cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A333473 a(n) = [x^n] ( S(x/(1 + x)) )^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 2, 12, 92, 752, 6352, 54768, 478928, 4231424, 37680320, 337622912, 3040354176, 27492359936, 249463806464, 2270319909632, 20714443816192, 189418898063360, 1735482632719360, 15928224355854336, 146414296847992832, 1347721096376573952, 12421053168197722112
Offset: 0

Views

Author

Peter Bala, Mar 23 2020

Keywords

Comments

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence s(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: s(n*p^k) == s(n*p^(k-1)) ( mod p^(k) ) for any prime p and positive integers n and k. For certain power series F(x) we may get stronger congruences. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = S(x/(1 + x)) = 1 + 2*x + 4*x^2 + 12*x^3 + 40*x^4 + 154*x^5 + 544*x^6 + ...(see A025227), where S(x) is the o.g.f. of the large Schröder numbers A006318. We conjecture that the associated sequence a(n) = [x^n] ( S(x/(1 + x)) )^n satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333472.
More generally, we conjecture that for a positive integer r and integer s, the sequence a(r,s;n) := [x^(r*n)] ( S(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note the sequence b(n) := [x^n] ( S(x) )^n = A103885(n) appears to satisfy the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*S(x/(1 + x)) )^n = A333090(n) also appears to satisfy the same congruences.

Examples

			Examples of congruences:
a(11) - a(1) = 3040354176 - 2 = 2*(11^2)*13*966419 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 12421053168197722112 - 92 = (2^2)*(3^7)*5*(7^2)* 5795401942927 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 90551762251592215396352 - 6352 = (2^4)*(5^4)*293* 30905038311123623 == 0 ( mod 5^4 ).
		

Crossrefs

Main diagonal of A378317.

Programs

  • Maple
    Sch := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
    G := x → Sch(x/(1+x));
    H := (x, n) -> series(G(x)^n, x, 51):
    seq(coeff(H(x, n), x, n), n = 0..25)
  • Mathematica
    Table[SeriesCoefficient[((1 - Sqrt[1- 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(n+2*k-1, 2*k)); \\ Seiichi Manyama, Nov 24 2024

Formula

a(n) = [x^n] ( (1 - sqrt(1- 4*x - 4*x^2))/(2*x) )^n.
a(n) ~ sqrt(((sqrt(2) + 1)^(2/3) + (sqrt(2) - 1)^(2/3) - 1)/3) * ((3*(71 + 8*sqrt(2))^(1/3) + 3*(71 - 8*sqrt(2))^(1/3) + 13))^n / (sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+2*k-1,2*k). - Seiichi Manyama, Nov 24 2024

A068767 Generalized Catalan numbers 5*x*A(x)^2 -A(x) +1 -4*x=0.

Original entry on oeis.org

1, 1, 10, 105, 1150, 13050, 152500, 1825625, 22293750, 276758750, 3483287500, 44352006250, 570333187500, 7396680812500, 96638930625000, 1270796364765625, 16806545339843750, 223400240246093750
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(5,5; n)/5 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-20*x*(1-4*x)])/(10*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (5^n) * p(n, -4/5) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 5*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-20*x*(1-4*x)))/(10*x).
(n+1)*a(n) = 80*(2-n)*a(n-2) + 10*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(10+10*sqrt(5)) * (10+2*sqrt(5))^n / (10*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014
Equivalently, a(n) ~ 2^(2*n) * 5^((n-1)/2) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

A068768 Generalized Catalan numbers 6*x*A(x)^2 -A(x) +1 -5*x =0.

Original entry on oeis.org

1, 1, 12, 150, 1944, 25992, 356832, 5008824, 71629920, 1040509152, 15315578496, 227981324736, 3426473187072, 51929043390720, 792725911280640, 12178706839758720, 188158789025809920, 2921622674591946240
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(6,6; n)/6 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-24*x*(1-5*x)])/(12*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (6^n) * p(n, -5/6) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 6*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-24*x*(1-5*x)))/(12*x).
D-finite with recurrence: (n+1)*a(n) = 120*(2-n)*a(n-2) + 12*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(3+3*sqrt(6)) * (12+2*sqrt(6))^n / (6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014

A068769 Generalized Catalan numbers 7*x*A(x)^2 -A(x) +1 -6*x=0.

