0, 6, 104, 239, 376, 122, 0, 13, 251, 618, 1022, 1492, 468, 0, 29, 535, 1403, 2452, 3416, 4988, 1686, 0, 60, 1076, 2828, 5400, 7803, 10112, 15028, 5807, 0, 122, 2090, 5482, 10570, 16875, 22106, 27635, 42252, 19338, 0

Offset: 1

R. H. Hardin, Jan 13 2017

Table starts
.0.....0......0......1......2.......6......13......29......60......122......241
.0.....1......6.....34....104.....251.....535....1076....2090.....3956.....7353
.0.....6.....68....239....618....1403....2828....5482...10342....19136....34907
.0....29....376...1022...2452....5400...10570...19892...36616....66354...118926
.0...122...1492...3416...7803...16875...32370...59669..107693...191953...339219
.0...468...4988..10112..22106...46610...87995..159317..282552...495655...864369
.0..1686..15028..27635..58005..119205..221212..394884..689397..1191475..2050466
.0..5807..42252..71419.144283..288527..526340..926077.1596265..2721514..4625355
.0.19338.113076.177320.344972..671451.1201939.2084979.3549039..5982952.10051523
.0.62731.291660.426696.800225.1515524.2661939.4546633.7645715.12749511.21211441

			Some solutions for n=4 k=4
..0..0..1..0. .0..0..0..1. .0..1..0..1. .0..1..1..0. .0..1..0..1
..1..1..0..1. .0..1..0..1. .1..0..1..0. .0..1..0..1. .1..0..1..1
..0..1..1..0. .0..1..1..0. .1..0..0..1. .0..1..0..1. .1..0..0..1
..0..0..1..1. .0..1..0..1. .0..1..0..0. .1..0..1..1. .1..0..1..0

R. H. Hardin, Table of n, a(n) for n = 1..338

Row 1 is A055243(n-4).

Empirical for column k:
k=1: a(n) = a(n-1)
k=2: a(n) = 9*a(n-1) -30*a(n-2) +45*a(n-3) -30*a(n-4) +9*a(n-5) -a(n-6)
k=3: a(n) = 5*a(n-1) -6*a(n-2) -4*a(n-3) +8*a(n-4) for n>6
k=4: [order 12] for n>14
k=5: [order 13] for n>16
k=6: [order 18] for n>21
k=7: [order 19] for n>23
Empirical for row n:
n=1: a(n) = 3*a(n-1) -5*a(n-3) +3*a(n-5) +a(n-6)
n=2: [order 8] for n>14
n=3: [same order 8] for n>15
n=4: [order 9] for n>17
n=5: [same order 9] for n>18
n=6: [same order 9] for n>19
n=7: [same order 9] for n>20

A210034 Triangle of coefficients of polynomials v(n,x) jointly generated with A210033; see the Formula section.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 7, 5, 2, 1, 12, 10, 6, 2, 1, 20, 20, 13, 7, 2, 1, 33, 38, 29, 16, 8, 2, 1, 54, 71, 60, 39, 19, 9, 2, 1, 88, 130, 122, 86, 50, 22, 10, 2, 1, 143, 235, 241, 187, 116, 62, 25, 11, 2, 1, 232, 420, 468, 392, 267, 150, 75, 28, 12, 2, 1, 376, 744, 894, 806

Offset: 1

Clark Kimberling, Mar 16 2012

For a discussion and guide to related arrays, see A208510.
From Gus Wiseman, Jun 29 2025: (Start)
This appears to be the number of subsets of {1..n} with k>0 maximal anti-runs (sequences of consecutive elements increasing by more than 1). For example, the subset {1,2,4,5} has maximal anti-runs ((1),(2,4),(5)) so is counted under T(5,3). Row n = 5 counts the following:
  {1}      {1,2}    {1,2,3}    {1,2,3,4}  {1,2,3,4,5}
  {2}      {2,3}    {2,3,4}    {2,3,4,5}
  {3}      {3,4}    {3,4,5}
  {4}      {4,5}    {1,2,3,5}
  {5}      {1,2,4}  {1,2,4,5}
  {1,3}    {1,2,5}  {1,3,4,5}
  {1,4}    {1,3,4}
  {1,5}    {1,4,5}
  {2,4}    {2,3,5}
  {2,5}    {2,4,5}
  {3,5}
  {1,3,5}
For runs instead of anti-runs we have A034839, with n A202064. For reversed partitions instead of subsets we have A268193. (End)

			First five rows:
  1
  2    1
  4    2    1
  7    5    2   1
  12   10   6   2   1
First three polynomials v(n,x): 1, 2 + x, 4 + 2*x + x^2.

