A025262 -id:A025262 - cp's OEIS frontend

A019497 Number of ternary search trees on n keys.

1, 1, 1, 3, 6, 16, 42, 114, 322, 918, 2673, 7875, 23457, 70551, 213846, 652794, 2004864, 6190612, 19207416, 59850384, 187217679, 587689947, 1850692506, 5845013538, 18509607753, 58759391013, 186958014766, 596108115402, 1904387243796, 6095040222192, 19540540075824

Offset: 0

James Fill (jimfill(AT)jhu.edu)

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1000
J. A. Fill and R. P. Dobrow, The number of m-ary search trees on n keys, Combin. Probab. Comput. 6 (1997), 435-453.
Jean-Marc Luck, Revisiting log-periodic oscillations, arXiv:2403.00432 [cond-mat.stat-mech], 2024. See p. 21.
Index entries for sequences related to rooted trees

Cf. A137954, A137959, A137966.

A:= proc(n) option remember; if n=0 then 1 else convert(series(1+x+x^2*A(n-1)^3, x=0,n+1), polynom) fi end: a:= n-> coeff(A(n), x,n): seq(a(n), n=0..27); # Alois P. Heinz, Aug 22 2008

a[0] = 1; a[n_] := Sum[Binomial[1*(n-k), k]/(n-k)*Binomial[3*k, n-k-1], {k, 0, n-1}]; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Apr 07 2015, after Paul D. Hanna *)

v=vector(50,j,1);for(n=3,50,A=sum(i=1,n,sum(j=1,n,sum(k=1,n,if(i+j+k-n,0,v[i]*v[j]*v[k]))));v[n]=A);a(n)=v[n+1];

{a(n)= local(A); if(n<0, 0, A= 1+O(x); forstep(k= 1, n, 2, A= 1+x+x*x*A^3); polcoeff(A, n))} /* Michael Somos, Mar 29 2007 */

{a(n)= if(n<0, 0, (-1)^n* polcoeff( serreverse((1-sqrt(1-4*x+4*x^3+x^2*O(x^n)))/2), n+1))} /* Michael Somos, Mar 29 2007 */

a(n)=if(n==0,1,sum(k=0,n-1,binomial(1*(n-k),k)/(n-k)*binomial(3*k,n-k-1))) \\ Paul D. Hanna, Jun 16 2009

a(0)=a(1)=1 and for n>=2 a(n)= sum( i+j+k=n-2, a(i)*a(j)*a(k) ) (i, j, k>=0). - Benoit Cloitre, Jun 14 2004
G.f. A(x) satisfies A(x)= 1+ x+ x^2*A(x)^3. - Michael Somos, Mar 29 2007
Given g.f. A(x), then x*A(-x) is series reversion of A025262(n-1). - Michael Somos, Mar 29 2007
a(n) = Sum_{k=0..n-1} C(n-k,k)/(n-k) * C(3*k,n-k-1) for n>0 with a(0)=1. - Paul D. Hanna, Jun 16 2009
a(n) ~ (8 + 3*sqrt(3))^(1/4) * 3^(n/2 - 3/8) * (3 + sqrt(9 + 8*sqrt(3)))^(n + 1/2) / (sqrt(Pi) * n^(3/2) * 2^(2*n + 2)). - Vaclav Kotesovec, Jul 31 2021

More terms from Olivier Gérard, Jul 1997

A056010 Number of words of length n in a simple grammar.

Original entry on oeis.org

1, 1, 3, 8, 23, 68, 207, 644, 2040, 6558, 21343, 70186, 232864, 778550, 2620459, 8872074, 30195288, 103246502, 354508628, 1221846856, 4225644866, 14659644348, 51002664023, 177909901566, 622093882290, 2180123564130, 7656055966092

