A084326 -id:A084326 - cp's OEIS frontend

A069429 Half the number of 3 X n binary arrays with no path of adjacent 1's or adjacent 0's from top row to bottom row.

3, 16, 84, 440, 2304, 12064, 63168, 330752, 1731840, 9068032, 47480832, 248612864, 1301753856, 6816071680, 35689414656, 186872201216, 978475548672, 5123364487168, 26826284728320, 140464250421248, 735480363614208, 3851025180000256, 20164229625544704, 105581277033267200

Offset: 1

R. H. Hardin, Mar 22 2002

			From _Andrew Howroyd_, Oct 27 2020: (Start)
Some of the 2*a(2) = 32 arrays are:
  0 0   0 0   0 0   0 1   0 1   0 0   0 1
  0 0   0 1   1 1   1 0   1 0   1 1   1 0
  1 1   1 1   1 1   1 1   0 1   0 0   1 1
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. 2 X n A000079, n X 1 A000225, vertical path of 1 A069361-A069395, vertical paths of 0+1 A069396-A069416, vertical path of 1 not 0 A069417-A069428, no vertical paths A069429-A069447, no horizontal or vertical paths A069448-A069452.

Cf. A084326.

LinearRecurrence[{6, -4}, {3, 16}, 100] (* Jean-François Alcover, Nov 01 2020 *)

Vec((3 - 2*x)/(1 - 6*x + 4*x^2) + O(x^30)) \\ Andrew Howroyd, Oct 27 2020

a(n) = 2^(n-1)*fibonacci(2*n+2) \\ Andrew Howroyd, Oct 27 2020

Empirical G.f.: x*(3-2*x)/(1-6*x+4*x^2).  - Colin Barker, Feb 22 2012
Empirical: a(n) = 3*A084326(n) - 2*A084326(n-1). - R. J. Mathar, Nov 09 2018
From Andrew Howroyd, Oct 27 2020: (Start)
The above conjectures are true and follow from formulas given in A069361 and A069396.
a(n) = (8^n)/2 - A069361(n) + A069396(n).
a(n) = 2^(n-1)*Fibonacci(2*n+2) = A084326(n+1)/2. (End)

Terms a(21) and beyond from Andrew Howroyd, Oct 27 2020

A069396 Half the number of 3 X n binary arrays with a path of adjacent 1's and a path of adjacent 0's from top row to bottom row.

Original entry on oeis.org

1, 25, 377, 4541, 48329, 476389, 4461489, 40306317, 354713977, 3060942133, 26020259201, 218626028573, 1820140085705, 15043088032837, 123602247055953, 1010793162739629, 8234370308667673, 66870924588036181

Offset: 2

R. H. Hardin, Mar 22 2002

nonn

G. C. Greubel, Table of n, a(n) for n = 2..1000

Cf. 1 X n A000225, 2 X n A016269, vertical path of 1 A069361-A069395, vertical paths of 0+1 A069396-A069416, vertical path of 1 not 0 A069417-A069428, no vertical paths A069429-A069447, no horizontal or vertical paths A069448-A069452.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x^2*(2*x+1)^2/(1-8*x)/(2*x^2-7*x+1)/(4*x^2-6*x+1))); // G. C. Greubel, Apr 22 2018

Drop[CoefficientList[Series[x^2*(2*x+1)^2/(1-8*x)/(2*x^2-7*x + 1)/(4*x^2 - 6*x + 1), {x, 0, 50}], x], 2] (* G. C. Greubel, Apr 22 2018 *)

x='x+O('x^30); Vec(x^2*(2*x+1)^2/(1-8*x)/(2*x^2-7*x+1)/(4*x^2 -6*x+1)) \\ G. C. Greubel, Apr 22 2018

G.f.: x^2*(2*x+1)^2/(1-8*x)/(2*x^2-7*x+1)/(4*x^2-6*x+1). - Vladeta Jovovic, Jul 02 2003

2*a(n) = 8^n+A084326(n+1) -2*A186446(n). - R. J. Mathar, May 09 2023

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

Original entry on oeis.org

1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565, 72723460248141, 308061521170129, 1304969544928657, 5527939700884757