Cf. A210033, A208510.
Column k = 1 is A000071.
Row sums are A000225.
Column k = 2 is A001629.
Column k = 3 is A055243.
The version including k = 0 is A384893.
A034839 counts subsets by number of maximal runs, see also A202023, A202064.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs of binary indices, firsts A384878.
Cf. A053538, A116674, A119900, A268193, A384177, A384879, A384889, A384905.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210033 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210034 *)

u(n,x)=u(n-1,x)+v(n-1,x)+1,
v(n,x)=u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A076791 Triangle a(n,k) giving number of binary sequences of length n containing k subsequences 00.

Original entry on oeis.org

1, 2, 3, 1, 5, 2, 1, 8, 5, 2, 1, 13, 10, 6, 2, 1, 21, 20, 13, 7, 2, 1, 34, 38, 29, 16, 8, 2, 1, 55, 71, 60, 39, 19, 9, 2, 1, 89, 130, 122, 86, 50, 22, 10, 2, 1, 144, 235, 241, 187, 116, 62, 25, 11, 2, 1, 233, 420, 468, 392, 267, 150, 75, 28, 12, 2, 1, 377, 744, 894, 806, 588, 363, 188, 89, 31, 13, 2, 1

Offset: 0

Roger Cuculière, Nov 16 2002

The triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 is A034867 because this number is C(n+1,2*k+1). I have not yet found a formula for subsequences 00.
The problem is equivalent to one encountered by David W. Wilson, Dept of Geography, University of Southampton, UK, in his work on Markov models for rainfall disaggregation. He asked for the number of ways in which there can be k instances of adjacent rainy days in a period of n consecutive days. Representing a rainy day by 0 and a fine day by 1, the problem is equivalent to that solved by this sequence. - E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004
Row n (n>=1) contains n terms.
Triangle, with zeros omitted, given by (2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
a(n-1,k) is also the number of permutations avoiding both 132 and 213 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

			a(5,2) = 6 because the binary sequences of length 5 with 2 subsequences 00 are 10001, 11000, 01000, 00100, 00010, 00011.
Triangle begins
   1;
   2;
   3,  1;
   5,  2, 1;
   8,  5, 2, 1;
  13, 10, 6, 2, 1;
  ...

Alois P. Heinz, Rows n = 0..150, flattened
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
Toufik Mansour and Armend Sh. Shabani, Bargraphs in bargraphs, Turkish Journal of Mathematics (2018) Vol. 42, Issue 5, 2763-2773.
Paul M. Rakotomamonjy, Sandrataniaina R. Andriantsoa, and Arthur Randrianarivony, Crossings over permutations avoiding some pairs of three length-patterns, arXiv:1910.13809 [math.CO], 2019.

Cf. a(n,1) = A001629, a(n,2) = A055243.

b:= proc(n, l) option remember; `if`(n=0, 1,
      expand(b(n-1, 1)*x^l)+b(n-1, 0))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..14);  # Alois P. Heinz, Sep 17 2019

f[list_] := Select[list, #>0&]; nn=10; a=1/(1-y x); b= x/(1-y x) +1; c=1/(1-x); Map[f, CoefficientList[Series[c b/(1-(a x^2 c)), {x,0,nn}], {x,y}]]//Flatten (* Geoffrey Critzer, Mar 05 2012 *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A053538 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A076791 *)
(* Clark Kimberling, Mar 08 2012 *)
T[ n_, k_] := If[n<2, (n+1)*Boole[n > -1 && k == 0], T[n, k] = T[n-1, k] + T[n-1, k-1] + T[n-2, k] - T[n-2, k-1] ]; (* Michael Somos, Sep 21 2024 *)

