A082761 -id:A082761 - cp's OEIS frontend

A033887 a(n) = Fibonacci(3*n + 1).

1, 3, 13, 55, 233, 987, 4181, 17711, 75025, 317811, 1346269, 5702887, 24157817, 102334155, 433494437, 1836311903, 7778742049, 32951280099, 139583862445, 591286729879, 2504730781961, 10610209857723, 44945570212853, 190392490709135, 806515533049393, 3416454622906707

Offset: 0

N. J. A. Sloane

Binomial transform of A063727, and second binomial transform of (1,1,5,5,25,25,...), which is A074872 with offset 0. - Paul Barry, Jul 16 2003
Equals INVERT transform of A104934: (1, 2, 8, 28, 100, 356, ...) and INVERTi transform of A005054: (1, 4, 20, 100, 500, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 3 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
F(3*n+1) = 3^n*a(n;2/3), where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers by Witula et al.). - Roman Witula, Jul 12 2012
We note that the remark above by Paul Barry can be easily obtained from the following scaling identity for delta-Fibonacci numbers y^n a(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) a(k;x) and the fact that a(n;2)=5^floor(n/2). Indeed, for x=y=2 we get 2^n a(n;1) = Sum_{k=0..n} binomial(n,k) a(k;2) and, by A000045: Sum_{k=0..n} binomial(n,k) 2^k a(k;1) = Sum_{k=0..n} binomial(n,k) F(k+1) 2^k = 3^n a(n;2/3) = F(3n+1). - Roman Witula, Jul 12 2012
Except for the first term, this sequence can be generated by Corollary 1 (iv) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Number of 1’s in the substitution system {0 -> 110, 1 -> 11100} at step n from initial string "1" (1 -> 11100 -> 111001110011100110110 -> ...). - Ilya Gutkovskiy, Apr 10 2017
The o.g.f. of {F(m*n + 1)}A000045%20and%20L%20=%20A000032.%20-%20_Wolfdieter%20Lang">{n>=0}, for m = 1, 2, ..., is G(m,x) = (1 - F(m-1)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and L = A000032. - _Wolfdieter Lang, Feb 06 2023

			a(5) = Fibonacci(3*5 + 1) = Fibonacci(16) = 987. - _Indranil Ghosh_, Feb 04 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1592
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, 7(38) (2012), 1871-1876.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), #10.8.2, Example 13.
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013), #13.4.5.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Mathematica, 46(1) (2013), 15-27.
Roman 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. A000032, A000045, A104934, A005054, A063727 (inverse binomial transform), A082761 (binomial transform), A001076, A001077.

Cf. A016777, A049310, A049651, A074872, A122070, A167808, A185384.

[Fibonacci(3*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Fibonacci[Range[1,5!,3]] (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)

a(n)=fibonacci(3*n+1) \\ Charles R Greathouse IV, Feb 03 2014

Vec((1-x)/(1-4*x-x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = A001076(n) + A001077(n) = A001076(n+1) - A001076(n).
a(n) = 2*A049651(n) + 1.
a(n) = 4*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=3;
G.f.: (1 - x)/(1 - 4*x - x^2).
a(n) = ((1 + sqrt(5))*(2 + sqrt(5))^n - (1 - sqrt(5))*(2 - sqrt(5))^n)/(2*sqrt(5)).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*F(n-j+1). - Paul Barry, May 19 2006
First differences of A001076. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = A167808(3*n+1). - Reinhard Zumkeller, Nov 12 2009
a(n) = Sum_{k=0..n} C(n,k)*F(n+k+1). - Paul Barry, Apr 19 2010
Let p[1]=3, p[i]=4, (i>1), and A be a 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) = det A. - Milan Janjic, Apr 29 2010
a(n) = Sum_{i=0..n} C(n,n-i)*A063727(i). - Seiichi Kirikami, Mar 06 2012
a(n) = Sum_{k=0..n} A122070(n,k) = Sum_{k=0..n} A185384(n,k). - Philippe Deléham, Mar 13 2012
a(n) = A000045(A016777(n)). - Michel Marcus, Dec 10 2015
a(n) = F(2*n)*L(n+1) + F(n-1)*(-1)^n for n > 0. - J. M. Bergot, Feb 09 2016
a(n) = Sum_{k=0..n} binomial(n,k)*5^floor(k/2)*2^(n-k). - Tony Foster III, Sep 03 2017
2*a(n) = Fibonacci(3*n) + Lucas(3*n). - Bruno Berselli, Oct 13 2017
a(n)^2 is the denominator of continued fraction [4,...,4, 2, 4,...,4], which has n 4's before, and n 4's after, the middle 2. - Greg Dresden and Hexuan Wang, Aug 30 2021
a(n) = i^n*(S(n, -4*i) + i*S(n-1, -4*i)), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified trisection formula. See the first entry above with A001076. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A147703 Triangle [1,1,1,0,0,0,...] DELTA [1,0,0,0,...] with Deléham DELTA defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 13, 27, 20, 7, 1, 34, 80, 73, 35, 9, 1, 89, 234, 252, 151, 54, 11, 1, 233, 677, 837, 597, 269, 77, 13, 1, 610, 1941, 2702, 2225, 1199, 435, 104, 15, 1, 1597, 5523, 8533, 7943, 4956, 2158, 657, 135, 17, 1