Offset: 0

Michael Somos, Aug 01 2000

nonn

The grammar defines a language L consisting of words from the alphabet S = {e, w, n, s}. If each letter in S is regarded as an integer lattice step, e = (1,0), w = (-1,0), n = (0,1), s = (0,-1), then each word is a path in the two-dimensional integer lattice starting from (0,0), never going below the x-axis and ending on the x-axis. Thus, this is a variant of Motzkin paths with two kinds of level steps. The algebraic definition is L = 1 + Le + Lw + LnLs - w where each word is regarded as a noncommutative monomial with variables in S. Replacing each letter in S by x and L by the g.f. A(x) leads to x + A(x) = (1 + x*A(x))^2. If we let y = x + x*x*A, then y^2 - y = x^3 - x which is an elliptic curve. - Michael Somos, Mar 28 2020

The Hankel number wall for the sequence L(0), L(1), ... has a zigzag diagonal sequence b(0) = 1, b(1) = 1. b(2) = e, b(3) = ew+ns, b(4) = na(ee-ew-ns), ... which is a generalized Somos-5 sequence with b(i)*b(i+5) = -n*s*b(i+1)*b(i+4) + e*n*s*b(i+2)*b(i+3). Define sequence c(0) = 0, c(1) = 1, c(i) = b(i-2) for i>1, and c(i) = -(-n*s)^(-i)*c(-i) if i<0. Then c(i)*c(i+5) = -n*s*c(i+1)*c(i+4) + e*n*s*c(i+2)*c(i+3) for all i in Z. If e=w=n=s=1, then c(i) = A006769(i) * (-1)^[mod(i,4)=3]. - Michael Somos, Oct 14 2024

			L(0) = 1, L(1) = e, L(2) = ee + ew + ns, L(3) = eee + ewe + nse + eew + eww + nsw + nes + ens.
G.f. = 1 + x + 3*x^2 + 8*x^3 + 23*x^4 + 68*x^5 + 207*x^6 + 644*x^7 + ...

Michael De Vlieger, Table of n, a(n) for n = 0..1000
Paul Barry, Integer sequences from elliptic curves, arXiv:2306.05025 [math.NT], 2023.
Hanna Mularczyk, Lattice Paths and Pattern-Avoiding Uniquely Sorted Permutations, arXiv:1908.04025 [math.CO], 2019.
Michael Somos, Number Walls in Combinatorics.

Cf. A001006, A006720, A006769, A025262.

CoefficientList[Series[(1 - 2 x - Sqrt[1 - 4 x + 4 x^3])/(2 x^2), {x, 0, 26}], x] (* Michael De Vlieger, Oct 30 2019 *)
a[ n_] := SeriesCoefficient[ (2 - 2*x)/(1 - 2*x + (1 - 4*x + 4*x^3)^(1/2)), {x, 0, n}]; (* Michael Somos, Oct 27 2024 *)
a[ n_] := If[ n<0, 0, SeriesCoefficient[Nest[(1 + x*#)^2 - x&, 1 + O[x], n], {x, 0, n}]]; (* Michael Somos, Oct 27 2024 *)

{a(n) = if( n<0, 0, polcoef( (1 - 2*x - sqrt( 1 - 4*x + 4*x^3 + x^3 * O(x^n)) ) / (2*x^2), n))};

{a(n) = if( n<0, 0, polcoef( (2 - 2*x)/(1 - 2*x + (1 - 4*x + 4*x^3 + x*O(x^n))^(1/2)), n))}; /* Michael Somos, Oct 27 2024 */

L = 1 + Le + Lw + LnLs - w.
a(n) = 2*a(n-1) + a(0)*a(n-2) + ... + a(n-2)*a(0) for n>1.
The Somos-4 sequence A006720(n+2) is the Hankel transform of a(n-1). See A001906 for definition of Hankel transform.
Let s(n)= A006769(n). Then 0 = f( -s(n-1) * s(n+1) / s(n)^2, -s(n) * s(n+2) / s(n+1)^2 ) where f(u, v) = u + v - (1 + u*v)^2.
G.f. A(x) satisfies 0 = f(x, A(x)) where f(u, v) = u + v - (1 + u*v)^2.
G.f.: (1 - 2*x - sqrt( 1 - 4*x + 4*x^3) ) / (2*x^2).
From Paul Barry, Mar 04 2010: (Start)
G.f.: ((1-x)/(1-2x))c(x^2(1-x)/(1-2x)^2), c(x) the g.f. of A000108;
a(n) = Sum_{k=0..floor(n/2)} (A000108(k) * Sum_{i=0..k+1} C(k+1,i)*C(n-i,n-2k-i)*(-1)^i*2^(n-2k-i)). (End)
a(n) = A025262(n+2) if n >= 0.
0 = a(n)*(+16*a(n+1) - 64*a(n+3) + 22*a(n+4)) + a(n+1)*(+32*a(n+2) - 14*a(n+3)) + a(n+2)*(+16*a(n+3) - 10*a(n+4)) + a(n+3)*(+2*a(n+3) + a(n+4)) if n>=0. - Michael Somos, Jan 18 2015