{T(n, k) = if(n<2, (n+1)*(n > -1 && k == 0), T(n-1, k) + T(n-1, k-1) + T(n-2, k) - T(n-2, k-1) )}; /* Michael Somos, Sep 21 2024 */

Recurrence: a(n, k) = (a(n-1, k) + a(n-2, k)) + (a(n-3, k-1) + a(n-4, k-2) + ... + a(n-k-2, 0)).
Special values: a(n, 0) = Fibonacci(n+1); a(n, n-1) = 1 for n >= 2; a(n, n-2) = 2 for n >= 3; a(n, n-3) = n + 1 for n >= 4, etc.
a(n, n-4) = 3*n - 5 for n >= 5, a(n, n-5) = (n^2 + 5*n - 26)/2 for n >= 6, a(n, n-6) = 2*n^2 - 8*n - 4, for n >= 7 etc.
Recurrence relation: a(n+1, k) = a(n, k) + a(n-1, k) + a(n, k-1) - a(n-1, k-1) for k >= 1, n >= 1.
Generating function: a(n, k) is coefficient of x^n in ((x^(k + 1))*((1 - x)^(k - 1)))/((1 - x - x^2)^(k + 1)) for k >= 1. - E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004
G.f.: (1 + (1 - t)*x)/(1 - (1 + t)*x - (1 - t)*x^2). [Carlitz-Scoville] - Emeric Deutsch, May 19 2006
A076791 is jointly generated with A053538 as an array of coefficients of polynomials u(n,x):  initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1)*x and v(n,x) = u(n-1,x) + v(n-1,x).  See the Mathematica section. - Clark Kimberling, Mar 08 2012

More terms from E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004

A129707 Number of inversions in all Fibonacci binary words of length n.

Original entry on oeis.org

0, 0, 1, 4, 12, 31, 73, 162, 344, 707, 1416, 2778, 5358, 10188, 19139, 35582, 65556, 119825, 217487, 392286, 703618, 1255669, 2230608, 3946020, 6954060, 12212280, 21377365, 37309288, 64935132, 112726771, 195224773, 337343034, 581700476

Offset: 0

Emeric Deutsch, May 12 2007

A Fibonacci binary word is a binary word having no 00 subword.

			a(3)=4 because the Fibonacci words 110,111,101,010,011 have a total of 2 + 0 + 1 + 1 + 0 = 4 inversions.

Michael De Vlieger, Table of n, a(n) for n = 0..4754
Kálmán Liptai, László Németh, Tamás Szakács, and László Szalay, On certain Fibonacci representations, arXiv:2403.15053 [math.NT], 2024. See p. 2.
Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See p. 2.
Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).

Cf. A129706.

Cf. A055243.

with(combinat): a[0]:=0: a[1]:=0: a[2]:=1: a[3]:=4: for n from 4 to 40 do a[n]:=2*a[n-1]+a[n-2]-2*a[n-3]-a[n-4]+fibonacci(n) od: seq(a[n],n=0..40);

CoefficientList[Series[x^2*(1 + x)/(1 - x - x^2)^3, {x,0,50}], x] (* G. C. Greubel, Mar 04 2017 *)

a(n) = sum(k*(k+1)*binomial(k,n-k-1),k,floor((n-1)/2),n-1)/2; /* Vladimir Kruchinin, Sep 17 2020 */

x='x+O('x^50); concat([0,0], Vec(x^2*(1 + x)/(1 - x - x^2)^3)) \\ G. C. Greubel, Mar 04 2017

a(n) = Sum_{k>=0} k*A129706(n,k).
G.f.: z^2*(1+z)/(1-z-z^2)^3.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) + F(n), a(0)=a(1)=0, a(2)=1, a(3)=4.
a(n-3) = ((5*n^2 - 37*n + 50)*F(n-1) + 4*(n-1)*F(n))/50 = (-1)^n*A055243(-n). - Peter Bala, Oct 25 2007
a(n) = A001628(n-3) + A001628(n-2). - R. J. Mathar, Dec 07 2011
a(n+1) = A123585(n+2,n). - Philippe Deléham, Dec 18 2011
a(n) = Sum_{k=floor((n-1)/2)..n-1}  k*(k+1)/2*C(k,n-k-1). - Vladimir Kruchinin, Sep 17 2020

