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.

Showing 1-5 of 5 results.

A088716 G.f. satisfies: A(x) = 1 + x*A(x)*d/dx[x*A(x)] = 1 + x*A(x)^2 + x^2*A(x)*A'(x).

Original entry on oeis.org

1, 1, 3, 14, 85, 621, 5236, 49680, 521721, 5994155, 74701055, 1003125282, 14437634276, 221727608284, 3619710743580, 62605324014816, 1143782167355649, 22014467470369143, 445296254367273457, 9444925598142843970
Offset: 0

Views

Author

Paul D. Hanna, Oct 12 2003

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(
          a(j)*a(n-j-1)*(j+1), j=0..n-1))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Aug 10 2017
  • Mathematica
    a=ConstantArray[0,21]; a[[1]]=1; a[[2]]=1; Do[a[[n+1]] = Sum[k*a[[n-k+1]]*a[[k]],{k,1,n}],{n,2,20}]; a (* Vaclav Kotesovec, Feb 21 2014 *)
    m = 20; A[_] = 0;
    Do[A[x_] = 1 + x A[x]^2 + x^2 A[x] A'[x] + O[x]^m // Normal, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Feb 18 2020 *)
    a[1]:=1; a[2]:=1; a[n_]:=a[n]=n/2 Sum[a[k] a[n-k], {k,1,n-1}];
    Map[a,Range[20]] (* Oliver Seipel, Nov 03 2024 ,after Schröder 1870 *)
  • PARI
    a(n)=if(n==0,1,sum(k=0,n-1,(k+1)*a(k)*a(n-k-1)))
    
  • PARI
    {a(n)=local(G=1+x);for(i=1,n,G=exp(x/(1 - x*deriv(G)/G+x*O(x^n))));polcoeff(log(G)/x,n)} \\ Paul D. Hanna, Jan 01 2011

Formula

a(n) = Sum_{k=1..n} k*a(k-1)*a(n-k) for n>=1 with a(0)=1.
Forms column 0 of triangle T=A112911, where the matrix inverse satisfies [T^-1](n,k) = -(k+1)*T(n-1,0) for n>k>=0.
Self-convolution is A112916, where a(n) = (n+1)/2*A112916(n-1) for n>0.
G.f.: A(x) = serreverse(x/f(x))/x where f(x) is the g.f. of A088715.
O.g.f.: A(x) = log(G(x))/x where G(x) is the e.g.f. of A182962 given by G(x) = exp( x/(1 - x*G'(x)/G(x)) ). [Paul D. Hanna, Jan 01 2011]
O.g.f. A(x) satisfies: [x^n] exp( n * x*A(x) ) / A(x) = 0 for n>0. - Paul D. Hanna, May 25 2018
O.g.f. A(x) satisfies [x^n] exp( n * x*A(x) ) * (1 - n*x) = 0 for n>0. - Paul D. Hanna, Jul 24 2019
From Paul D. Hanna, Jul 20 2018 (Start):
O.g.f. A(x) satisfies:
* [x^n] exp(-n * x*A(x)) * (2 - 1/A(x)) = 0 for n >= 1.
* [x^n] exp(-n^2 * x*A(x)) * (n + 1 - n/A(x)) = 0 for n >= 1.
* [x^n] exp(-n^(p+1) * x*A(x)) * (n^p + 1 - n^p/A(x)) = 0 for n>=1 and for fixed integer p >= 0. (End)
a(n) ~ c * n! * n^2, where c = 0.21795078944715106549282282244231982088... (see A238223). - Vaclav Kotesovec, Feb 21 2014

A112911 Triangle T, read by rows, such that the matrix inverse satisfies: [T^-1](n,k) = -(k+1)*T(n-1,0) for n>k>=0, with T(n,n)=1 for n>=0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 14, 8, 3, 1, 85, 44, 15, 4, 1, 621, 298, 96, 24, 5, 1, 5236, 2358, 729, 176, 35, 6, 1, 49680, 21154, 6327, 1492, 290, 48, 7, 1, 521721, 211100, 61380, 14220, 2725, 444, 63, 8, 1, 5994155, 2313030, 655944, 149812, 28425, 4590, 644, 80, 9, 1
Offset: 0

Views

Author

Paul D. Hanna, Oct 06 2005

Keywords

Comments

Matrix inverse square satisfies: [T^-2](3*n+2,n) = 0 for n>=0.

