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A004212 Shifts one place left under 3rd-order binomial transform.

%I A004212 M3557 #81 Jul 01 2024 13:15:12
%S A004212 1,1,4,19,109,742,5815,51193,498118,5296321,60987817,754940848,
%T A004212 9983845261,140329768789,2087182244308,32725315072135,539118388883449,
%U A004212 9304591246975030,167804098493079547,3155000165773280893
%N A004212 Shifts one place left under 3rd-order binomial transform.
%C A004212 Equals the eigensequence of triangle A027465, the cube of Pascal's triangle. - _Gary W. Adamson_, Apr 10 2009
%C A004212 Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+3 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=3, otherwise F(k+1)=F(k); see example and Fxtbook link. - _Joerg Arndt_, Apr 30 2011
%D A004212 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A004212 Vincenzo Librandi, <a href="/A004212/b004212.txt">Table of n, a(n) for n = 0..200</a>
%H A004212 Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>, section 17.3.5, pp. 366-368
%H A004212 I. M. Gessel, <a href="http://arxiv.org/abs/math.CO/0108121">Applications of the classical umbral calculus</a>. arXiv:math/0108121v3 [math.CO], 2001.
%H A004212 Adalbert Kerber, <a href="http://dx.doi.org/10.1016/0012-365X(78)90163-2">A matrix of combinatorial numbers related to the symmetric groups</a>, Discrete Math., 21 (1978), 319-321.
%H A004212 A. Kerber, <a href="/A004211/a004211.pdf">A matrix of combinatorial numbers related to the symmetric groups<</a>, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
%H A004212 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F A004212 a_n = Sum_{k=0..n} 3^(n-k)*Stirling2(n, k). - _Emeric Deutsch_, Feb 11 2002
%F A004212 E.g.f.: exp((exp(3*x)-1)/3).
%F A004212 O.g.f. A(x) satisfies A'(x)/A(x) = e^(3*x).
%F A004212 E.g.f.: exp(Integral_{t = 0..x} exp(3*t)). - _Joerg Arndt_, Apr 30 2011
%F A004212 O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-3*j*x). - _Joerg Arndt_, Apr 30 2011
%F A004212 Hankel transform is A000178(n)*3^C(n+1,2). - _Paul Barry_, Mar 31 2008
%F A004212 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
%F A004212 a(n) = the upper left term in M^n, M = the following infinite square production matrix:
%F A004212   1, 3, 0, 0, 0, ...
%F A004212   1, 1, 3, 0, 0, ...
%F A004212   1, 2, 1, 3, 0, ...
%F A004212   1, 3, 3, 1, 3, ...
%F A004212   ... (in which a diagonal of (3,3,3,...) is appended to the right of Pascal's triangle). - _Gary W. Adamson_, Jul 29 2011
%F A004212 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]
%F A004212 From _Peter Bala_, May 16 2012: (Start)
%F A004212 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.
%F A004212 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)
%F A004212 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
%F A004212 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
%F A004212 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
%F A004212 a(n) = exp(-1/3)*Sum_{n >= 0} (3*n)^k/(n!*3^n). - _Peter Bala_, Jun 29 2024
%e A004212 Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
%e A004212 a(2)=3 to [00], [01], [02], and [03], a(3) = 19 to
%e A004212         RGS           F
%e A004212 01:  [ 0 0 0 ]    [ 0 0 0 ]
%e A004212 02:  [ 0 0 1 ]    [ 0 0 0 ]
%e A004212 03:  [ 0 0 2 ]    [ 0 0 0 ]
%e A004212 04:  [ 0 0 3 ]    [ 0 0 3 ]
%e A004212 05:  [ 0 1 0 ]    [ 0 0 0 ]
%e A004212 06:  [ 0 1 1 ]    [ 0 0 0 ]
%e A004212 07:  [ 0 1 2 ]    [ 0 0 0 ]
%e A004212 08:  [ 0 1 3 ]    [ 0 0 3 ]
%e A004212 09:  [ 0 2 0 ]    [ 0 0 0 ]
%e A004212 10:  [ 0 2 1 ]    [ 0 0 0 ]
%e A004212 11:  [ 0 2 2 ]    [ 0 0 0 ]
%e A004212 12:  [ 0 2 3 ]    [ 0 0 3 ]
%e A004212 13:  [ 0 3 0 ]    [ 0 3 3 ]
%e A004212 14:  [ 0 3 1 ]    [ 0 3 3 ]
%e A004212 15:  [ 0 3 2 ]    [ 0 3 3 ]
%e A004212 16:  [ 0 3 3 ]    [ 0 3 3 ]
%e A004212 17:  [ 0 3 4 ]    [ 0 3 3 ]
%e A004212 18:  [ 0 3 5 ]    [ 0 3 3 ]
%e A004212 19:  [ 0 3 6 ]    [ 0 3 6 ]
%e A004212 - _Joerg Arndt_, Apr 30 2011
%t A004212 Table[Sum[StirlingS2[n,k] 3^(-k+n),{k,n}],{n,20}] (* _Vincenzo Librandi_, May 21 2012 *)
%t A004212 Table[3^n BellB[n, 1/3], {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 20 2015 *)
%o A004212 (PARI) x='x+O('x^66); /* that many terms */
%o A004212 egf=exp(intformal(exp(3*x))); /* =  1 + x + 2*x^2 + 19/6*x^3 + 109/24*x^4 + ... */
%o A004212 /* egf=exp(1/3*(exp(3*x)-1)) */ /* alternative computation */
%o A004212 Vec(serlaplace(egf)) /* show terms */ /* _Joerg Arndt_, Apr 30 2011 */
%o A004212 (Maxima)
%o A004212 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 */
%Y A004212 Cf. A075498 (row sums).
%Y A004212 Cf. A027465. - _Gary W. Adamson_, Apr 10 2009
%Y A004212 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
%Y A004212 Cf. A009235.
%K A004212 nonn,easy,eigen
%O A004212 0,3
%A A004212 _N. J. A. Sloane_