Offset: 0

Paul Barry, Nov 10 2008

Equal to A062110*A007318 when A062110 is regarded as a triangle read by rows.

			Triangle begins
   1;
   1,   1;
   2,   3,   1;
   5,   9,   5,   1;
  13,  27,  20,   7,  1;
  34,  80,  73,  35,  9,  1;
  89, 234, 252, 151, 54, 11, 1;

Indranil Ghosh, Rows 0..100, flattened

Row sums are A006012. Diagonal sums are A147704.

Cf. A062110, A007318, A206800, A001519, A321620.

# The function RiordanSquare is defined in A321620:
RiordanSquare(1 / (1 - x / (1 - x / (1 - x))), 10); # Peter Luschny, Jan 26 2020

nmax=9; Flatten[CoefficientList[Series[CoefficientList[Series[(1 - 2*x)/(1 - (3 + y)*x + (1 + y)*x^2), {x, 0, nmax}], x], {y, 0, nmax}], y]] (* Indranil Ghosh, Mar 11 2017 *)

Riordan array ((1-2x)/(1-3x+x^2), x(1-x)/(1-3x+x^2)).
Sum_{k=0..n} T(n,k)*x^k = A152239(n), A152223(n), A152185(n), A152174(n), A152167(n), A152166(n), A152163(n), A000007(n), A001519(n), A006012(n), A081704(n), A082761(n), A147837(n), A147838(n), A147839(n), A147840(n), A147841(n), for x = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Dec 01 2008
G.f.: (1-2*x)/(1-(3+y)*x+(1+y)*x^2). - Philippe Deléham, Nov 26 2011
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), for n > 1. - Philippe Deléham, Feb 12 2012
The Riordan square of the odd indexed Fibonacci numbers A001519. - Peter Luschny, Jan 26 2020

A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).

Original entry on oeis.org

1, 5, 15, 55, 197, 727, 2705, 10165, 38455, 146301, 559131, 2145025, 8255575, 31861025, 123256495, 477823895, 1855782325, 7219352975, 28125910825, 109720617995, 428537256445, 1675561707275, 6557869020325, 25689734662775

Offset: 0

Clark Kimberling

nonn

p divides a((p-3)/2) for p in A045468 (primes congruent to {1, 4} mod 5). - Alexander Adamchuk, Jul 05 2006
The sequence 1,1,5,15,55,... has general term sum{k=0..n, (-1)^(n-k)*C(2k,k)}. Its Hankel transform is A082761. - Paul Barry, Apr 10 2007
From Paul Barry, Mar 29 2010: (Start)
The sequence 1,1,5,15,... has g.f. 1/((1+x)*sqrt(1-4x)).
The doubled sequence 1,1,1,1,5,5,... has e.g.f. dif(int((sin(x-t)+cos(x-t))*Bessel_I(0,2t),t,0,x),x). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

T(2n, n), array T as in A054106.

Cf. A066796, A000984, A054109, A006134, A045468.

