

A033321


Binomial transform of Fine's sequence A000957: 1, 0, 1, 2, 6, 18, 57, 186, ...


28



1, 1, 2, 6, 21, 79, 311, 1265, 5275, 22431, 96900, 424068, 1876143, 8377299, 37704042, 170870106, 779058843, 3571051579, 16447100702, 76073821946, 353224531663, 1645807790529, 7692793487307, 36061795278341, 169498231169821
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OFFSET

0,3


COMMENTS

Number of permutations avoiding the patterns {2431,4231,4321}; number of weak sorting class based on 2431.  Len Smiley, Nov 01 2005
Number of permutations avoiding the patterns {2413, 3142, 2143}.  Vincent Vatter, Aug 16 2006
Number of permutations avoiding the patterns {2143, 3142, 4132}.  Alexander Burstein and Jonathan Bloom, Aug 03 2013
Number of unimodal Lehmer codes. Those are exactly the inversion sequences for permutations avoiding the patterns {2143, 3142, 4132}.  Alexander Burstein, Jun 16 2015
Number of skew Dyck paths of semilength n ending with a down step (1,1). A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the xaxis, consists of steps U=(1,1)(up), D=(1,1)(down) and L=(1,1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. Number of skew Dyck paths of semilength n and ending with a left step is A128714(n).  Emeric Deutsch, May 11 2007
Number of permutations sortable by a pop stack followed directly by a stack. Equivalently, the number of permutations avoiding {2431, 3142, 3241}.  Vincent Vatter, Mar 06 2013
Starting with offset 1, Hankel transform = oddindexed Fibonacci numbers.  Gary W. Adamson, Dec 27 2008
Starting with offset 1 = INVERT transform of A002212: (1, 1, 3, 10, 36, 137, ...).  Gary W. Adamson, May 19 2009
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j) < e(k). [Martinez and Savage, 2.20]  Eric M. Schmidt, Jul 17 2017
a(n) is the number of permutations of [n] in which the excedances and subcedances are both increasing. (For example, the 3 permutations of [4] NOT counted by a(4)=21 are 3421, 4312, 4321 with excedances/subcedances 34/21, 43/12, 43/21 respectively.)
Proof. It suffices to show that (*) the number of such permutations of [n] containing k fixed points is binomial(n,k)*F(nk), where F is the Fine number A000957. Since F(n) is the number of 321avoiding derangements of [n] and because inserting or deleting a fixed point in a permutation does not change the excedance/fixed point/subcedance status of any other entry, (*) is an immediate consequence of the following claim: The excedances and subcedances of a permutation p are both increasing if and only if p avoids 321. The claim is a nice exercise utilizing the cycles of p for the "if" direction and the pigeonhole principle for the "only if" direction. (End)
Conjectured to be the number of permutations of length n that are sorted to the identity by a consecutive231avoiding stack followed by a classical21avoiding stack.  Colin Defant, Aug 30 2020
a(n) is the number of permutations of length n avoiding the partially ordered pattern (POP) {3>1, 3>4, 1>2} of length 4. That is, the number of length n permutations having no subsequences of length 4 in which the third element is the largest and the first element is larger than the second element.  Sergey Kitaev, Dec 10 2020


LINKS

M. Albert, R. Aldred, M. Atkinson, C Handley, D. Holton, D. McCaughan and H. van Ditmarsch, Sorting Classes, Elec. J. of Comb. 12 (2005) R31.
E. Deutsch, E. Munarini, and S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 21912203.


FORMULA

G.f.: 2/(1 + x + sqrt(1  6*x + 5*x^2)).
Dfinite with recurrence a(n) = ((13*n5)*a(n1)  (16*n23)*a(n2) + 5*(n2)*a(n3))/(2*(n+1)) (n>=3); a(0)=a(1)=1, a(2)=2.  Emeric Deutsch, Mar 21 2004
Binomial transform of Fine's sequence: a(n) = Sum_{k=0..n} binomial(n, k)*A000957(nk).
G.f.: 1/(1xx^2/(13xx^2/(13xx^2/(13xx^2/(1... (continued fraction).  Paul Barry, Jun 15 2009
a(n) = Sum_{m=1..n1} (Sum_(k=1..nm} (binomial(nm1, k1)*(m/(k+m))*binomial(2*k+m1, k+m1) ) ) + 1.  Vladimir Kruchinin, May 12 2011
a(n) = upper left term in M^n, M = the production matrix:
1, 1, 0, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, 0, ...
1, 2, 1, 1, 0, 0, 0, ...
1, 2, 1, 2, 1, 0, 0, ...
1, 2, 1, 2, 1, 1, 0, ...
1, 2, 1, 2, 1, 2, 1, ...
...


MAPLE

a[0] := 1: a[1] := 1: a[2] := 2: for n from 3 to 23 do a[n] := ((13*n5)*a[n1](16*n23)*a[n2]+5*(n2)*a[n3])/2/(n+1) od;


MATHEMATICA

f[n_] := Sum[Binomial[n, k]*g[n  k], {k, 0, n}]; g[n_] := Sum[(1)^(m + n)(n + m)!/n!/m!(n  m + 1)/(n + 1), {m, 0, n}]; Table[ f[n], {n, 24}] (* Robert G. Wilson v, Nov 04 2005 *)


PROG

(Maxima)
a(n):=sum(sum(binomial(nm1, k1)*m/(k+m)*binomial(2*k+m1, k+m1), k, 1, nm), m, 1, n1)+1; /* Vladimir Kruchinin, May 12 2011 */
(PARI) a(n)=1+sum(m=1, n1, sum(k=1, nm, binomial(nm1, k1)/(k+m)* binomial(2*k+m1, k+m1)*m)) \\ Charles R Greathouse IV, Mar 06 2013
(PARI) x='x+O('x^50); Vec(2/(1+x+sqrt(16*x+5*x^2))) \\ Altug Alkan, Oct 22 2015


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