A160702 Sequence such that the Hankel transform of a(n+1) satisfies a generalized Somos-4 recurrence.

Original entry on oeis.org

1, 1, 1, 5, 19, 79, 333, 1441, 6351, 28451, 129185, 593373, 2752427, 12876343, 60684533, 287857209, 1373286375, 6584979659, 31719337353, 153416338549, 744777567043, 3627787084319

Offset: 0

Paul Barry, May 24 2009

The Hankel transform of a(n+1) satisfies a generalized Somos-4 Hankel determinant recurrence.
Hankel transform of a(n+1) is A160703. In general, we can conjecture that the Hankel transform of
h(n) of a(n+1), where a(n)=if(n=0,1,if(n=1,1,if(n=2,1,r*a(n-1)+s*sum{k=0..n-2, a(k)*a(n-1-k)}))),
satisfies the generalized Somos-4 recurrence
h(n)=(s^2*h(n-1)*h(n-3)+s^3*(2*s+r-2)*h(n-2)^2)/h(n-4).
The case r=s=1 is proved in the Xin reference.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
Gouce Xin, Proof of the Somos-4 Hankel determinants conjecture, Advances in Applied Mathematics, Volume 42, Issue 2, February 2009, Pages 152-156.

Cf.: A025262, A056010, A006720.

CoefficientList[Series[1/4+(1-Sqrt[16*x^3+x^2-6*x+1])/(4*x),{x,0,20}],x] (* Vaclav Kotesovec, Nov 20 2012 *)

a(n) = if(n=0,1,if(n=1,1,if(n=2,1,a(n-1)+2*sum{k=0..n-2, a(k)*a(n-1-k)})))
Recurrence: (n+1)*a(n) = 3*(2*n-1)*a(n-1) - (n-2)*a(n-2) - 8*(2*n-7)*a(n-3). - Vaclav Kotesovec, Nov 20 2012
G.f.: 1/4+(1-sqrt(16*x^3+x^2-6*x+1))/(4*x). - Vaclav Kotesovec, Nov 20 2012

A157002 Transform of Catalan numbers whose Hankel transform gives the Somos-4 sequence.

Original entry on oeis.org

1, 0, 1, 2, 6, 17, 51, 156, 488, 1552, 5006, 16337, 53849, 179015, 599535, 2020924, 6851150, 23344138, 79902364, 274606264, 947240592, 3278404274, 11381240074, 39621423949, 138288477617, 483805404673, 1696318159457, 5959737806635

Offset: 0

Paul Barry, Feb 20 2009

Image of the Catalan numbers A000108 by the Riordan array (1-x,x(1-x^2)). Hankel transform is A006720(n+1).
The sequence a(n)+a(n+1) begins 1,1,3,8,23,68,... which is A056010. The sequence a(n)+a(n-1) begins 1,1,1,3,8,23,68,... which is A025262. This is obtained by applying (1-x^2,x(1-x^2)) to the Catalan numbers.
Hankel transform of a(n+1) is -A051138(n). - Michael Somos, Feb 10 2015

			G.f. = 1 + x^2 + 2*x^3 + 6*x^4 + 17*x^5 + 51*x^6 + 156*x^7 + 488*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Gouce Xin, Proof of the Somos-4 Hankel determinants conjecture, Advances in Applied Mathematics, Volume 42, Issue 2, February 2009, Pages 152-156.