Table[Sum[(-1)^(k+n)*((2k)!/(k!)^2),{k,0,n}], {n,1,50}] (* Alexander Adamchuk, Jul 05 2006 *)
CoefficientList[Series[(1/Sqrt[1-4*x]/(1+x)-1)/x, {x, 0, 20}], x]
(* or *)
Table[(-1)^(n+1)*Sum[(-1)^k*Binomial[2*k, k], {k, 0, n+1}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)
Round@Table[Binomial[2 (n + 2), n + 2] Hypergeometric2F1[1, n + 5/2, n + 3, -4] - (-1)^n/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

a(n)=(-1)^(n+1)*sum(k=0,n+1,(-1)^k*binomial(2*k,k))

from math import comb
def A054108(n): return (1 if n % 2 else -1)*sum((-1 if k % 2 else 1)*comb(2*k,k) for k in range(n+2)) # Chai Wah Wu, Jan 19 2022

a(n) = C(2n, n) - a(n-1) with a(0)=1. - Labos Elemer, Apr 26 2003
C(2n,n) - C(2n-2,n-1) + ... +(-1)^(k+n)*C(2k,k)+ ... + (-1)^(1+n)*C(2,1) + (-1)^n*C(0,0), where C(2k,k)=(2k)!/(k!)^2 - central binomial coefficients A000984[k]. - Alexander Adamchuk, Jul 05 2006
a(n) = Sum_{k=0..n} (-1)^(k+n)*((2k)!/(k!)^2). - Alexander Adamchuk, Jul 05 2006
G.f.: (1/sqrt(1-4*x)/(1+x)-1)/x = (-1 + 2/(U(0)-2*x))/(1+x) where U(k)= 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 27 2012
a(n) ~ 2^(2*n+4)/(5*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
Recurrence: (n+1)*a(n) = (3*n+1)*a(n-1) + 2*(2*n+1)*a(n-2). - Vaclav Kotesovec, Nov 06 2012

Formula from Benoit Cloitre, Sep 29 2002

Definition corrected by Vaclav Kotesovec, Nov 06 2012

A046748 Row sums of triangle A046521.

Original entry on oeis.org

1, 3, 13, 61, 295, 1447, 7151, 35491, 176597, 880125, 4390901, 21920913, 109486993, 547018941, 2733608905, 13662695645, 68294088535, 341399727335, 1706739347095, 8532741458075, 42660172763995, 213287735579135, 1066389745361635, 5331765761680895

Offset: 0

Wolfdieter Lang, Dec 11 1999

Hankel transform is A082761. - Paul Barry, Apr 14 2010

			G.f. = 1 + 3*x + 13*x^2 + 61*x^3 + 295*x^4 + 1447*x^5 + 7151*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
T.-X. He, L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, p 35.

Cf. A000108, A046521, A046714, A082761.

R:=PowerSeriesRing(Rationals(), 40);
Coefficients(R!( Sqrt(1-4*x)/(1-5*x) )); // G. C. Greubel, Jul 28 2024

a[ n_] := SeriesCoefficient[ Sqrt[ 1 - 4 x] / (1 - 5 x), {x, 0, n}]; (* Michael Somos, May 25 2014 *)
a[ n_] := Binomial[ 2 n, n] Hypergeometric2F1[ -n, 1, 1/2, -1/4]; (* Michael Somos, May 25 2014 *)

{a(n) = if( n<0, 0, polcoeff( sqrt( 1 - 4*x + x * O(x^n)) / (1 - 5*x), n))}; /* Michael Somos, May 25 2014 */

def A046748_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( sqrt(1-4*x)/(1-5*x) ).list()
A046748_list(40) # G. C. Greubel, Jul 28 2024

a(n) = binomial(2*n, n)*Sum_{k=0..n} binomial(n, k)/binomial(2*k, k).
a(n) = 5^n - 2*A046714(n-1), A046714(-1) := 0.
a(n) = 5*a(n-1) - 2*A000108(n-1).
G.f.: sqrt(1-4*x)/(1-5*x).
a(n) = (3*(3*n-2)/n)*a(n-1) - (10*(2*n-3)/n)*a(n-2), n >= 1, a(-1) := 0, a(0)=1 (homogeneous recursion).
a(n) = binomial(2*n,n)*hypergeom([ -n,1 ],[ 1/2 ],-1/4) (hypergeometric 2F1 form).
0 = a(n)*(+400*a(n+1) - 330*a(n+2) + 50*a(n+3)) + a(n+1)*(-30*a(n+1) + 71*a(n+2) - 15*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, May 25 2014
a(n) ~ 5^(n - 1/2). - Vaclav Kotesovec, Jul 07 2016
D-finite with recurrence n*a(n) +3*(-3*n+2)*a(n-1) +10*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jul 23 2017

