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.

A004213 Shifts one place left under 4th-order binomial transform.

Original entry on oeis.org

1, 1, 5, 29, 201, 1657, 15821, 170389, 2032785, 26546673, 376085653, 5736591885, 93614616409, 1625661357673, 29905322979421, 580513190237573, 11850869542405409, 253669139947767777, 5678266212792053029, 132607996474971041789, 3224106929536557918697
Offset: 0

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Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+4 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=4, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011

Examples

			Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
a(2)=3 to [00], [01], [02], [03], and [04], a(3) = 29 to
       RGS          F
.1:  [ 0 0 0 ]    [ 0 0 0 ]
.2:  [ 0 0 1 ]    [ 0 0 0 ]
.3:  [ 0 0 2 ]    [ 0 0 0 ]
.4:  [ 0 0 3 ]    [ 0 0 0 ]
.5:  [ 0 0 4 ]    [ 0 0 4 ]
.6:  [ 0 1 0 ]    [ 0 0 0 ]
.7:  [ 0 1 1 ]    [ 0 0 0 ]
.8:  [ 0 1 2 ]    [ 0 0 0 ]
.9:  [ 0 1 3 ]    [ 0 0 0 ]
10:  [ 0 1 4 ]    [ 0 0 4 ]
11:  [ 0 2 0 ]    [ 0 0 0 ]
12:  [ 0 2 1 ]    [ 0 0 0 ]
13:  [ 0 2 2 ]    [ 0 0 0 ]
14:  [ 0 2 3 ]    [ 0 0 0 ]
15:  [ 0 2 4 ]    [ 0 0 4 ]
16:  [ 0 3 0 ]    [ 0 0 0 ]
17:  [ 0 3 1 ]    [ 0 0 0 ]
18:  [ 0 3 2 ]    [ 0 0 0 ]
19:  [ 0 3 3 ]    [ 0 0 0 ]
20:  [ 0 3 4 ]    [ 0 0 4 ]
21:  [ 0 4 0 ]    [ 0 4 4 ]
22:  [ 0 4 1 ]    [ 0 4 4 ]
23:  [ 0 4 2 ]    [ 0 4 4 ]
24:  [ 0 4 3 ]    [ 0 4 4 ]
25:  [ 0 4 4 ]    [ 0 4 4 ]
26:  [ 0 4 5 ]    [ 0 4 4 ]
27:  [ 0 4 6 ]    [ 0 4 4 ]
28:  [ 0 4 7 ]    [ 0 4 4 ]
29:  [ 0 4 8 ]    [ 0 4 8 ]
[_Joerg Arndt_, Apr 30 2011]
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075499 (row sums).
A004211 (RGS where s(k)<=F(k)+2), A004212 (s(k)<=F(k)+3), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011

Programs

  • Maple
    A004213 := proc(n)
        add(4^(n-m)*combinat[stirling2](n,m),m=0..n) ;
    end proc:
    seq(A004213(n),n=0..30) ; # R. J. Mathar, Aug 20 2022
  • Mathematica
    Table[4^n BellB[n, 1/4], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(4^(n-k)*binomial(n-1, k-1)*a(k-1), k, 1, n); /* Vladimir Kruchinin, Nov 28 2011 */
  • PARI
    x='x+O('x^66);
    egf=exp(intformal(exp(4*x))); /* =  1 + x + 5/2*x^2 + 29/6*x^3 + 67/8*x^4 + ... */
    /* egf=exp(1/4*(exp(4*x)-1)) */ /* alternative computation */
    Vec(serlaplace(egf)) /* Joerg Arndt, Apr 30 2011 */
    

Formula

a(n) = Sum_{m=0..n} 4^(n-m)*Stirling2(n, m).
E.g.f.: exp((exp(4*x)-1)/4).
O.g.f. A(x) satisfies A'(x)/A(x) = e^(4x).
E.g.f.: exp(Integral_{t = 0..x} exp(4*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-4*j*x). - Joerg Arndt, Apr 30 2011
Define f_1(x), f_2(x), ... such that f_1(x) = e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n = 2, 3, .... Then a(n) = e^(-1/4)*4^{n-1}*f_n(1/4). - Milan Janjic, May 30 2008
a(n) = upper left term in M^n, M = an infinite square production matrix in which a diagonal of (4,4,4,...) is appended to the right of Pascal's triangle:
1, 4, 0, 0, 0, ...
1, 1, 4, 0, 0, ...
1, 2, 1, 4, 0, ...
1, 3, 3, 1, 4, ...
... - Gary W. Adamson, Jul 29 2011
G.f. satisfies A(x) = 1 + x/(1 - 4*x)*A(x/(1 - 4*x)). a(n) = Sum_{k = 1..n} 4^(n-k)*binomial(n-1,k-1)*a(k-1), n > 0, a(0) = 1. - Vladimir Kruchinin, Nov 28 2011 [corrected by Ilya Gutkovskiy, May 02 2019]
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-4*k*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 24 2013
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-4*k*x)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 31 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)/( 4*x^2*(k+1) - (1-x-4*x*k)*(1-5*x-4*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) = exp(-1/4) * Sum_{k>=0} 4^(n-k) * k^n / k!. - Vaclav Kotesovec, Jul 15 2021
a(n) ~ 4^n * n^n * exp(n/LambertW(4*n) - 1/4 - n) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^n). - Vaclav Kotesovec, Jul 15 2021
From Peter Bala, Jun 29 2024: (Start)
a(n) = exp(-1/4)*Sum_{n >= 0} (4*n)^k/(n!*4^n).
Touchard's congruence holds: for odd prime p, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,.... In particular, a(p) == 2 (mod p) for odd prime p. See A004211. (End)