Cf. A000108, A006720, A025262, A051138, A056010.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (1 -Sqrt(1-4*x*(1-x^2)))/(2*x*(1+x)) )); // G. C. Greubel, Feb 26 2019

CoefficientList[Series[(1-Sqrt[1-4x(1-x^2)])/(2x(1+x)), {x,0,30}], x] (* G. C. Greubel, Feb 26 2019 *)

{a(n) = if( n<0, -(-1)^n / 2 * (n<-1), polcoeff( (1 - sqrt(1 - 4*x * (1 - x^2) + x^2 * O(x^n))) / (2 * x * (1 + x)), n))}; /* Michael Somos, Feb 10 2015 */

((1-sqrt(1-4*x*(1-x^2)))/(2*x*(1+x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

G.f.: (1 - sqrt(1-4*x*(1-x^2)))/(2*x*(1+x)).
a(n) = Sum_{k=0..n} (-1)^floor((n-k+1)/2)*C(k,floor((n-k)/2))*A000108(k).
Conjecture: (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2) +2*(2*n-7)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Nov 19 2014
0 = a(n)*(+16*a(n+1) + 16*a(n+2) - 64*a(n+3) - 42*a(n+4) + 22*a(n+5)) + a(n+1)*(+16*a(n+1) + 48*a(n+2) - 46*a(n+3) - 56*a(n+4) + 22*a(n+5)) + a(n+2)*(+32*a(n+2) + 34*a(n+3) - 8*a(n+4) - 10*a(n+5)) + a(n+3)*(+18*a(n+3) + 11*a(n+4) - 9*a(n+5)) + a(n+4)*(+3*a(n+4) + a(n+5)) for all n in Z. - Michael Somos, Feb 10 2015

A025268 a(n) = a(1)a(n-1) + a(2)a(n-2) + ...+ a(n-1)*a(1) for n >= 5, with initial values 1,1,1,1.

Original entry on oeis.org

1, 1, 1, 1, 4, 11, 32, 95, 284, 860, 2630, 8115, 25242, 79080, 249342, 790719, 2520546, 8072216, 25961150, 83814536, 271538192, 882527618, 2876712308, 9402284815, 30806948110, 101172278362, 332965892290, 1097990333320, 3627433618396

Offset: 1

Clark Kimberling

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..1857

Cf. A000108, A025262.

Phi:=proc(t,u,M) local i,a; a:=Array(0..M);
for i from 0 to t-1 do a[i]:=u[i+1]; od:
for i from t to M do a[i]:=a[i-1]+add(a[j]*a[i-1-j],j=0..i-2); od:
[seq(a[i],i=0..M)]; end;
Phi(4,[1,1,1,1],30);
# N. J. A. Sloane, Oct 29 2008

nmax = 30; aa = ConstantArray[0,nmax]; aa[[1]] = 1; aa[[2]] = 1; aa[[3]] = 1; aa[[4]] = 1; Do[aa[[n]] = Sum[aa[[k]]*aa[[n-k]],{k,1,n-1}],{n,5,nmax}]; aa (* Vaclav Kotesovec, Jan 25 2015 *)

Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([1,1,1,1]). - Gary W. Adamson, Oct 27 2008
Conjecture: n*a(n) +(n+1)*a(n-1) +10*(-2*n+5)*a(n-2) +2*(2*n-9)*a(n-3) +2*(14*n-79)*a(n-4) +40*(n-7)*a(n-5)=0. - R. J. Mathar, Jan 25 2015
G.f.: 1/2 - sqrt(8*x^4+4*x^3-4*x+1)/2. - Vaclav Kotesovec, Jan 25 2015
Recurrence: n*a(n) = 2*(2*n-3)*a(n-1) - 2*(2*n-9)*a(n-3) - 8*(n-6)*a(n-4). - Vaclav Kotesovec, Jan 25 2015

A157100 Transform of Catalan numbers whose Hankel transform satisfies the Somos-4 recurrence.

Original entry on oeis.org

1, 2, 3, 6, 14, 37, 105, 312, 956, 2996, 9554, 30897, 101083, 333947, 1112497, 3732956, 12605030, 42800318, 146046820, 500555448, 1722402304, 5948047170, 20607691518, 71610355541, 249520257107, 871614139397, 3051737703527

Offset: 0

Paul Barry, Feb 22 2009

Hankel transform is A157101.

