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A051295 a[ 0 ]=1, a[ m+1 ]=sum{k=0 to m} [ a[ m-k ] k! ]. +0
10
1, 1, 2, 5, 15, 54, 235, 1237, 7790, 57581, 489231, 4690254, 49986715, 585372877, 7463687750, 102854072045, 1522671988215, 24093282856182 (list; graph; listen)
OFFSET

0,3

COMMENT

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 (malbert(AT)cs.otago.ac.nz), Dec 14 2004

REFERENCES

David Callan, A Combinatorial Interpretation of the Eigensequence for Composition, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.4.

LINKS

David Callan, A combinatorial interpretation of the eigensequence for composition

FORMULA

It appears that the INVERT transform of factorial numbers A000142 gives 1, 2, 5, 15, 54, 235, 1237, ... - Antti Karttunen (Antti.Karttunen(AT)iki.fi), May 30 2003

a(n) = Sum_{k>=0} A084938(n, k) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Feb 05 2004

Regarding Karttunen's conjecture: 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.

EXAMPLE

a[ 4 ]=15=a[ 3 ]*0!+a[ 2 ]*1!+a[ 1 ]*2!+a[ 0 ]*3!=5*1+2*1+1*2+1*6

CROSSREFS

Cf. A051296.

Row sums of A084938.

Sequence in context: A006966 A056841 A107112 this_sequence A009383 A104429 A109319

Adjacent sequences: A051292 A051293 A051294 this_sequence A051296 A051297 A051298

KEYWORD

easy,nonn

AUTHOR

Leroy Quet (qq-quet(AT)mindspring.com)

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Last modified July 25 07:41 EDT 2008. Contains 142293 sequences.


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