Offset: 0

Olivier Gérard

If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
For n >= 1, row sums of triangle
  m |k=0     1     2     3     4     5     6     7
====+=============================================
  0 |  1
  1 |  1     4
  2 |  1     4    16
  3 |  1     8    16    64
  4 |  1     8    48    64   256
  5 |  1    12    48   256   256  1024
  6 |  1    12    96   256  1280  1024  4096
  7 |  1    16    96   640  1280  6144  4096 16384
which is triangle for numbers 4^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) = a(n;-2) = 3^n*Sum_{k=0..n} binomial(n,k)*F(k+1)*(-2/3)^k, where a(n;d), n=0,1,...,d, denotes the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). We note that (see A033887) F(3n+1) = 3^n*a(n,2/3) = Sum_{k=0..n} binomial(n,k)*F(k-1)*(-2/3)^k, which implies F(3n+1) + 3^(-n)*a(n) = Sum_{k=0..n} binomial(n,k)*L(k)*(-2/3)^k, where L(k) denotes the k-th Lucas number. - Roman Witula, Jul 12 2012
a(n+1) is (for n >= 0) the number of length-n strings of 5 letters {0,1,2,3,4} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 4, 4*3, 4*3^2, 4*3^3, ...); i.e., of A003946: (1, 4, 12, 36, 108, ...). - Gary W. Adamson, Aug 06 2016
a(n+1) equals the number of quinary sequences of length n such that no two consecutive terms differ by 3. - David Nacin, May 31 2017

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), pp. 313-315.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (see Corollary 1 (vi)).
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Amelia Gilson, Hadley Killen, Steven J. Miller, Nadia Razek, Joshua M. Siktar, and Liza Sulkin, Zeckendorf's Theorem Using Indices in an Arithmetic Progression, arXiv:2005.10396 [math.NT], 2020.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
R. Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A147722 (INVERT transform), A109499 (INVERTi transform), A154626 (Binomial transform), A086344 (inverse binomial transform), A003946, A049310.

Cf. A000032, A000045, A014445, A014448, A033887, A046854, A055830, A055870, A084326, A167808.

[Fibonacci(3*n-1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

with(combinat): a:=n->fibonacci(n,4)-3*fibonacci(n-1,4): seq(a(n), n=1..23); # Zerinvary Lajos, Apr 04 2008

Fibonacci/@(3*Range[0,30]-1) (* Vladimir Joseph Stephan Orlovsky, Mar 01 2010 *)
LinearRecurrence[{4,1},{1,1},30] (* Harvey P. Dale, May 15 2019 *)

a[0]:1$
a[1]:1$
a[n]:=4*a[n-1]+a[n-2]$
A015448(n):=a[n]$
makelist(A015448(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = fibonacci(3*n-1); \\ Altug Alkan, Dec 10 2015