The ratio of this generating function by the generating function of A025262 is x*(1-x), which means this sequence is the partial sums of A025262. - Sean A. Irvine, R. J. Mathar, Jun 27 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A157101.

Partial sums of A025262.

a[n_]:= Sum[(-1)^Binomial[k, 2]*Binomial[n-k, Floor[k/2]]*CatalanNumber[n-k], {k,0,n}];
Table[a[n], {n, 0, 40}] (* G. C. Greubel, Jan 11 2022 *)

def A157100(n): return sum((-1)^binomial(k,2)*binomial(n-k, k//2)*catalan_number(n-k) for k in (0..n))
[A157100(n) for n in (0..40)] # G. C. Greubel, Jan 11 2022

G.f.: (1+x)*c(x*(1-x^2)), c(x) the g.f. of A000108;
a(n) = Sum_{k=0..n} (-1)^binomial(n-k,2)*binomial(k,floor((n-k)/2))*A000108(k).
Conjecture: (n+1)*a(n) +(-5*n+1)*a(n-1) +2*(2*n-1)*a(n-2) +2*(2*n-7)*a(n-3) +2*(-2*n+7)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015

A168151 Riordan array (1/u,(1-u)/2), u=sqrt(1-4x+4*x^3).

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 18, 9, 4, 1, 58, 29, 13, 5, 1, 192, 96, 44, 18, 6, 1, 650, 325, 151, 64, 24, 7, 1, 2232, 1116, 524, 228, 90, 31, 8, 1, 7746, 3873, 1833, 813, 333, 123, 39, 9, 1, 27096, 13548, 6452, 2904, 1222, 473, 164, 48, 10, 1

Offset: 0

Philippe Deléham, Nov 19 2009

T(n,0) = A157004(n).

			Triangle begins:
  1
  2 1
  6 3 1
  18 9 4 1
  58 29 13 5 1
  192 96 44 18 6 1
  650 325 151 64 24 7 1
  ...

Baccherini, D.; Merlini, D.; Sprugnoli, R. Binary words excluding a pattern and proper Riordan arrays. Discrete Math. 307 (2007), no. 9-10, 1021--1037. MR2292531 (2008a:05003). See Example 5.6. - N. J. A. Sloane, Mar 25 2014

Cf. A025262, A157004.

T(n,0) = 2*T(n,1) for n>0, T(0,0) = 1, T(n,k) = T(n-1,k-1)-T(n-3,k-1)+T(n,k+1).

A176703 Coefficients of a recursive polynomial based on Chaitin's S expressions: a(0)=1; a(1)=x; a(2)=1; a(n)=vector(a(n-1)).reverse(a(n-1)).

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 4, 2, 2, 9, 8, 4, 2, 22, 24, 14, 8, 56, 70, 52, 24, 5, 146, 208, 176, 84, 30, 388, 624, 574, 320, 120, 14, 1048, 1876, 1868, 1184, 470, 112, 2869, 5648, 6088, 4236, 1900, 560, 42, 7942, 17040, 19804, 14928, 7560, 2492, 420, 22192, 51526, 64232

Offset: 0

Roger L. Bagula, Apr 24 2010

The result is an alternative way to expand s expressions as a binary rooted tree recursion.

			{1},
{0, 1},
{1, 0},
{2, 0, 1},
{4, 2, 2},
{9, 8, 4, 2},
{22, 24, 14, 8},
{56, 70, 52, 24, 5},
{146, 208, 176, 84, 30},
{388, 624, 574, 320, 120, 14},
{1048, 1876, 1868, 1184, 470, 112},
{2869, 5648, 6088, 4236, 1900, 560, 42},
{7942, 17040, 19804, 14928, 7560, 2492, 420},
{22192, 51526, 64232, 52208, 29190, 10864, 2520, 132},
{62510, 156128, 207808, 181320, 110260, 46256, 12684, 1584},
{177308, 473952, 670966, 625408, 410400, 190932, 59976, 11088, 429}