A062110 A(n,k) is the coefficient of x^k in (1-x)^n/(1-2*x)^n for n, k >= 0; Table A read by descending antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 5, 3, 1, 0, 8, 12, 9, 4, 1, 0, 16, 28, 25, 14, 5, 1, 0, 32, 64, 66, 44, 20, 6, 1, 0, 64, 144, 168, 129, 70, 27, 7, 1, 0, 128, 320, 416, 360, 225, 104, 35, 8, 1, 0, 256, 704, 1008, 968, 681, 363, 147, 44, 9, 1, 0, 512, 1536, 2400, 2528, 1970

Offset: 0

Henry Bottomley, May 30 2001

The triangular version of this square array is defined by T(n,k) = A(k,n-k) for 0 <= k <= n. Conversely, A(n,k) = T(n+k,n) for n,k >= 0. We have [o.g.f of T](x,y) = [o.g.f. of A](x*y, x) and [o.g.f. of A](x,y) = [o.g.f. of T](y,x/y). - Petros Hadjicostas, Feb 11 2021
From Paul Barry, Nov 10 2008: (Start)
As number triangle, Riordan array (1, x(1-x)/(1-2x)). A062110*A007318 is A147703.
[0,1,1,0,0,0,....] DELTA [1,0,0,0,.....]. (Philippe Deléham's DELTA is defined in A084938.) (End)
Modulo 2, this triangle T becomes triangle A106344. - Philippe Deléham, Dec 18 2008

			Table A(n,k) (with rows n >= 0 and columns k >= 0) begins:
  1, 0,  0,   0,   0,    0,    0,     0,     0,     0, ...
  1, 1,  2,   4,   8,   16,   32,    64,   128,   256, ...
  1, 2,  5,  12,  28,   64,  144,   320,   704,  1536, ...
  1, 3,  9,  25,  66,  168,  416,  1008,  2400,  5632, ...
  1, 4, 14,  44, 129,  360,  968,  2528,  6448, 16128, ...
  1, 5, 20,  70, 225,  681, 1970,  5500, 14920, 39520, ...
  1, 6, 27, 104, 363, 1182, 3653, 10836, 31092, 86784, ...
  ... - _Petros Hadjicostas_, Feb 15 2021
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,   1;
  0,   4,   5,   3,   1;
  0,   8,  12,   9,   4,   1;
  0,  16,  28,  25,  14,   5,   1;
  0,  32,  64,  66,  44,  20,   6,   1;
  0,  64, 144, 168, 129,  70,  27,   7,   1;
  0, 128, 320, 416, 360, 225, 104,  35,   8,   1;
  ... - _Philippe Deléham_, Nov 30 2008

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened.)
Eunice Y. S. Chan, Robert M. Corless, Laureano Gonzalez-Vega, J. Rafael Sendra, and Juana Sendra, Upper Hessenberg and Toeplitz Bohemians, arXiv:1907.10677 [cs.SC], 2019.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq., 21 (2018), #18.1.4.

Rows of A include A000007, A011782, A045623, A058396, A062109, A169792, A169793, A169794, A169795, A169796, A169797.
Columns of A include A000012, A001477, A000096, A000297.
Main diagonal of A is A002002.
Table A(n, k) is a multiple of 2^(k-n); dividing by this gives a table similar to A050143 except at the edges.
Essentially the same array as A105306, A160232.

t[n_, n_] = 1; t[n_, k_] := 2^(n-2*k)*k*Hypergeometric2F1[1-k, n-k+1, 2, -1]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 30 2013, after Philippe Deléham + symbolic sum *)

a(i,j)=if(i<0 || j<0,0,polcoeff(((1-x)/(1-2*x)+x*O(x^j))^i,j))