a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley, Dec 09 2001
a(n) = Sum_{k=0..n} 3^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n. - Gary W. Adamson, Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009
a(n) = Fibonacci(3n+1) mod Fibonacci(3n), n > 0.
a(n) = (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n))/2.
For n >= 2, a(n) = F_n(4) + F_(n+1)(4), where F_n(x) is a Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A001076(n+1) - 3*A001076(n). - R. J. Mathar, Jul 12 2012
From Gary Detlefs and Wolfdieter Lang, Aug 20 2012: (Start)
a(n) = (5*F(n)^3 + 5*F(n-1)^3 + 3*(-1)^n*F(n-2))/2,
a(n) = (F(n+1)^3 + 2*F(n)^3 - F(n-2)^3)/2, n >= 0, with F(-1) = 1 and F(-2) = -1. Second line from first one with 3*(-1)^n* F(n-2) = F(n-1)^3 - 4*F(n-2)^3 - F(n-3)^3 (in Koshy's book, p. 89, 32. (with a - sign) and 33. For the Koshy reference see A000045) and the F^3 recurrence (see row n=4 of A055870, or Koshy p. 87, 1.). First line from the preceding R. J. Mathar formula with F(3*n) = 5*F(n)^3 + 3*(-1)^n*F(n) (Koshy p. 89, 46.) and the above mentioned formula, Koshy's 32. and 33., with n -> n+2 in order to eliminate F(n+1)^3. (End)
For n > 0, a(n) = L(n-1)*L(n)*F(n) + F(n+1)*(-1)^n with L(n)=A000032(n). - J. M. Bergot, Dec 10 2015
For n > 1, a(n)^2 is the denominator of continued fraction [4,4,...,4, 6, 4,4,...4], which has n-1 4's before, and n-1 4's after, the middle 6. - Greg Dresden, Sep 18 2019
From Gary Detlefs and Wolfdieter Lang, Mar 06 2023: (Start)
a(n) = A001076(n) + A001076(n-1), with A001076(-1) = 1. See the R. J. Mathar formula above.
a(n+1) = i^n*(S(n-1,-4*i) - i*S(n-2,-4*i)), for n >= 0, with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified Fibonacci trisection formula for {F(3*n+2)}_{n>=0}. (End)
a(n) = Sum_{k=0..n} A046854(n-1,k)*4^k. - R. J. Mathar, Feb 10 2024
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Jun 03 2024

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A053538 Triangle: a(n,m) = ways to place p balls in n slots with m in the rightmost p slots, 0<=p<=n, 0<=m<=n, summed over p, a(n,m)= Sum_{k=0..n} binomial(k,m)*binomial(n-k,k-m), (see program line).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 5, 4, 1, 1, 8, 10, 7, 5, 1, 1, 13, 18, 16, 9, 6, 1, 1, 21, 33, 31, 23, 11, 7, 1, 1, 34, 59, 62, 47, 31, 13, 8, 1, 1, 55, 105, 119, 101, 66, 40, 15, 9, 1, 1, 89, 185, 227, 205, 151, 88, 50, 17, 10, 1, 1, 144, 324, 426, 414, 321, 213, 113, 61, 19, 11, 1, 1

Offset: 0

Wouter Meeussen, May 23 2001

Riordan array (1/(1-x-x^2), x(1-x)/(1-x-x^2)). Row sums are A000079. Diagonal sums are A006053(n+2). - Paul Barry, Nov 01 2006
Subtriangle of the triangle given by (0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 05 2012
Mirror image of triangle in A208342. - Philippe Deléham, Mar 05 2012
A053538 is jointly generated with A076791 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 at A076791. - Clark Kimberling, Mar 08 2012
The matrix inverse starts
    1;
   -1,   1;
   -1,  -1,   1;
    1,  -2,  -1,  1;
    3,   1,  -3, -1,  1;
    1,   6,   1, -4, -1,  1;
   -7,   4,  10,  1, -5, -1,  1;
  -13, -13,   8, 15,  1, -6, -1,  1;
    3, -31, -23, 13, 21,  1, -7, -1, 1; - R. J. Mathar, Mar 15 2013
Also appears to be the number of subsets of {1..n} containing n with k maximal anti-runs of consecutive elements increasing by more than 1. For example, the subset {1,3,6,7,11,12} has maximal anti-runs ((1,3,6),(7,11),(12)) so is counted under a(12,3). For runs instead of anti-runs we get A202064. - Gus Wiseman, Jun 26 2025

			n=4; Table[binomial[k, j]binomial[n-k, k-j], {k, 0, n}, {j, 0, n}] splits {1, 4, 6, 4, 1} into {{1, 0, 0, 0, 0}, {3, 1, 0, 0, 0}, {1, 4, 1, 0, 0}, {0, 0, 3, 1, 0}, {0, 0, 0, 0, 1}} and this gives summed by columns {5, 5, 4, 1, 1}
Triangle begins :
   1;
   1,  1;
   2,  1,  1;
   3,  3,  1, 1;
   5,  5,  4, 1, 1;
   8, 10,  7, 5, 1, 1;
  13, 18, 16, 9, 6, 1, 1;
...
(0, 1, 1, -1, 0, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, 0, ...) begins :
  1;
  0,  1;
  0,  1,  1;
  0,  2,  1,  1;
  0,  3,  3,  1, 1;
  0,  5,  5,  4, 1, 1;
  0,  8, 10,  7, 5, 1, 1;
  0, 13, 18, 16, 9, 6, 1, 1;

Alois P. Heinz, Rows n = 0..140, flattened
R. P. Grimaldi, Extraordinary subsets: a generalization, Fib. Quart., 55 (No. 3, 2017), 114-122. See Table 1.

