OFFSET
0,3
COMMENTS
a(n) = number of permutations on [n] that contain a 132 pattern only as part of a 4132 pattern. For example, a(4) = 15 counts the 14 132-avoiding permutations on [4] (Catalan numbers A000108) and 4132.
a(n) is the number of permutations on [n] that contain a (scattered) 342 pattern only as part of a 1342 pattern. For example, 412635 fails because 463 is an offending 342 pattern (= 231 pattern).
This sequence gives the number of permutations of {1,2,...,n} such that the elements of each cycle of the permutation form an interval. - Michael Albert, Dec 14 2004
Starting (1, 2, 5, 15, ...) = row sums of triangle A143965. - Gary W. Adamson, Apr 10 2009
Number of compositions of n where there are (k-1)! sorts of part k. - Joerg Arndt, Aug 04 2014
LINKS
Vincenzo Librandi and Vaclav Kotesovec, Table of n, a(n) for n = 0..448 (first 100 terms from Vincenzo Librandi)
David Callan, A combinatorial interpretation of the eigensequence for composition, arXiv:math/0507169 [math.CO], 2005.
David Callan, A Combinatorial Interpretation of the Eigensequence for Composition, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.4.
Stefan Forcey, Aaron Lauve and Frank Sottile, Cofree compositions of coalgebras, Annals of Combinatorics 17 (1) pp. 105-130 March, 2013.
Richard J. Martin, and Michael J. Kearney, Integral representation of certain combinatorial recurrences, Combinatorica: 35:3 (2015), 309-315.
Lara Pudwell, Enumeration Schemes for Permutations Avoiding Barred Patterns, El. J. Combinat. 17 (1) (2010) R29.
FORMULA
It appears that the INVERT transform of factorial numbers A000142 gives 1, 2, 5, 15, 54, 235, 1237, ... - Antti Karttunen, May 30 2003
This is true: translating the defining recurrence to a generating function identity yields A(x)= 1/(1 - (0!x + 1!x^2 + 2!x^3 + ...)) which is the INVERT formula.
In other words: let F(x) = Sum_{n>=0} n!*x^n then the g.f. is 1/(1-x*F(x)), cf. A052186 (g.f. F(x)/(1+x*F(x))). - Joerg Arndt, Apr 25 2011
a(n) = Sum_{k>=0} A084938(n, k). - Philippe Deléham, Feb 05 2004
G.f. A(x) satisfies: A(x) = (1-x)*A(x)^2 - x^2*A'(x). - Paul D. Hanna, Aug 02 2008
G.f.: A(x) = 1/(1-x/(1-1*x/(1-1*x/(1-2*x/(1-2*x/(1-3*x/(1-3*x...))))))))) (continued fraction). - Paul Barry, Sep 25 2008
From Gary W. Adamson, Jul 22 2011: (Start)
a(n) = upper left term in M^n, M = an infinite square production matrix in which a column of 1's is prepended to Pascal's triangle, as follows:
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 2, 1, 0, ...
1, 1, 3, 3, 1, ...
...
Also, a(n+1) = sum of top row terms of M^n. (End)
G.f.: 1+x/(U(0)-x) where U(k) = 1 + x*k - x*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 10 2012
G.f.: 1/(U(0) - x) where U(k) = 1 - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Nov 12 2012
a(n) ~ (n-1)! * (1 + 2/n + 7/n^2 + 31/n^3 + 165/n^4 + 1025/n^5 + 7310/n^6 + 59284/n^7 + 543702/n^8 + 5618267/n^9 + 65200918/n^10), for coefficients see A260532. - Vaclav Kotesovec, Jul 28 2015
EXAMPLE
a[ 4 ]=15=a[ 3 ]*0!+a[ 2 ]*1!+a[ 1 ]*2!+a[ 0 ]*3!=5*1+2*1+1*2+1*6.
As to matrix M, a(3) = 5 since the top row of M^n = (5, 5, 4, 1), with a(4) = 15 = (5 + 5 + 4 + 1).
MAPLE
a := proc(n) option remember; `if`(n<2, 1, add(a(n-j-1)*j!, j=0..n-1)) end proc: seq(a(n), n=0..30); # Vaclav Kotesovec, Jul 28 2015
MATHEMATICA
Table[Coefficient[Series[E^x/(E^x-ExpIntegralEi[x]), {x, Infinity, 20}], x, -n], {n, 0, 20}] (* Vaclav Kotesovec, Feb 22 2014 *)
PROG
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x^2*deriv(A)/A)/(1-x)); polcoeff(A, n)} \\ Paul D. Hanna, Aug 02 2008
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
EXTENSIONS
More terms from Vincenzo Librandi, Feb 23 2013
STATUS
approved