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A004211
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Shifts one place left under 2nd-order binomial transform.
(Formerly M2900)
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49
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1, 1, 3, 11, 49, 257, 1539, 10299, 75905, 609441, 5284451, 49134923, 487026929, 5120905441, 56878092067, 664920021819, 8155340557697, 104652541401025, 1401572711758403, 19546873773314571, 283314887789276721, 4259997696504874817, 66341623494636864963
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), section 17.3.5, pp. 366-368
P. Barry, Constructing Exponential Riordan Arrays from Their A and Z Sequences, Journal of Integer Sequences, 17 (2014), #14.2.6.
D. Callan (Proposer), Problem 11567, Amer. Math. Monthly, 118 (2011), 371-378.
I. M. Gessel, Applications of the classical umbral calculus, arXiv:math/0108121v3 [math.CO], 2001.
Adam M. Goyt, 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.
A. Kerber, A matrix of combinatorial numbers related to the symmetric groups<, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Huyile Liang, Jeffrey Remmel, 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
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FORMULA
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O.g.f. A(x) satisfies A'(x)/A(x) = e^(2x).
E.g.f.: exp(sinh(x)*exp(x)) = exp(int(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_{l=1..k} (1-2*l*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]
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. - Peter Bala, May 16 2012
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 + 1/2) / (exp(n + 1/2 - (n+1)/LambertW(2*n+2)) * sqrt(n+1) * sqrt(1 + 1/LambertW(2*n+2)) * LambertW(2*n+2)^(n + 1/2)). - Vaclav Kotesovec, Jan 07 2019
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EXAMPLE
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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)
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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 */
(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 */
(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
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CROSSREFS
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Cf. A075497 (row sums).
Cf. A038207. - Gary W. Adamson, Apr 10 2009
Cf. A004212 (RGS where s(k)<=F(k)+3), A004213 (s(k)<=F(k)+4), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1), A009235. - Joerg Arndt, Apr 30 2011
Main diagonal of A261275.
Cf. A034691, A075729.
Sequence in context: A246985 A326521 A340813 * A180869 A001339 A307030
Adjacent sequences: A004208 A004209 A004210 * A004212 A004213 A004214
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KEYWORD
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nonn,easy,nice,eigen
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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