A004212 Shifts one place left under 3rd-order binomial transform.
1, 1, 4, 19, 109, 742, 5815, 51193, 498118, 5296321, 60987817, 754940848, 9983845261, 140329768789, 2087182244308, 32725315072135, 539118388883449, 9304591246975030, 167804098493079547, 3155000165773280893
Offset: 0
Examples
Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
a(2)=3 to [00], [01], [02], and [03], a(3) = 19 to
RGS F
01: [ 0 0 0 ] [ 0 0 0 ]
02: [ 0 0 1 ] [ 0 0 0 ]
03: [ 0 0 2 ] [ 0 0 0 ]
04: [ 0 0 3 ] [ 0 0 3 ]
05: [ 0 1 0 ] [ 0 0 0 ]
06: [ 0 1 1 ] [ 0 0 0 ]
07: [ 0 1 2 ] [ 0 0 0 ]
08: [ 0 1 3 ] [ 0 0 3 ]
09: [ 0 2 0 ] [ 0 0 0 ]
10: [ 0 2 1 ] [ 0 0 0 ]
11: [ 0 2 2 ] [ 0 0 0 ]
12: [ 0 2 3 ] [ 0 0 3 ]
13: [ 0 3 0 ] [ 0 3 3 ]
14: [ 0 3 1 ] [ 0 3 3 ]
15: [ 0 3 2 ] [ 0 3 3 ]
16: [ 0 3 3 ] [ 0 3 3 ]
17: [ 0 3 4 ] [ 0 3 3 ]
18: [ 0 3 5 ] [ 0 3 3 ]
19: [ 0 3 6 ] [ 0 3 6 ]
- _Joerg Arndt_, Apr 30 2011
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Joerg Arndt, Matters Computational (The Fxtbook), section 17.3.5, pp. 366-368
- I. M. Gessel, Applications of the classical umbral calculus. arXiv:math/0108121v3 [math.CO], 2001.
- Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321.
- A. Kerber, A matrix of combinatorial numbers related to the symmetric groups<, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
- N. J. A. Sloane, Transforms
Crossrefs
Cf. A075498 (row sums).
Cf. A027465. - Gary W. Adamson, Apr 10 2009
Cf. A004211 (RGS where s(k)<=F(k)+2), A004213 (s(k)<=F(k)+4), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011
Cf. A009235.
Programs
-
Mathematica
Table[Sum[StirlingS2[n,k] 3^(-k+n),{k,n}],{n,20}] (* Vincenzo Librandi, May 21 2012 *) Table[3^n BellB[n, 1/3], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *) -
Maxima
a(n):=if n=0 then 1 else sum(3^(n-k)*binomial(n-1,k-1)*a(k-1),k,1,n); /* Vladimir Kruchinin, Nov 28 2011 */
-
PARI
x='x+O('x^66); /* that many terms */ egf=exp(intformal(exp(3*x))); /* = 1 + x + 2*x^2 + 19/6*x^3 + 109/24*x^4 + ... */ /* egf=exp(1/3*(exp(3*x)-1)) */ /* alternative computation */ Vec(serlaplace(egf)) /* show terms */ /* Joerg Arndt, Apr 30 2011 */
Formula
a_n = Sum_{k=0..n} 3^(n-k)*Stirling2(n, k). - Emeric Deutsch, Feb 11 2002
E.g.f.: exp((exp(3*x)-1)/3).
O.g.f. A(x) satisfies A'(x)/A(x) = e^(3*x).
E.g.f.: exp(Integral_{t = 0..x} exp(3*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-3*j*x). - Joerg Arndt, Apr 30 2011
Hankel transform is A000178(n)*3^C(n+1,2). - Paul Barry, Mar 31 2008
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^(-1/2)*3^{n-1}*f_n(1/3). - Milan Janjic, May 30 2008
a(n) = the upper left term in M^n, M = the following infinite square production matrix:
1, 3, 0, 0, 0, ...
1, 1, 3, 0, 0, ...
1, 2, 1, 3, 0, ...
1, 3, 3, 1, 3, ...
... (in which a diagonal of (3,3,3,...) is appended to the right of Pascal's triangle). - Gary W. Adamson, Jul 29 2011
G.f. satisfies A(x) = 1+x/(1-3*x)*A(x/(1-3*x)). a(n) = Sum_{k=1..n} 3^(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]
From Peter Bala, May 16 2012: (Start)
Recurrence equation: a(n+1) = Sum_{k = 0..n} 3^(n-k)*C(n,k)*a(k). Written umbrally this is a(n+1) = (a + 3)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a+3) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a-3)*(a-6)*...*(a-3*(n-1)) = 1 with a(0) = 1. Compare with the Bell numbers B(n) = A000110(n), which satisfy the umbral recurrence B*(B-1)*...*(B-(n-1)) = 1 with B(0) = 1.
Touchard's congruence holds: for prime p not equal to 3, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,... (adapt the proof of Theorem 10.1 in Gessel). In particular, a(p) == 2 (mod p) for prime p <> 3. (End)
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-x*3*k)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 3*x^2*(k+1)/( 3*x^2*(k+1) - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) ~ 3^n * n^n * exp(n/LambertW(3*n) - 1/3 - n) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^n). - Vaclav Kotesovec, Jul 15 2021
a(n) = exp(-1/3)*Sum_{n >= 0} (3*n)^k/(n!*3^n). - Peter Bala, Jun 29 2024
Comments