Column k = 1 is A000045.
Row sums are A000079.
Column k = 2 is A010049.
For runs instead of anti-runs we have A202064.
For integer partitions see A268193, strict A384905, runs A116674.
A034839 counts subsets by number of maximal runs.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
A384893 counts subsets by number of maximal anti-runs.
Cf. A000071, A001629, A010027, A208342, A384177, A384879, A384889, A384890.

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j->  Binomial(j,k)*Binomial(n-j,j-k)) ))); # G. C. Greubel, May 16 2019

[[(&+[Binomial(j,k)*Binomial(n-j,j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, May 16 2019

a:= (n, m)-> add(binomial(k, m)*binomial(n-k, k-m), k=0..n):
seq(seq(a(n,m), m=0..n), n=0..12);  # Alois P. Heinz, Sep 19 2013

Table[Sum[Binomial[k, m]*Binomial[n-k, k-m], {k,0,n}], {n,0,12}, {m,0,n}]

{T(n,k) = sum(j=0,n, binomial(j,k)*binomial(n-j,j-k))}; \\ G. C. Greubel, May 16 2019

[[sum(binomial(j,k)*binomial(n-j,j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, May 16 2019

From Philippe Deléham, Mar 05 2012: (Start)
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if k>n.
G.f.: 1/(1-(1+y)*x-(1-y)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A077957(n), A000045(n+1), A000079(n), A001906(n+1), A007070(n), A116415(n), A084326(n+1), A190974(n+1), A190978(n+1), A190984(n+1), A190990(n+1), A190872(n+1) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively. (End)

A116415 a(n) = 5a(n-1) - 3a(n-2).

Original entry on oeis.org

1, 5, 22, 95, 409, 1760, 7573, 32585, 140206, 603275, 2595757, 11168960, 48057529, 206780765, 889731238, 3828313895, 16472375761, 70876937120, 304967558317, 1312206980225, 5646132226174, 24294040190195, 104531804272453

Offset: 0

Paul Barry, Feb 13 2006

Row sums of A116414.
Partial sums of A018902. - Greg Dresden and Mulong Xu, Aug 31 2024
Binomial transform of the sequence A006190. - Sergio Falcon, Nov 23 2007
a(n+1) equals the number of words of length n over {0,1,2,3,4} avoiding 01, 02 and 03. - Milan Janjic, Dec 17 2015

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Sergio Falcón, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Sergio Falcón and Ángel Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals, Volume 39, Issue 3, 15 February 2009, Pages 1005-1019.
Index entries for linear recurrences with constant coefficients, signature (5,-3).

Cf. A001906, A007070, A018902, A084326.

I:=[1,5]; [n le 2 select I[n] else 5*Self(n-1)-3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 16 2015

Join[{a=1,b=5},Table[c=5*b-3*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
LinearRecurrence[{5,-3},{1,5},40] (* Harvey P. Dale, Jun 19 2012 *)

Vec(1/(1-5*x+3*x^2) + O(x^100)) \\ Altug Alkan, Dec 17 2015

[lucas_number1(n,5,3) for n in range(1, 24)] # Zerinvary Lajos, Apr 22 2009

G.f.: 1/(1 - 5*x + 3*x^2).
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n-j,k)*C(k+j,j)*3^j.
a(n) = (1/sqrt(13))*(((5+sqrt(13))/2)^n - ((5-sqrt(13))/2)^n). - Sergio Falcon, Nov 23 2007
If p[i] = (3^i-1)/2, and if A is the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(n) = 4*a(n-1) + a(n-2) + a(n-3) + ... + a(0) + 1. These expansions with the partial sums on one side can be generated en masse by taking the g.f. of the partial sum and its partial fraction, 1/(1-x)/(1 - 5*x + 3*x^2) = -1/(1-x)+(2-3*x)/(1 - 5*x + 3*x^2) and reading this as a(0) + a(1) + ... + a(n) = -1 + 2*a(n)- 3*a(n-1). - Gary W. Adamson, Feb 18 2011