Original entry on oeis.org

1, 1, 14, 203, 3038, 46746, 736764, 11853051, 194053622, 3224557406, 54265836548, 923218762270, 15854602773100, 274500192707860, 4786546243533432, 83989334625037947, 1481965556616225702
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(7,7; n)/7 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 1; a[2] = 14; a[n_] := (168 (2 - n) a[n - 2] + 14 (2 n - 1) a[n - 1])/(n + 1); Table[a[n], {n, 0, 20}] (* Wesley Ivan Hurt, Mar 04 2014 *)
    CoefficientList[Series[(1-Sqrt[1-28*x*(1-6*x)])/(14*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (7^n) * p(n, -6/7) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 7*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-28*x*(1-6*x)))/(14*x).
Recurrence: (n+1)*a(n) = 168*(2-n)*a(n-2) + 14*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(14+14*sqrt(7)) * (14+2*sqrt(7))^n / (14*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014

A068770 Generalized Catalan numbers 8*x*A(x)^2 -A(x) +1 -7*x=0.

Original entry on oeis.org

1, 1, 16, 264, 4480, 77952, 1386496, 25135616, 463233024, 8658673664, 163829383168, 3132565553152, 60446638866432, 1175715287400448, 23028562592268288, 453848132868898816, 8993594212565909504
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(8,8; n)/8 with K(a,b; n) defined in a comment to A068763.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-32*x*(1-7*x)])/(16*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)
  • PARI
    a(n) = if(n, (4^(n-1)*14^(1/2*n+1/2)*pollegendre(n+1,2/7*14^(1/2)) - pollegendre(n,2/7*14^(1/2))*4^n*14^(n/2))\/n, 1) \\ Charles R Greathouse IV, Mar 19 2017

Formula

a(n) = (8^n) * p(n, -7/8) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 8*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-32*x*(1-7*x)))/(16*x).
a(n) = (4^(n-1)*14^(1/2*n+1/2)*LegendreP(n+1,2/7*14^(1/2)) - LegendreP(n,2/7*14^(1/2))*4^n*14^(1/2*n))/n for n > 0. - Mark van Hoeij, Apr 23 2010
Recurrence: (n+1)*a(n) = 224*(2-n)*a(n-2) + 16*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(1+2*sqrt(2)) * (16+4*sqrt(2))^n / (4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014

A068771 Generalized Catalan numbers 9*x*A(x)^2 -A(x) +1 -8*x=0.

Original entry on oeis.org

1, 1, 18, 333, 6318, 122634, 2429028, 48974949, 1002875094, 20814628158, 437088964860, 9272342710962, 198456435657036, 4280758166952756, 92972201833888200, 2031520673763657621, 44630859892110807654
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := (288 (2 - n) a[n - 2] + 18 (2 n - 1) a[n - 1])/(n + 1); Table[a[n], {n, 0, 20}](* Wesley Ivan Hurt, Mar 04 2014 *)
    CoefficientList[Series[(1-Sqrt[1-36*x*(1-8*x)])/(18*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (9^n) * p(n, -8/9) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 9*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-36*x*(1-8*x)))/(18*x).
Recurrence: (n+1)*a(n) = 288*(2-n)*a(n-2) + 18*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(2) * 24^n / (3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014

A100238 G.f. A(x) satisfies: 2^n + 1 = Sum_{k=0..n} [x^k] A(x)^n for n>=1.

Original entry on oeis.org

1, 2, -2, 4, -12, 40, -144, 544, -2128, 8544, -35008, 145792, -615296, 2625792, -11311616, 49124352, -214838528, 945350144, -4182412288, 18593224704, -83015133184, 372090122240, -1673660915712, 7552262979584, -34178799378432, 155096251351040, -705533929816064
Offset: 0

Views

Author

Paul D. Hanna, Nov 30 2004

Keywords

Examples

			From the table of powers of A(x), we see that
2^n+1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1: [1, 2], -2, 4, -12, 40, -144, 544, -2128, 8544, ...;
A^2: [1, 4, 0], 0, -4, 16, -64, 256, -1040, 4288, ...;
A^3: [1, 6, 6, -4], 0, 0, -8, 48, -240, 1120, -5088, ...;
A^4: [1, 8, 16, 0, -8], 0, 0, 0, -16, 128, -768, ...;
A^5: [1, 10, 30, 20, -20, -8], 0, 0, 0, 0, -32, ...;
A^6: [1, 12, 48, 64, -12, -48, 0], 0, 0, 0, 0, 0, ...;
A^7: [1, 14, 70, 140, 56, -112, -56, 16], 0, 0, 0, ...;
A^8: [1, 16, 96, 256, 240, -128, -256, 0, 32], 0, 0, ...; ...
In the above table of coefficients in A(x)^n, the main diagonal satisfies:
[x^n] A(x)^(n+1) = (n+1)*A009545(n+1) for n>=0.
		