Examples

			Triangle T begins:
1;
1,1;
3,2,1;
14,8,3,1;
85,44,15,4,1;
621,298,96,24,5,1;
5236,2358,729,176,35,6,1;
49680,21154,6327,1492,290,48,7,1; ...
Matrix inverse T^-1 begins:
1;
-1,1;
-1,-2*1,1;
-3,-2*1,-3*1,1;
-14,-2*3,-3*1,-4*1,1;
-85,-2*14,-3*3,-4*1,-5*1,1;
-621,-2*85,-3*14,-4*3,-5*1,-6*1,1; ...
where [T^-1](n,k) = -(k+1)*T(n-1,0) for n>k>=0.
		

Crossrefs

Cf. A088716 (column 0), A112912 (column 1), A112913 (column 2), A112914 (column 3).

Programs

  • PARI
    {T(n,k)=local(A=Mat(1),B); for(m=2,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=-j*(A^-1)[i-j,1] );));A=B);return((A^-1)[n+1,k+1])}

A112913 Column 2 of triangle A112911.

Original entry on oeis.org

1, 3, 15, 96, 729, 6327, 61380, 655944, 7642557, 96338925, 1306003803, 18947038404, 292956751236, 4810423131828, 83620543389804, 1534493081643552, 29649997606132197, 601828286370137865, 12804746026265284149
Offset: 0

Views

Author

Paul D. Hanna, Oct 06 2005

Keywords

Crossrefs

Cf. A112911 (triangle), A088716 (column 0), A112912 (column 1), A112914 (column 3).

Programs

  • PARI
    {a(n)=local(A=Mat(1),B);for(m=2,n+3,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=-j*(A^-1)[i-j,1]);));A=B);return((A^-1)[n+3,3])}

Formula

a(n) = Sum_{k=0..n-1} (n-k+2)*A088716(k)*a(n-k-1) for n>0 with a(0)=1.

A112914 Column 3 of triangle A112911.

Original entry on oeis.org

1, 4, 24, 176, 1492, 14220, 149812, 1724760, 21519108, 289101836, 4160750192, 63873461064, 1042061108096, 18008563271632, 328712410879632, 6320649807989400, 127724091041379492, 2706397280528315148
Offset: 0

Views

Author

Paul D. Hanna, Oct 06 2005

Keywords

Crossrefs

Cf. A112911 (triangle), A088716 (column 0), A112912 (column 1), A112913 (column 2).

Programs

  • PARI
    {a(n)=local(A=Mat(1),B);for(m=2,n+4,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=-j*(A^-1)[i-j,1]);));A=B);return((A^-1)[n+4,4])}

Formula

a(n) = Sum_{k=0..n-1} (n-k+3)*A088716(k)*a(n-k-1) for n>0 with a(0)=1.

A122913 Minimum numbers k such that (k^2*2^n + 1) is prime.

Original entry on oeis.org

1, 1, 3, 1, 6, 2, 3, 1, 6, 5, 3, 4, 12, 2, 6, 1, 3, 10, 15, 5, 9, 5, 18, 25, 9, 13, 9, 14, 12, 7, 6, 9, 3, 17, 9, 9, 15, 12, 9, 6, 6, 3, 3, 11, 42, 18, 21, 9, 66, 10, 33, 5, 27, 7, 48, 80, 24, 40, 12, 20, 6, 10, 3, 5, 3, 7, 3, 79, 75, 63, 96, 40, 48, 20, 24, 10, 12, 5, 6, 15, 3, 22, 72, 11
Offset: 1

Views

Author

Alexander Adamchuk, Sep 18 2006

Keywords

Comments

3 divides a(2k+1) for k>0. Corresponding primes of the form (k^2*2^n + 1) are listed in A122912[n] = {3,5,73,17,1153,257,1153,257,18433,25601,18433,65537,1179649,65537,1179649,65537,1179649,26214401,117964801,...}. There are repeating patterns in a(n) such that for many n a(n) = 2*a(n+2) and a(n+1) = 2*a(n+3). For example, {6,2,3,1}, {12,2,6,1}, {42,18,21,9}, {96,40,48,20,24,10,12,5,6}, {66,10,33,5}, {48,80,24,40,12,20,6,10,3}, {366,38,183,19}. These patterns correspond to identical twin runs in A122912[n] such that A122912[n] = A122912[n+2] and A122912[n+1] = A122912[n+3]. The final index of many such twin runs is perfect power such as {8,16,25,64,81,100,...}.

Crossrefs

Cf. A112912.

Programs

  • Mathematica
    mnk[n_]:=Module[{k=1,c=2^n},While[!PrimeQ[k^2 c+1],k++];k]; Array[mnk,90] (* Harvey P. Dale, Jul 22 2025 *)

Formula

a(n) = Sqrt[ (A122912[n] - 1) / 2^n ].
Showing 1-5 of 5 results.