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 39, 252, 1629, 10530, 68067, 439992, 2844153, 18384894, 118841823, 768205620, 4965759189, 32099171994, 207492309531, 1341251373168, 8669985167601, 56043665125110, 362271946253463, 2341762672896108, 15137391876137037, 97849639275510546, 632510011281474387

Offset: 0

Paul Barry, Nov 19 2003

From Johannes W. Meijer, Aug 09 2010: (Start)

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner or side square on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032. The central square leads to A180028. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A141041, A180028, A180032.

Sequences with g.f. of the form 1/(1 - 6*x - k*x^2): A106392 (k=-10), A027471 (k=-9), A006516 (k=-8), A081179 (k=-7), A030192 (k=-6), A003463 (k=-5), A084326 (k=-4), A138395 (k=-3), A154244 (k=-2), A001109 (k=-1), A000400 (k=0), A005668 (k=1), A135030 (k=2), this sequence (k=3), A135032 (k=4), A015551 (k=5), A057089 (k=6), A015552 (k=7), A189800 (k=8), A189801 (k=9), A190005 (k=10), A015553 (k=11).

[n le 2 select 6^(n-1) else 6*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

a:= n-> (<<0|1>, <3|6>>^n. <<1,6>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 17 2011

Join[{a=1,b=6},Table[c=6*b+3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{6,3}, {1,6}, 41] (* G. C. Greubel, Oct 10 2022 *)

my(x='x+O('x^30)); Vec(1/(1-6*x-3*x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,6,-3) for n in range(1, 31)] # Zerinvary Lajos, Apr 24 2009

a(n) = (3+2*sqrt(3))^n*(sqrt(3)/4+1/2) + (1/2-sqrt(3)/4)*(3-2*sqrt(3))^n.
a(n) = (-i*sqrt(3))^n * ChebyshevU(n, isqrt(3)), i^2=-1.
From Johannes W. Meijer, Aug 09 2010: (Start)
G.f.: 1/(1 - 6*x - 3*x^2).
Limit_{k->oo} a(n+k)/a(k) = A141041(n) + A090018(n-1)*sqrt(12) for n >= 1.
Limit_{n->oo} A141041(n)/A090018(n-1) = sqrt(12). (End)
a(n) = Sum_{k=0..n} A099089(n,k)*3^k. - Philippe Deléham, Nov 21 2011
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(3)*x) + sqrt(3)*sinh(2*sqrt(3)*x))/2. - Stefano Spezia, Apr 23 2025

Typo in Mathematica program corrected by Vincenzo Librandi, Nov 15 2011

A098648 Expansion of (1-3x)/(1 - 6x + 4*x^2).

Original entry on oeis.org

1, 3, 14, 72, 376, 1968, 10304, 53952, 282496, 1479168, 7745024, 40553472, 212340736, 1111830528, 5821620224, 30482399232, 159607914496, 835717890048, 4375875682304, 22912382533632, 119970792472576, 628175224700928, 3289168178315264, 17222308171087872

Offset: 0

Paul Barry, Sep 18 2004

Binomial transform of A001077. Second binomial transform of A084057. Third binomial transform of 1/(1-5*x^2). Let A=[1,1,1,1;3,1,-1,-3;3,-1,-1,3;1,-1,1,-1], the 4 X 4 Krawtchouk matrix. Then a(n)=trace((16(A*A`)^(-1))^n)/4.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A098647.

a[n_]:=(MatrixPower[{{5,1},{1,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
CoefficientList[Series[(1-3x)/(1-6x+4x^2),{x,0,30}],x] (* or *) LinearRecurrence[{6,-4},{1,3},31] (* Harvey P. Dale, Jun 06 2011 *)
Table[2^(n - 1) LucasL[2 n], {n, 0, 20}] (* Eric W. Weisstein, Mar 31 2017 *)