Formulas for the square array (A(n,k): n,k >= 0):
A(n, k) = A(n-1, k) + Sum_{0 <= j < k} A(n, j) for n >= 1 and k >= 0 with A(0, k) = 0^k for k >= 0.
G.f.: 1/(1-x*(1-y)/(1-2*y)) = Sum_{i, j >= 0} A(i, j) x^i*y^j.
From Petros Hadjicostas, Feb 15 2021: (Start)
A(n,k) = 2^(k-n)*n*hypergeom([1-n, k+1], [2], -1) for n >= 0 and k >= 1.
A(n,k) = 2*A(n,k-1) + A(n-1,k) - A(n-1,k-1) for n,k >= 1 with A(n,0) = 1 for n >= 0 and A(0,k) = 0 for k >= 1. (End)
Formulas for the triangle (T(n,k): 0 <= k <= n):
From Philippe Deléham, Aug 01 2006: (Start)
T(n,k) = A121462(n+1,k+1)*2^(n-2*k) for 0 <= k < n.
T(n,k) = 2^(n-2*k)*k*hypergeom([1-k, n-k+1], [2], -1) for 0 <= k < n. (End)
Sum_{k=0..n} T(n,k)*x^k = A152239(n), A152223(n), A152185(n), A152174(n), A152167(n), A152166(n), A152163(n), A000007(n), A001519(n), A006012(n), A081704(n), A082761(n), A147837(n), A147838(n), A147839(n), A147840(n), A147841(n), for x = -7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Dec 09 2008
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) for 1 <= k <= n-1 with T(0,0) = T(1,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, and T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Oct 30 2013
G.f.: Sum_{n.k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/(x^2*y - x*y - 2*x + 1). - Petros Hadjicostas, Feb 15 2021

Various sections edited by Petros Hadjicostas, Feb 15 2021

A083076 Third row of number array A083075.

Original entry on oeis.org

1, 5, 33, 229, 1601, 11205, 78433, 549029, 3843201, 26902405, 188316833, 1318217829, 9227524801, 64592673605, 452148715233, 3165041006629, 22155287046401, 155087009324805, 1085609065273633, 7599263456915429, 53194844198408001

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A067411. Inverse binomial transform of A082412.

Trinomial transform of Jacobsthal numbers A001045. - Paul Barry, Sep 10 2007

Nathaniel Johnston, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (8,-7).

Cf. A034478, A066443, A082761, A083077.

[(2*7^n+1)/3: n in [0..30]]; // Vincenzo Librandi, Nov 06 2011

seq((2*7^n+1)/3,n=0..20); # Nathaniel Johnston, Jun 26 2011

a(n) = (2*7^n + 1)/3.
G.f.: (1-3*x)/((1-x)*(1-7*x)).
E.g.f.: (2*exp(7*x) + exp(x))/3.
a(n) = Sum_{k=0..2*n} trinomial(n,k)*Fibonacci(k+1), where trinomial(n,k) are the trinomial coefficients (A027907). - Paul Barry, Sep 10 2007
a(n) = 7*a(n-1) - 2, a(n) = 8*a(n-1) - 7*a(n-2). - Vincenzo Librandi, Nov 06 2011

A375354 T(m,n) is the number of suitably connected m X n Legendrian mosaics read by rows, with 1<=n<=m.

Original entry on oeis.org

1, 1, 2, 1, 4, 20, 1, 8, 104, 1504, 1, 16, 544, 22208, 948032, 1, 32, 2848, 329216, 40930304, 5204262912, 1, 64, 14912, 4883968, 1772261888, 666548862976, 254112496082944, 1, 128, 78080, 72464384, 76795762688, 85575149027328, 97392800416399360, 111879597850371293184

Offset: 1

Margaret Kipe, Samantha Pezzimenti, Leif Schaumann, Luc Ta, Wing Hong Tony Wong, Aug 13 2024

An m X n Legendrian mosaic is an m X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection. The condition of being suitably connected means that the connection points of each tile coincide with those of the contiguous tiles.
The Mathematica program below is based on the algorithm given in Theorem 1 of Oh, Hong, Lee, and Lee, adapted to the Legendrian setting: since Legendrian mosaic tiles omit the crossing tile T_9 used in general knot mosaics, the bottom-right submatrix of O_(k+1) is 3*O_k rather than 4*O_k. See Theorem 6 of Kipe et al.
T(m,2) = A375353(m,2) = A000079(m-1) for all m >= 2 since neither classical nor Legendrian link mosaics with only 2 columns or rows can use T_9 tiles.

