A004211 Shifts one place left under 2nd-order binomial transform.

1, 1, 3, 11, 49, 257, 1539, 10299, 75905, 609441, 5284451, 49134923, 487026929, 5120905441, 56878092067, 664920021819, 8155340557697, 104652541401025, 1401572711758403, 19546873773314571, 283314887789276721, 4259997696504874817, 66341623494636864963

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice
• eigen

Comments

Equals the eigensequence of A038207, the square of Pascal's triangle. - Gary W. Adamson, Apr 10 2009

The g.f. of the second binomial transform is 1/(1-2x-x/(1-2x/(1-2x-x/(1-4x/(1-2x-x/(1-6x/(1-2x-x/(1-8x/(1-... (continued fraction). - Paul Barry, Dec 04 2009

Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+2 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=2, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011

It appears that the infinite set of "Shifts 1 place left under N-th order binomial transform" sequences has a production matrix of:

1, N, 0, 0, 0, ...

1, 1, N, 0, 0, ...

1, 2, 1, N, 0, ...

1, 3, 3, 1, N, ...

... (where a diagonal of (N,N,N,...) is appended to the right of Pascal's triangle). a(n) in each sequence is the upper left term of M^n such that N=1 generates A000110, then (N=2 - A004211), (N=3 - A004212), (N=4 - A004213), (N=5 - A005011), ... - Gary W. Adamson, Jul 29 2011

Number of "unlabeled" hierarchical orderings on set partitions of {1..n}, see comments on A034691. - Gus Wiseman, Mar 03 2016

From Lorenzo Sauras Altuzarra, Jun 17 2022: (Start)

Number of n-variate noncontradictory conjunctions of logical equalities of literals (modulo logical equivalence).

Equivalently, number of n-variate noncontradictory Krom formulas with palindromic truth-vector (modulo logical equivalence).

a(n) <= A109457(n). (End)

Examples

			From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings: a(0)=1 corresponds to the empty string;
a(1)=1 to [0]; a(2)=3 to [00], [01], and [02]; a(3) = 11 to
        RGS           F
[ 1]  [ 0 0 0 ]    [ 0 0 0 ]
[ 2]  [ 0 0 1 ]    [ 0 0 0 ]
[ 3]  [ 0 0 2 ]    [ 0 0 2 ]
[ 4]  [ 0 1 0 ]    [ 0 0 0 ]
[ 5]  [ 0 1 1 ]    [ 0 0 0 ]
[ 6]  [ 0 1 2 ]    [ 0 0 2 ]
[ 7]  [ 0 2 0 ]    [ 0 2 2 ]
[ 8]  [ 0 2 1 ]    [ 0 2 2 ]
[ 9]  [ 0 2 2 ]    [ 0 2 2 ]
[10]  [ 0 2 3 ]    [ 0 2 2 ]
[11]  [ 0 2 4 ]    [ 0 2 4 ]. (End)
From _Lorenzo Sauras Altuzarra_, Jun 17 2022: (Start)
The 11 trivariate noncontradictory conjunctions of logical equalities of literals are (x <-> y) /\ (y <-> z), (~ x <-> y) /\ (y <-> z), (x <-> ~ y) /\ (~ y <-> z), (x <-> y) /\ (y <-> ~ z), (x <-> y) /\ (z <-> z), (~ x <-> y) /\ (z <-> z), (x <-> z) /\ (y <-> y), (~ x <-> z) /\ (y <-> y), (y <-> z) /\ (x <-> x), (~ y <-> z) /\ (x <-> x), and (x <-> x) /\ (y <-> y) /\ (z <-> z) (modulo logical equivalence).
The third complete Bell polynomial is x^3 + 3 x y + z; and note that (2^0)^3 + 3*2^0*2^1 + 2^2 = 11.
The truth-vector of (x <-> y) /\ (y <-> z), 10000001, is palindromic. (End)
		

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
• Paul Barry, Constructing Exponential Riordan Arrays from Their A and Z Sequences, Journal of Integer Sequences, 17 (2014), #14.2.6.
• Zhanar Berikkyzy, Pamela E. Harris, Anna Pun, Catherine Yan, and Chenchen Zhao, On the Limiting Vacillating Tableaux for Integer Sequences, arXiv:2208.13091 [math.CO], 2022.
• Zhanar Berikkyzy, Pamela E. Harris, Anna Pun, Catherine Yan, and Chenchen Zhao, Combinatorial Identities for Vacillating Tableaux, arXiv:2308.14183 [math.CO], 2023. See pp. 22, 29.
• David Callan (Proposer), Problem 11567, Amer. Math. Monthly, 118 (2011), 371-378.
• Agustina Czenky, Unoriented 2-dimensional TQFTs and the category Rep(S_t wr Z_2), arXiv:2306.08826 [math.QA], 2023. See p. 40.
• I. M. Gessel, Applications of the classical umbral calculus, arXiv:math/0108121v3 [math.CO], 2001.
• Adam M. Goyt and Lara K. Pudwell, Solution to Monthly Problem 11567, 14 May 2011.
• Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321.
• Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups<, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
• Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 20.
• Janusz Milek, Quantum Implementation of Risk Analysis-relevant Copulas, arXiv:2002.07389 [stat.ME], 2020.
• N. J. A. Sloane, Transforms

Crossrefs

Cf. A075497 (row sums).

Cf. A038207.

Cf. A000110 (RGS where s(k) <= F(k) + 1), A004212 (RGS where s(k) <= F(k) + 3), A004213 (s(k) <= F(k) + 4), A005011 (s(k) <= F(k) + 5), A005012 (s(k) <= F(k) + 6), A075506 (s(k) <= F(k) + 7), A075507 (s(k) <= F(k) + 8), A075508 (s(k) <= F(k) + 9), A075509 (s(k) <= F(k) + 10).