Vec((1-3*x)/(1 - 6*x + 4*x^2) + O(x^25)) \\ Jinyuan Wang, Jul 24 2021

E.g.f.: exp(3*x)*cosh(sqrt(5)*x).
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2.
a(n) = 2*(3*a(n-1) - 2*a(n-2)). - Lekraj Beedassy, Oct 22 2004
a(n) = A084326(n+1) - 3*A084326(n). - R. J. Mathar, Nov 10 2013
a(n) = 2^(n-1)*Lucas(2*n) = 2^(n-1)*A005248(n), n>0. - Eric W. Weisstein, Mar 31 2017

A154626 a(n) = 2^n*A001519(n).

Original entry on oeis.org

1, 2, 8, 40, 208, 1088, 5696, 29824, 156160, 817664, 4281344, 22417408, 117379072, 614604800, 3218112512, 16850255872, 88229085184, 461973487616, 2418924584960, 12665653559296, 66318223015936, 347246723858432, 1818207451086848, 9520257811087360

Offset: 0

Paul Barry, Jan 13 2009

Hankel transform of 1,1,3,11,45,... (see A026375). Binomial transform of A015448.
From Gary W. Adamson, Jul 22 2016: (Start)
A production matrix for the sequence is M =
  1, 1, 0, 0, 0, ...
  1, 0, 5, 0, 0, ...
  1, 0, 0, 5, 0, ...
  1, 0, 0, 0, 5, ...
  ...
Take powers of M, extracting the upper left terms; getting
the sequence starting (1, 1, 2, 8, 40, 208, ...). (End)
The sequence is N=5 in an infinite set of INVERT transforms of powers of N prefaced with a "1". (1, 2, 8, 40, ...) is the INVERT transform of (1, 1, 5, 25, 125, ...). The first six of such sequences are shown in A006012 (N=3). - Gary W. Adamson, Jul 24 2016
From Gary W. Adamson, Jul 27 2016: (Start)
The sequence is the first in an infinite set in which we perform the operation for matrix M (Cf. Jul 22 2016), but change the left border successively from (1, 1, 1, 1, ...) then to (1, 2, 2, 2, ...), then (1, 3, 3, 3, ...) ...; generally (1, N, N, N, ...). Extracting the upper left terms of each matrix operation, we obtain the infinite set beginning:
  N=1  (A154626): 1,  2,  8,  40,  208,  1088, ...
  N=2  (A084120): 1,  3, 15,  81,  441,  1403, ...
  N=3  (A180034): 1,  4, 22, 124,  700,  3952, ...
  N=4  (A001653): 1,  5, 29, 169,  985,  5741, ...
  N=5  (A000400): 1,  6, 36, 216, 1296,  7776, ...
  N=6  (A015451): 1,  7, 43, 265, 1633, 10063, ...
  N=7  (A180029): 1,  8, 50, 316, 1996, 12608, ...
  N=8  (A180028): 1,  9, 57, 369, 1285, 15417, ...
  N=9  (.......): 1, 10, 64, 424, 2800, 18496, ...
  N=10 (A123361): 1, 11, 71, 481, 3241, 21851, ...
  N=11 (.......): 1, 12, 78, 540, 3708, 25488, ...
... Each of the sequences begins (1, (N+1), (7*N + 1),
  (40*N + (N-1)^2), ... (End)
The set of infinite sequences shown (Cf. comment of Jul 27 2016), can be generated from the matrices P = [(1,N; 1,5]^n, (N=1,2,3,...) by extracting the upper left terms. Example: N=6 sequence (A015451): (1, 7, 43, 265, ...) can be generated from the matrix P = [(1,6); (1,5)]^n. - Gary W. Adamson, Jul 28 2016

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
Bishal Deb, Cyclic sieving phenomena via combinatorics of continued fractions, arXiv:2508.13709 [math.CO], 2025. See p. 42.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A006012, A084120, A180034, A001653, A000400, A015451, A180029, A180028, A123362.