			Triangle begins:
  1;
  1,  2;
  1,  4,   20;
  1,  8,  104,   1504;
  1, 16,  544,  22208,   948032;
  1, 32, 2848, 329216, 40930304, 5204262912;
  ...
T(2,2) = 2 since the only suitably connected 2 X 2 Legendrian mosaics are the empty mosaic and the mosaic depicting the Legendrian unknot with maximal Thurston-Bennequin invariant.
For all n >= 1, we have T(n,1) = 1 since the only suitably connected Legendrian mosaic with one column is empty.

Luc Ta, First 11 rows of the triangle, flattened
Margaret Kipe, Samantha Pezzimenti, Leif Schaumann, Luc Ta, and Wing Hong Tony Wong, Bounds on the mosaic number of Legendrian knots, arXiv: 2410.08064 [math.GT], 2024.
Seungsang Oh, Kyungpyo Hong, Ho Lee, and Hwa Jeong Lee, Quantum knots and the number of knot mosaics, arXiv: 1412.4460 [math.GT], 2014.
S. Pezzimenti and A. Pandey, Geography of Legendrian knot mosaics, Journal of Knot Theory and its Ramifications, 31 (2022), article no. 2250002, 1-22.
Index entries for sequences related to knots

The main diagonal T(n,n) is A374947.

Cf. A000079, A082761, A374946, A375353, A261400.

x[0] = o[0] = {{1}};
x[n_] := ArrayFlatten[{{x[n - 1], o[n - 1]}, {o[n - 1], x[n - 1]}}];
o[n_] := ArrayFlatten[{{o[n - 1], x[n - 1]}, {x[n - 1], 3*o[n - 1]}}];
legendrian[m_, n_] := If[m > 1 && n > 1, 2*Total[MatrixPower[x[m - 2] + o[m - 2], n - 2], 2], 1];
Flatten[ParallelTable[legendrian[m, n], {m, 1, 11}, {n, 1, m}]] (* Luc Ta, Aug 13 2024 *)

T(m,3) = A082761(m-1) for all m >= 1. - Luc Ta, Aug 20 2024

A080878 a(n)a(n+3) - a(n+1)a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=3.

Original entry on oeis.org

1, 1, 3, 4, 14, 20, 72, 104, 376, 544, 1968, 2848, 10304, 14912, 53952, 78080, 282496, 408832, 1479168, 2140672, 7745024, 11208704, 40553472, 58689536, 212340736, 307302400, 1111830528, 1609056256, 5821620224, 8425127936, 30482399232

Offset: 0

Paul D. Hanna, Feb 22 2003

nonn

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 14*x^4 + 20*x^5 + 72*x^6 + 104*x^7 + 376*x^8 + ...

Index entries for linear recurrences with constant coefficients, signature (0, 6, 0, -4).

Cf. A080876, A080877, A080879, A080880, A080881, A080882.

Cf. A000079, A082761, A098648.

a[ n_] := If[ n < 0, 2^n, 1] SeriesCoefficient[ (1 + x - 3*x^2 - 2*x^3)/(1 - 6*x^2 + 4*x^4), {x, 0, Abs@n}]; (* Michael Somos, May 25 2014 *)
a[ n_] := 2^Quotient[ n - 1, 2] If[ OddQ@n, Fibonacci@n, LucasL@n]; (* Michael Somos, May 25 2014 *)
LinearRecurrence[{0,6,0,-4},{1,1,3,4},40] (* Harvey P. Dale, Dec 07 2014 *)

{a(n) = if( n<0, 2^n, 1) * polcoeff( (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, May 25 2014 */

{a(n) = 2^((n - 1)\2) * if( n%2, fibonacci(n), fibonacci(n-1) + fibonacci(n+1))}; /* Michael Somos, May 25 2014 */

G.f.: (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4). a(n) = 6*a(n-2) - 4*a(n-4). - Michael Somos, Mar 05 2003
a(2n) = A080877(2n+1), a(2n+1) = A080877(2n+2)/2.
a(n) = (1/20*10^(1/2) + 1/4)*(sqrt(3 + sqrt(5)))^n + (1/20*10^(1/2) + 1/4)*(sqrt(3 - sqrt(5)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 + sqrt(5))))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 - sqrt(5))))^n. - Richard Choulet, Dec 07 2008
a(-n) = a(n) / 2^n. a(2*n) = A098648(n). a(2*n + 1) = A082761(n). - Michael Somos, May 25 2014
0 = a(n)*(+2*a(n+2)) + a(n+1)*(+2*a(n+1) - 7*a(n+2) + a(n+3)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, May 25 2014

A163073 a(n) = ((5+sqrt(5))(4+sqrt(5))^n + (5-sqrt(5))(4-sqrt(5))^n)/10.