Crossrefs

a(n) = -(-1)^n * A025227(n), if n>1.

Programs

  • PARI
    {a(n)=if(n==0,1,(2^n+1-sum(k=0,n,polcoeff(sum(j=0,min(k,n-1),a(j)*x^j)^n+x*O(x^k),k)))/n)}
    
  • PARI
    {a(n)=if(n==0,1,if(n==1,2,if(n==2,-2,(-2*(2*n-3)*a(n-1)+4*(n-3)*a(n-2))/n)))}
    
  • PARI
    {a(n)=polcoeff( (1+2*x+sqrt(1+4*x-4*x^2+x^2*O(x^n)))/2,n)}
    
  • PARI
    a(n)=polcoeff((1+2*x+sqrt(1+4*x-4*x^2+x*O(x^n)))/2,n)

Formula

a(n) = (-2*(2*n-3)*a(n-1) + 4*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=2, a(2)=-2.
G.f.: A(x) = (1+2*x + sqrt(1+4*x-4*x^2))/2.
G.f. satisfies: (2+z)^n + (1+z)^n - z^n = Sum_{k=0..n} [x^k] (A(x)+z*x)^n for all z, where [x^k] F(x) denotes the coefficient of x^k in F(x).
Given g.f. A(x), then B(x)=A(x)-1-x series reversion is -B(-x). - Michael Somos, Sep 07 2005
Given g.f. A(x) and C(x) = g.f. of A025225, then B(x)=A(x)-1-x satisfies B(x)=x-C(x*B(x)). - Michael Somos, Sep 07 2005
G.f.: 4x^2/(1+2x - sqrt(1+4x-4x^2)). - Michael Somos, Sep 08 2005

A182399 G.f. A(x) satisfies: A(A(x)) - A(A(x))^2 = x + x^2.

Original entry on oeis.org

1, 1, 1, 3, 7, 21, 61, 187, 583, 1837, 5885, 19027, 62167, 204917, 680621, 2275211, 7648519, 25852573, 87812093, 299349795, 1023570647, 3515918501, 12140103149, 41894710427, 143835281351, 501071173901, 1808088546557, 6212411239539, 17720665594455
Offset: 1

Views

Author

Paul D. Hanna, Apr 27 2012

Keywords

Comments

a(33) is the first negative term.
If B(x) = x + 2*x^2 + 8*x^3 + 36*x^4 + 160*x^5 + 736*x^6 + 3648*x^7 + ..., then g.f. A(x) = x + B(x * A(x)). - Michael Somos, Jun 27 2017

Examples

			G.f.: A(x) = x + x^2 + x^3 + 3*x^4 + 7*x^5 + 21*x^6 + 61*x^7 + 187*x^8 +...
Related expansions:
A(A(x)) = x + 2*x^2 + 4*x^3 + 12*x^4 + 40*x^5 + 144*x^6 + 544*x^7 + 2128*x^8 +...
A(A(x))^2 = x^2 + 4*x^3 + 12*x^4 + 40*x^5 + 144*x^6 + 544*x^7 + 2128*x^8 +...
where A(A(x)) - A(A(x))^2 = x + x^2.
Let C(x) satisfy C(x-x^2) = x, where C(x) begins:
C(x) = x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 +...+ A000108(n-1)*x^n +...
then
A(-C(-x)) = x + x^3 + 4*x^5 + 21*x^7 + 122*x^9 + 758*x^11 + 4958*x^13 +...+ (-1)^(n-1)*A179270(2*n-1)*x^(2*n-1) +...
		

