0,3

Comment from Axel Kohnert (Axel.Kohnert(AT)uni-bayreuth.de): This is the number of permutations of degree n with at most one fall; called Grassmannian permutations by L. and S.

Number of different permutations of a deck of n cards that can be produced by a single shuffle [DeSario]

Number of Dyck paths of semilength n having at most one long ascent (i.e. ascent of length at least two). Example: a(4)=12 because among the 14 Dyck paths of semilength 4, the only paths that have more than one long ascent are UUDDUUDD and UUDUUDDD (each with two long ascents). Here U=(1,1) and D=(1,-1). Also number of ordered trees with n edges having at most one branch node (i.e. vertex of outdegree at least two). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 22 2004

Number of {12,1*2*,21*}-avoiding signed permutations in the hyperoctahedral group.

Number of 1342-avoiding circular permutations on [n+1].

2^n-n is the number of ways to partition {1,2,...,n} into arithmetic progressions, where in each partition all the progressions have the same common difference and have lengths at least 1. - Marty Getz (ffmpg1(AT)uaf.edu) and Dixon Jones (fndjj(AT)uaf.edu), May 21 2005

A107907(a(n+2)) = A000051(n+2) for n>0. - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), May 28 2005

if b(0)=x, and b(n)=b(n-1)+b(n-1)^2*x^(n-2) for n>0, then b(n) is a polynomial of degree a(n). - Michael Somos Nov 04 2006

R. DeSario et al., Invertible shuffles, Problem 10931, Amer. Math. Monthly, 111 (No. 2, 2004), 169-170.

Lascoux and Schutzenberger, Schubert polynomials and the Littlewood Richardson rule, Letters in Math. Physics 10 (1985) 111-124.

Problem 11005, American Math. Monthly, Vol. 112, Jan. 2005, p. 89. (The published solution is incomplete. Letting d be the common difference of the arithmetic progressions, the solver's expression q_1(n,d)=2^(n-d) must be summed over all d=1,...,n, and duplicate partitions must be removed.)

D. Callan, Pattern avoidance in circular permutations.

T. Mansour and J. West, Avoiding 2-letter signed patterns.

a(n+1) = 2*a(n) + n - 1, a(0) = 1. - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Apr 12 2003

Binomial transform of 1, 0, 1, 1, 1, .... The sequence starting 1, 2, 5, ...has a(n)=1+n+2*sum{k=2..n, binom(n, k)}=2^(n+1)-n-1. This is the binomial transform of 1, 1, 2, 2, 2, 2, .... a(n)=1+sum{k=2..n, C(n, k)}. - Paul Barry (pbarry(AT)wit.ie), Jun 06 2003

G.f. = (1-3x+3x^2)/[(1-2x)(1-x)^2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 22 2004

a(n+1) = sum of n-th row for the triangle in A109128. - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Jun 20 2005

Row sums of triangle A133116 - Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 14 2007

A000325 := proc(n) option remember; if n <=1 then n+1 else 2*A000325(n-1)+n-1; fi; end;

(PARI) {a(n)=if(n<0, 0, 2^n-n)} /* Michael Somos Nov 04 2006 */

Cf. A000108.

Column 1 of triangle A008518.

Cf. A133116.

Adjacent sequences: A000322 A000323 A000324 this_sequence A000326 A000327 A000328

Sequence in context: A078410 A096766 A111000 this_sequence A076878 A129983 A083378

nonn

Rosario Salamone (Rosario.Salamone(AT)risc.uni-linz.ac.at)