A280554 T(n,k)=Number of nXk 0..1 arrays with no element equal to more than one of its horizontal and vertical neighbors, with the exception of exactly two elements, and with new values introduced in order 0 sequentially upwards.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 9, 9, 1, 2, 34, 50, 34, 2, 6, 87, 144, 144, 87, 6, 13, 194, 325, 382, 325, 194, 13, 29, 400, 670, 832, 832, 670, 400, 29, 60, 790, 1316, 1666, 1764, 1666, 1316, 790, 60, 122, 1511, 2502, 3182, 3438, 3438, 3182, 2502, 1511, 122, 241, 2830, 4654

Offset: 1

R. H. Hardin, Jan 05 2017

Table starts
...0....0....0.....1.....2.....6....13.....29.....60....122....241....468
...0....0....9....34....87...194...400....790...1511...2830...5213...9484
...0....9...50...144...325...670..1316...2502...4654...8521..15412..27612
...1...34..144...382...832..1666..3182...5886..10680..19122..33920..59754
...2...87..325...832..1764..3438..6386..11506..20389..35757..62317.108150
...6..194..670..1666..3438..6502.11697..20440..35226..60300.102974.175746
..13..400.1316..3182..6386.11697.20334..34360..57389..95548.159360.266756
..29..790.2502..5886.11506.20440.34360..56102..90614.146264.237392.388378
..60.1511.4654.10680.20389.35226.57389..90614.141398.220709.347310.553046
.122.2830.8521.19122.35757.60300.95548.146264.220709.332810.506359.781682

			Some solutions for n=4 k=4
..0..1..0..1. .0..1..0..0. .0..1..1..0. .0..1..0..1. .0..1..0..1
..1..0..1..0. .0..1..0..1. .1..1..0..1. .0..1..0..1. .1..0..1..1
..0..1..1..0. .1..0..1..0. .0..0..1..0. .1..0..1..0. .0..1..0..0
..1..0..1..0. .0..0..1..0. .1..1..0..1. .1..1..0..0. .0..1..0..1

R. H. Hardin, Table of n, a(n) for n = 1..391

Column 1 is A055243(n-4).

Column 2 is A279128.

Empirical for column k:
k=1: a(n) = 3*a(n-1) -5*a(n-3) +3*a(n-5) +a(n-6)
k=2: a(n) = 5*a(n-1) -7*a(n-2) -2*a(n-3) +10*a(n-4) -2*a(n-5) -5*a(n-6) +a(n-7) +a(n-8) for n>9
k=3: a(n) = 5*a(n-1) -7*a(n-2) -2*a(n-3) +10*a(n-4) -2*a(n-5) -5*a(n-6) +a(n-7) +a(n-8) for n>10
k=4: a(n) = 6*a(n-1) -12*a(n-2) +5*a(n-3) +12*a(n-4) -12*a(n-5) -3*a(n-6) +6*a(n-7) -a(n-9) for n>11
k=5: a(n) = 6*a(n-1) -12*a(n-2) +5*a(n-3) +12*a(n-4) -12*a(n-5) -3*a(n-6) +6*a(n-7) -a(n-9) for n>12
k=6: a(n) = 6*a(n-1) -12*a(n-2) +5*a(n-3) +12*a(n-4) -12*a(n-5) -3*a(n-6) +6*a(n-7) -a(n-9) for n>12
k=7: a(n) = 6*a(n-1) -12*a(n-2) +5*a(n-3) +12*a(n-4) -12*a(n-5) -3*a(n-6) +6*a(n-7) -a(n-9) for n>12

A136531 Coefficients of polynomials B(x,n) = ((1+a+b)x - c)B(x,n-1) - abB(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, -1, 3, -2, 1, 2, -5, 6, -3, 1, -3, 10, -13, 10, -4, 1, 5, -18, 29, -26, 15, -5, 1, -8, 33, -60, 65, -45, 21, -6, 1, 13, -59, 122, -151, 125, -71, 28, -7, 1, -21, 105, -241, 338, -321, 217, -105, 36, -8, 1, 34, -185, 468, -730, 784, -609, 350, -148, 45, -9, 1

Offset: 0

Roger L. Bagula, Mar 23 2008

			Triangle begins
        k=0  k=1  k=2  k=3  k=4  k=5  k=6
  n=0:   1;
  n=1:   0,   1;
  n=2:   1,  -1,   1;
  n=3:  -1,   3,  -2,   1;
  n=4:   2,  -5,   6,  -3,   1;
  n=5:  -3,  10, -13,  10,  -4,   1;
  n=6:   5, -18,  29, -26,  15,  -5,   1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000027, A000217, A008728, A010049, A039834.