Cf. A000587, A009235, A317996, A318179 - A318181.

Main diagonal of A261275.

Cf. A034691, A075729, A109457.

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n-1, j-1)*2^(j-1), j=1..n))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, May 30 2021
# second Maple program:
a:= n -> CompleteBellB(n, seq(2^k, k=0..n)):
seq(a(n), n=0..23);  # Lorenzo Sauras Altuzarra, Jun 17 2022

Mathematica

Table[ Sum[ StirlingS2[ n, k ] 2^(-k+n), {k, n} ], {n, 16} ] (* Wouter Meeussen *)
Table[2^n BellB[n, 1/2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)

Maxima

a(n):=if n=0 then 1 else sum(2^(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(2*x))); /* = 1 + x + 3/2*x^2 + 11/6*x^3 + ... */
/* egf=exp(1/2*(exp(2*x)-1)) */ /* alternative e.g.f. */
Vec(serlaplace(egf))  /* Joerg Arndt, Apr 30 2011 */

SageMath

def A004211(n): return sum(2^(n-k)*stirling_number2(n, k) for k in (0..n))
print([A004211(n) for n in range(21)]) # Peter Luschny, Apr 15 2020

Formula

E.g.f. A(x) satisfies A'(x)/A(x) = e^(2x).

E.g.f.: exp(sinh(x)*exp(x)) = exp(Integral_{t = 0..x} exp(2*t)) = exp((exp(2*x)-1)/2). - Joerg Arndt, Apr 30 2011 and May 13 2011

a(n) = Sum_{k=0..n} 2^(n-k)*Stirling2(n, k). - Emeric Deutsch, Feb 11 2002

G.f.: Sum_{k >= 0} x^k/Product_{j = 1..k} (1-2*j*x). - Ralf Stephan, Apr 18 2004

Stirling transform of A000085. - Vladeta Jovovic May 14 2004

O.g.f.: A(x) = 1/(1-x-2*x^2/(1-3*x-4*x^2/(1-5*x-6*x^2/(1-... -(2*n-1)*x-2*n*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006

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)*2^{n-1}*f_n(1/2). - Milan Janjic, May 30 2008

G.f.: 1/(1-x/(1-2x/(1-x/(1-4x/(1-x/(1-6x/(1-x/(1-8x/(1-... (continued fraction). - Paul Barry, Dec 04 2009

a(n) = upper left term in M^n, M = an infinite square production matrix with an appended diagonal of (2,2,2,...) to the right of Pascal's triangle:

1, 2, 0, 0, 0, ...

1, 1, 2, 0, 0, ...

1, 2, 1, 2, 0, ...

1, 3, 3, 1, 2, ...

... - Gary W. Adamson, Jul 29 2011

a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+2*x)*d/dx. Cf. A000110. - Peter Bala, Nov 25 2011

G.f. A(x) satisfies A(x)=1+x/(1-2*x)*A(x/(1-2*x)), a(n) = Sum_{k=1..n} binomial(n-1,k-1)*2^(n-k)*a(k-1), 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} 2^(n-k)*C(n,k)*a(k).

Written umbrally this is a(n+1) = (a + 2)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a+2) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a-2)*(a-4)*...*(a-2*(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. Cf. A009235.

Touchard's congruence holds: for odd prime p, 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 odd prime p. (End)

G.f.: (2/E(0) - 1)/x where E(k) = 1 + 1/(1 + 2*x/(1 - 4*(k+1)*x/E(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 20 2012

G.f.: (1/E(0)-1)/x where E(k) = 1 - x/(1 + 2*x - 2*x*(k+1)/E(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Sep 21 2012

a(n) = -1 + 2*Sum_{k=0..n} C(n,k)*A166922(k). - Peter Luschny, Nov 01 2012

G.f.: G(0)- 1/x where G(k) = 1 - (4*x*k-1)/(x - x^4/(x^3 - (4*x*k-1)*(4*x*k+2*x-1)*(4*x*k+4*x-1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 08 2013.

G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*k*x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013

G.f.: -G(0) where G(k) = 1 + 2*(1-k*x)/(2*k*x - 1 - x*(2*k*x - 1)/(x - 2*(1-k*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 29 2013

G.f.: 1/Q(0), where Q(k) = 1 - x/(1 - 2*x*(2*k+1)/( 1 - x/(1 - 4*x*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Apr 15 2013

G.f.: 1 + x/Q(0), where Q(k)= 1 - x*(2*k+3) - x^2*(2*k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 05 2013

For n > 0, a(n) = exp(-1/2)*Sum_{k > 0} (2*k)^n/(k!*2^k). - Vladimir Reshetnikov, May 09 2013

G.f.: -(1+(2*x+1)/G(0))/x, where G(k)= 2*x*k - x - 1 - 2*(k+1)*x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 20 2013

G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013

Sum_{k=0..n} C(n,k)*a(k)*a(n-k) = 2^n*Bell(n) = A055882(n). - Vaclav Kotesovec, Apr 03 2016

a(n) ~ 2^n * n^n * exp(n/LambertW(2*n) - n - 1/2) / (sqrt(1 + LambertW(2*n)) * LambertW(2*n)^n). - Vaclav Kotesovec, Jan 07 2019, simplified Oct 01 2022

a(n) = B_n(2^0, ..., 2^(n - 1)), where B_n(x_1, ..., x_n) is the n-th complete Bell polynomial. - Lorenzo Sauras Altuzarra, Jun 17 2022