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 1-3 of 3 results.

A068102 a(n) = n! * 2^n * Sum_{i=1..n} 1/(i*2^i).

Original entry on oeis.org

1, 5, 32, 262, 2644, 31848, 446592, 7150512, 128749536, 2575353600, 56661408000, 1359913708800, 35358235430400, 990036819072000, 29701191750451200, 950439443688806400, 32314962008209305600, 1163338987982963097600, 44206887945726303436800, 1768275639474152546304000
Offset: 1

Views

Author

Benoit Cloitre, Apr 14 2002

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,5]; [n le 2 select I[n]  else (3*n-1)*Self(n-1)-2*(n-1)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    seq(add(n!/i*2^(n-i), i=1..n), n=1..100); # Robert Israel, Aug 14 2014
  • Mathematica
    a[n_] := FullSimplify[n! (2^n Log[2] - LerchPhi[1/2, 1, 1 + n]/2)]; Array[a,10] (* Vladimir Reshetnikov, Jan 21 2011 *)
  • Maxima
    a(b):=n!*sum(binomial(k,n-k)*(-1)^(n-k)*binomial(2*k,k)*(H(2*k)-H(k)),k,floor(n/2),n); /* Vladimir Kruchinin, Feb 04 2023 */
  • PARI
    first(n)=my(v=vector(n),t=1); v[1]=1; for(k=2,n, v[k]=2*k*v[k-1] + t; t*=k); v \\ Charles R Greathouse IV, Aug 22 2022
    

Formula

E.g.f.: -log(1-x)/(1-2*x). - Vladeta Jovovic, Feb 07 2003
a(n+1) = 2*(n+1)*a(n) + n!, a(0)=0. - Jaume Oliver Lafont, Sep 15 2009
a(n) = 2^n*n!*(log(2) - 2*Integral_{x=0..1} x^(2*n+1)/(1+x^2)^(n+1) dx). Thus a(n)/(2^n*n!) -> log(2) as n -> inf. Cf. A087547. - Peter Bala, Jun 21 2013
a(n) = (3*n-1)*a(n-1) - 2*(n-1)^2*a(n-2). - Vaclav Kotesovec, Aug 13 2013
The sequence b(n) = 2^n*n! = A000165(n) also satisfies the above second-order recurrence of Kotesovec. This leads to the generalized continued fraction expansion lim_{n->oo} a(n)/b(n) = log(2) = 1/(2 - 2/(5 - 8/(8 - 18/(11 - ... - 2*(n - 1)^2/((3*n - 1) - ... ))))). - Peter Bala, Feb 18 2015
a(n)/n! is the linear term of the sum of the n-th row of a Pascal-like triangle T in which T(n,k) = binomial(x+n, k). - Greg Dresden and Ivan Kuznetsov, Aug 22 2022
a(n) = n!*Sum_{k=floor(n/2)..n} (-1)^(n-k)*C(k,n-k)*C(2*k,k)*(H(2*k)-H(k)), where H(n) are the harmonic numbers. - Vladimir Kruchinin, Feb 04 2023

A384199 Expansion of e.g.f. log(1 + x)/(1 - 3*x).

Original entry on oeis.org

0, 1, 5, 47, 558, 8394, 150972, 3171132, 76102128, 2054797776, 61643570400, 2034241452000, 73232652355200, 2856073920854400, 119955098448864000, 5397979517377171200, 259103015526429849600, 13214253812770712217600, 713569705533931031654400
Offset: 0

Views

Author

Seiichi Manyama, May 22 2025

Keywords

Crossrefs

Programs

  • Magma
    [0] cat [n le 1 select 1  else 3 * n * Self(n-1) - (-1)^n * Factorial(n-1): n in [1..20]]; // Vincenzo Librandi, May 22 2025
  • Mathematica
    a[n_]:= n! * Sum[(-1)^(k-1)*3^(n-k)/k,{k,1,n}];Table[a[n],{n,0,18}] (* Vincenzo Librandi, May 22 2025 *)
  • PARI
    a(n) = n!*sum(k=1, n, (-1)^(k-1)*3^(n-k)/k);
    

Formula

a(n) = n! * Sum_{k=1..n} (-1)^(k-1) * 3^(n-k)/k.
a(n) = 3 * n * a(n-1) - (-1)^n * (n-1)!.
a(n) = (2*n+1) * a(n-1) + 3 * (n-1)^2 * a(n-2).
a(n) ~ log(4/3) * 3^n * n!. - Vaclav Kotesovec, May 23 2025

A300489 a(n) = n! * [x^n] -log(1 - x)/(1 - n*x).

Original entry on oeis.org

0, 1, 5, 65, 1766, 83674, 6124584, 639826452, 90328291248, 16558780949136, 3823322392154880, 1085461798576638240, 371610484248792556800, 150961314165968542273920, 71790302154674639506682880, 39506878580692178250399571200, 24909116615180033772524150937600
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 07 2018

Keywords

Examples

			The table of coefficients of x^k in expansion of e.g.f. -log(1 - x)/(1 - n*x) begins:
n = 0: (0), 1,   1,    2,     6,      24,  ...
n = 1:  0, (1),  3,   11,    50,     274,  ...
n = 2:  0,  1,  (5),  32,   262,    2644,  ...
n = 3:  0,  1,   7,  (65),  786,   11814,  ...
n = 4:  0,  1,   9,  110, (1766),  35344,  ...
n = 5:  0,  1,  11,  167,  3346,  (83674), ...
...
This sequence is the main diagonal of the table.
		

Crossrefs

Programs

  • Mathematica
    Table[n! SeriesCoefficient[-Log[1 - x]/(1 - n x), {x, 0, n}], {n, 0, 16}]
    Join[{0}, Table[n! n^n Sum[1/(k n^k), {k, 1, n}], {n, 1, 16}]]
  • PARI
    a(n) = n!*n^n*sum(i=1, n, 1/(i*n^i)); \\ Altug Alkan, Mar 08 2018

Formula

a(n) = n!*n^n*Sum_{k=1..n} 1/(k*n^k).
Showing 1-3 of 3 results.