Cf. A055243, A078008, A136526, A151575, A212804.

C := ComplexField(); // T = A136531
T:= func< n,k | k eq n select 1 else Round(i^(k-n-1)*(i*Evaluate(GegenbauerPolynomial(n-k, k+1), 1/(2*i)) - Evaluate(GegenbauerPolynomial(n-k-1, k+1), 1/(2*i)))) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 26 2022

Mathematica

(* First program *) a = -b; c = 1; b = 1; B[x_, n_]:= B[x, n]= If[n<2, x^n, ((1+a+b)*x -c)*B[x, n-1] -a*b*B[x, n-2]]; Table[CoefficientList[B[x,n], x], {n,0,10}]//Flatten (* Second program *) B[x_, n_]:= (-1)^n*(Fibonacci[n+1, 1-x] - Fibonacci[n, 1-x]); Table[CoefficientList[B[x, n], x], {n,0,16}]//Flatten (* G. C. Greubel, Sep 22 2022 *)

SageMath

def T(n,k): # T = A136531 if k==n: return 1 else: return i^(k-n-1)*(i*gegenbauer(n-k, k+1, 1/(2*i)) - gegenbauer(n-k-1, k+1, 1/(2*i))) flatten([[T(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 26 2022

Formula

G.f.: (1+y) / (1 + (1-x)*y - y^2). - Kevin Ryde, Sep 21 2022

From G. C. Greubel, Sep 22 2022: (Start)

T(n, k) = coefficients of i^n*(ChebyshevU(n, (x-1)/(2*i)) - i*ChebyshevU(n-1, (x-1)/(2*i))).

T(n, k) = coefficients of (-1)^n*( Fibonacci(n+1, 1-x) - Fibonacci(n, 1-x) ).

T(n, k) = i^(k-n-1)*(i*GegenbauerC(n-k, k+1, 1/(2*i)) - GegenbauerC(n-k-1, k+1, 1/(2*i))).

T(n, 0) = Fibonacci(1-n) = (-1)^n*A212804(n) = A039834(n-1).

T(n, 1) = (-1)^(n-1)*A010049(n), n >= 1.

T(n, 2) = (-1)^n*A055243(n-2), n >= 2.

T(n, n) = 1.

T(n, n-1) = -(n-1).

T(n, n-2) = A000217(n-1), n >= 2.

T(n, n-3) = -A008728(n-3), n >= 3.

Sum_{k=0..n-2} T(n, k) = A000027(n-1), n >= 2.

Sum_{k=0..n} T(n, k) = 1.

Sum_{k=0..floor(n/2)} T(n-k, k) = A151575(n) = (-1)^n*A078008(n). (End)

Extensions

Offset corrected by Kevin Ryde, Sep 21 2022

cp's OEIS Frontend

A001628 Convolved Fibonacci numbers.

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A281056 T(n,k)=Number of nXk 0..1 arrays with no element equal to more than one of its horizontal and antidiagonal neighbors, with the exception of exactly two elements, and with new values introduced in order 0 sequentially upwards.

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A210034 Triangle of coefficients of polynomials v(n,x) jointly generated with A210033; see the Formula section.

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A076791 Triangle a(n,k) giving number of binary sequences of length n containing k subsequences 00.

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A129707 Number of inversions in all Fibonacci binary words of length n.

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A280554 T(n,k)=Number of nXk 0..1 arrays with no element equal to more than one of its horizontal and vertical neighbors, with the exception of exactly two elements, and with new values introduced in order 0 sequentially upwards.

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A136531 Coefficients of polynomials B(x,n) = ((1+a+b)x - c)B(x,n-1) - abB(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.