G. J. Chaitin, Algorithmic Information Theory, Cambridge Press, 1987, page 169

Cf. A025227, A025262, A072851, A124027

a[0] := 1; a[1] := x; a[2] = 1;
a[n_] := a[n] = Table[a[i], {i, 0, n - 1}].Table[a[n - 1 - i], {i, 0, n - 1}];
Table[ CoefficientList[a[n], x], {n, 0, 15}];
Flatten[%]

{T(n, k) = if( 2*k-1  > n, 0, polcoeff( polcoeff( ( 1 - sqrt( (1 - 2*x)^2 - 4*x^2 * (x + y - 2*x*y) + x^2*O(x^n))) / (2*x), n), k))} /* Michael Somos, Jan 09 2012 */

a(0)=1;a(1)=x;a(2)=1;
a(n)=vector(a(n-1)).reverse(a(n-1));
t(n,m)=coefficients(a(n) in x)
Let b(0) = b(2) = 1, b(1) = x, and b(n) = Sum_{i=1..n} b(i-1) * b(n-i) if n>2. Then T(n, k) = coefficient of x^k in b(n) where 0 <= k <= (n+1)/2.
G.f. A(x,y) satisfies A(x,y) = 1 - x * (1 - x - y + 2*x*y) + x * A(x,y)^2. - Michael Somos, Jan 09 2012
G.f.: ( 1 - sqrt( (1 - 2*x)^2 - 4*x^2 * (x + y - 2*x*y) )) / (2*x). - Michael Somos, Jan 09 2012
Row sums are A025262 if offset 0.

A367044 G.f. satisfies A(x) = 1 - x^2 + x*A(x)^3.

Original entry on oeis.org

1, 1, 2, 9, 40, 192, 963, 5000, 26649, 144990, 802023, 4497150, 25504380, 146037955, 843134220, 4902661503, 28686940053, 168785282241, 997968554037, 5926617173205, 35335723342962, 211433954924955, 1269252184538408, 7642065274626855, 46137678521488140

Offset: 0

Seiichi Manyama, Nov 03 2023

nonn

Cf. A025262, A367045.

Cf. A367040.

a(n) = sum(k=0, n\2, (-1)^k*binomial(2*(n-2*k)+1, k)*binomial(3*(n-2*k), n-2*k)/(2*(n-2*k)+1));

a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2*(n-2*k)+1,k) * binomial(3*(n-2*k),n-2*k)/(2*(n-2*k)+1).

A367045 G.f. satisfies A(x) = 1 - x^2 + x*A(x)^4.

Original entry on oeis.org

1, 1, 3, 18, 112, 755, 5348, 39302, 296916, 2291861, 17997052, 143319918, 1154728056, 9395809374, 77099733884, 637298480966, 5301568498768, 44351526986704, 372890978840156, 3149155955471690, 26702387443603200, 227238745573918511, 1940201017862028108

Offset: 0

Seiichi Manyama, Nov 03 2023

nonn

Cf. A025262, A367044.

Cf. A367041.

a(n) = sum(k=0, n\2, (-1)^k*binomial(3*(n-2*k)+1, k)*binomial(4*(n-2*k), n-2*k)/(3*(n-2*k)+1));

a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(3*(n-2*k)+1,k) * binomial(4*(n-2*k),n-2*k)/(3*(n-2*k)+1).

cp's OEIS Frontend

A019497 Number of ternary search trees on n keys.

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A056010 Number of words of length n in a simple grammar.

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Mathematica

PARI

PARI

Formula

A160702 Sequence such that the Hankel transform of a(n+1) satisfies a generalized Somos-4 recurrence.

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Mathematica

Formula

A157002 Transform of Catalan numbers whose Hankel transform gives the Somos-4 sequence.

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Magma

Mathematica

PARI

Sage

Formula

A025268 a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ...+ a(n-1)*a(1) for n >= 5, with initial values 1,1,1,1.

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Maple

Mathematica

Formula

A157100 Transform of Catalan numbers whose Hankel transform satisfies the Somos-4 recurrence.

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Mathematica

Sage

Formula

A168151 Riordan array (1/u,(1-u)/2), u=sqrt(1-4x+4*x^3).

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A025268 a(n) = a(1)a(n-1) + a(2)a(n-2) + ...+ a(n-1)*a(1) for n >= 5, with initial values 1,1,1,1.