[n le 2 select (n) else 6*Self(n-1)-4*Self(n-2): n in [1..25]]; // Vincenzo Librandi, May 15 2015

LinearRecurrence[{6, -4}, {1, 2}, 30] (* Vincenzo Librandi, May 15 2015 *)

Vec((1-4*x) / (1-6*x+4*x^2) + O(x^30)) \\ Colin Barker, Sep 22 2017

G.f.: (1 - 4*x) / (1 - 6*x + 4*x^2).
a(n) = A084326(n+1) - 4*A084326(n). - R. J. Mathar, Jul 19 2012
From Colin Barker, Sep 22 2017: (Start)
a(n) = (((3-sqrt(5))^n*(1+sqrt(5)) + (-1+sqrt(5))*(3+sqrt(5))^n)) / (2*sqrt(5)).
a(n) = 6*a(n-1) - 4*a(n-2) for n>1. (End)
E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Aug 26 2025

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

Original entry on oeis.org

1, 6, 37, 227, 1394, 8559, 52553, 322678, 1981261, 12165051, 74694082, 458625767, 2815987409, 17290317414, 106163498933, 651849716563, 4002393075346, 24574913392671, 150891318490777, 926480986202582, 5688644160448349

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner or side square (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white chess queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the central square (we assume here that a red queen might behave like a white queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the corner and side squares the 512 red queens lead to 17 red queen sequences, see the cross-references for the complete set.
The sequence above corresponds to 8 red queen vectors, i.e., A[5] vectors, with decimal values 239, 367, 431, 463, 487, 491, 493 and 494. The central square leads for these vectors to A152240.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 5*x - k*x^2). The members of this family that are red queen sequences are A180030 (k=8), A180032 (k=7; this sequence), A000400 (k=6), A180033 (k=5), A126501 (k=4), A180035 (k=3), A180037 (k=2) A015449 (k=1) and A003948 (k=0). Other members of this family are A030221 (k=-1), A109114 (k=-3), A020989 (k=-4), A166060 (k=-6).
Inverse binomial transform of A054413.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wikipedia, Alice in Wonderland (2010 film)
Index entries for linear recurrences with constant coefficients, signature (5, 7).

Cf. A180028 (Central square).

Cf. Red queen sequences corner and side squares [decimal value A[5]]: A090018 [511], A135030 [255], A180030 [495], A005668 [127], A180032 [239], A000400 [63], A180033 [47], A001109 [31], A126501 [15], A154244 [23], A180035 [7], A138395 [19], A180037 [3], A084326 [17], A015449 [1], A003463 [16], A003948 [0].

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [1,1,1,1,0,1,1,1,0]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): 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);

LinearRecurrence[{5,7},{1,6},40] (* Vincenzo Librandi, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-5x-7x^2),{x,0,30}],x] (* Harvey P. Dale, Apr 04 2024 *)

G.f.: (1+x)/(1 - 5*x - 7*x^2).
a(n) = 5*a(n-1) + 7*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+9*A)*A^(-n-1) + (7+9*B)*B^(-n-1))/53 with A = (-5+sqrt(53))/14 and B = (-5-sqrt(53))/14.

cp's OEIS Frontend

A069429 Half the number of 3 X n binary arrays with no path of adjacent 1's or adjacent 0's from top row to bottom row.

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A069396 Half the number of 3 X n binary arrays with a path of adjacent 1's and a path of adjacent 0's from top row to bottom row.

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A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

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A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A053538 Triangle: a(n,m) = ways to place p balls in n slots with m in the rightmost p slots, 0<=p<=n, 0<=m<=n, summed over p, a(n,m)= Sum_{k=0..n} binomial(k,m)*binomial(n-k,k-m), (see program line).

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A116415 a(n) = 5*a(n-1) - 3*a(n-2).

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A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

A116415 a(n) = 5a(n-1) - 3a(n-2).

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

A098648 Expansion of (1-3x)/(1 - 6x + 4*x^2).

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).