Original entry on oeis.org

1, 5, 29, 177, 1097, 6829, 42565, 265401, 1654993, 10320533, 64359341, 401348865, 2502838169, 15607867837, 97331722837, 606967236489, 3785088940705, 23604071924261, 147196597046333, 917927985203793, 5724261314120681, 35696882675723725, 222608186950462309, 1388199786170737497

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 20 2009

Binomial transform of A082761. Fourth binomial transform of A074872.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-11).

Cf. A082761, A074872 (1,1,5,5,25,25,...).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-5); S:=[ ((5+r)*(4+r)^n+(5-r)*(4-r)^n)/10: n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 24 2009

LinearRecurrence[{8,-11},{1,5},30] (* Harvey P. Dale, Dec 11 2017 *)

x='x+O('x^30); Vec((1-3*x)/(1-8*x+11*x^2)) \\ G. C. Greubel, Jan 08 2018

a(n) = 8*a(n-1)-11*a(n-2) for n > 1; a(0) = 1, a(1) = 5.
G.f.: (1-3*x)/(1-8*x+11*x^2).
E.g.f.: exp(4*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Oct 25 2023

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 24 2009

A176281 Hankel transform of A176280.

Original entry on oeis.org

1, 3, 12, 56, 280, 1440, 7488, 39104, 204544, 1070592, 5604864, 29345792, 153653248, 804532224, 4212572160, 22057287680, 115493404672, 604731211776, 3166413520896, 16579556016128, 86811681488896, 454551863820288

Offset: 0

Paul Barry, Apr 14 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-16,8).

Cf. A000045, A082761, A176280.

List([0..30], n-> 2^(n-1)*(Fibonacci(2*n+1) + 1)); # G. C. Greubel, Nov 24 2019

[2^(n-1)*(Fibonacci(2*n+1) + 1): n in [0..30]]; // G. C. Greubel, Nov 24 2019

with(combinat); seq(2^(n-1)*(fibonacci(2*n+1) + 1), n=0..30); # G. C. Greubel, Nov 24 2019

CoefficientList[Series[(1-5x+4x^2)/((1-2x)(1-6x+4x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-16,8},{1,3,12},40] (* Harvey P. Dale, Aug 14 2013 *)

vector(31, n, 2^(n-2)*(fibonacci(2*n-1) + 1)) \\ G. C. Greubel, Nov 24 2019

[2^(n-1)*(fibonacci(2*n+1) + 1) for n in (0..30)] # G. C. Greubel, Nov 24 2019

G.f.: (1-5*x+4*x^2)/(1-8*x+16*x^2-8*x^3) = (1-5*x+4*x^2)/((1-2*x)*(1-6*x+4*x^2)).
a(n) = 2^(n-1) + (3-sqrt(5))^n*((5-sqrt(5))/20) + (3+sqrt(5))^n*((5+sqrt(5))/20).
a(n) = 2^(n-1) + A082761(n)/2. - R. J. Mathar, Sep 30 2012
a(0)=1, a(1)=3, a(2)=12, a(n) = 8*a(n-1) - 16*a(n-2) + 8*a(n-3). - Harvey P. Dale, Aug 14 2013
a(n) = 2^(n-1)*(Fibonacci(2*n+1) + 1). - G. C. Greubel, Nov 24 2019

cp's OEIS Frontend

A033887 a(n) = Fibonacci(3*n + 1).

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A375354 T(m,n) is the number of suitably connected m X n Legendrian mosaics read by rows, with 1<=n<=m.

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A054108 a(n) = (-1)^(n+1)Sum_{k=0..n+1}(-1)^kbinomial(2*k,k).

A080878 a(n)a(n+3) - a(n+1)a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=3.

A163073 a(n) = ((5+sqrt(5))(4+sqrt(5))^n + (5-sqrt(5))(4-sqrt(5))^n)/10.