Crossrefs

Programs

  • Maxima
    T(n, m):= if n=m then 1 else ((sum((binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m),k,0,n-m))*m -sum(T(n, i) *T(i, m), i, m+1, n-1))/2;
    makelist(T(n, 1), n, 1, 10); /* Vladimir Kruchinin, Apr 28 2012 */
  • PARI
    {a(n)=local(A=x+x^2,G);for(i=1,n,G=subst(A,x,A+x*O(x^n));A=A+(x+x^2-G+G^2)/2);polcoeff(A,n)}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    /* Faster vectorized version: */
    {MM=100;A=[1];B=x;C=(1-sqrt(1-4*(x+x^2+x*O(x^MM))))/2; for(n=1,oo,A=concat(A,0);B=x*Ser(A); A[n]=Vec((B+subst(C+x*O(x^n),x,serreverse(B)))/2)[n]; print1(A[n],", "))}
    
  • PARI
    /* PARI/GP Version of Vladimir Kruchinin's formula: */
    {T(n, m)=if(n==m,1, if(n>m, (sum(k=0,n-m,(binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m))*m - sum(i=m+1,n-1,T(n, i) *T(i, m)))/2 ))}
    {a(n)=T(n,1)}
    

Formula

G.f. satisfies: A(-A(-x)) = x.
G.f. satisfies: A(A(x)) = (1 - sqrt(1-4*(x+x^2)))/2 is the g.f. of A025227; thus, A(A(x)) = C(x+x^2) where C(x-x^2) = x.
G.f. satisfies: A(-C(-x)) = -I*G(I*x) where C(x-x^2) = x and G(x) is the g.f. of A179270 such that the inverse of function G(x) + I*G(x)^2 equals the complex conjugate: G(x) - I*G(x)^2.
a(n) = T(n,1), with T(n, m) = (sum((binomial(k+m,n-k-m)*binomial(2*k+m-1,k+m-1))/(k+m),k,0,n-m)*m -sum(T(n, i) *T(i, m), i, m+1, n-1))/2, n>m, T(n,n) = 1. - Vladimir Kruchinin, Apr 28 2012

A260774 Certain directed lattice paths.

Original entry on oeis.org

1, 6, 33, 189, 1107, 6588, 39663, 240894, 1473147, 9058554, 55954395, 346934745, 2157989445, 13459891500, 84152389833, 527224251861, 3309194474451, 20804569738218, 130987600581699, 825796890644895, 5212349717906889, 32935490120006604, 208316726580941037
Offset: 0

Views

Author

N. J. A. Sloane, Jul 30 2015

Keywords

Comments

See Dziemianczuk (2014) for precise definition.

Crossrefs

Programs

  • Maple
    b:= proc(x, y) option remember; `if`([x, y]=[0$2], 1,
          `if`(x>0, add(b(x-1, y+j), j=-1..1), 0)+
          `if`(y>0, b(x, y-1), 0)+`if`(y<0, b(x, y+1), 0))
        end:
    a:= n-> b(n, 1):
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 21 2021
  • Mathematica
    b[x_, y_] := b[x, y] = If[{x, y} == {0, 0}, 1,
         If[x > 0, Sum[b[x - 1, y + j], {j, -1, 1}], 0] +
         If[y > 0, b[x, y - 1], 0] + If[y < 0, b[x, y + 1], 0]];
    a[n_] := b[n, 1];
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, May 02 2022, after Alois P. Heinz *)

Formula

See Dziemianczuk (2014) Equation (33a) with m=1.
From Vaclav Kotesovec, Jul 15 2022: (Start)
Recurrence: (n+1)*(4*n - 3)*a(n) = 6*(4*n^2 - n - 1)*a(n-1) + 3*(n-1)*(4*n + 1)*a(n-2).
a(n) ~ (3 + 2*sqrt(3))^(n+1) / sqrt(6*Pi*n). (End)

Extensions

More terms from Lars Blomberg, Aug 01 2015

A366554 G.f. A(x) satisfies A(x) = 1 + x + x^4*A(x)^2.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 1, 0, 2, 6, 6, 2, 5, 20, 30, 20, 19, 70, 140, 140, 112, 266, 630, 840, 762, 1176, 2814, 4620, 5049, 6204, 12936, 24156, 31460, 36894, 63492, 123552, 185471, 228800, 338910, 634920, 1050686, 1411410, 1944800, 3354780, 5820256, 8513804, 11644490
Offset: 0

Views

Author

Seiichi Manyama, Oct 13 2023

Keywords

Crossrefs

Programs

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

Formula

G.f.: A(x) = 2*(1+x) / (1+sqrt(1-4*x^4*(1+x))).
a(n) = Sum_{k=0..floor(n/4)} binomial(k+1,n-4*k) * binomial(2*k,k)/(k+1).
a(n) = A366589(n) + A366589(n-1).
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