1,3

Or, number of permutations with no fixed points avoiding 213 and 132.

Number of derangements of {1,2,...,n} having ascending runs consisting of consecutive integers. Example: a(4)=6 because we have 234/1, 34/12, 34/2/1, 4/123, 4/3/12, 4/3/2/1, the ascending runs being as indicated. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 08 2004

Let c be twice sequence A002450 interlaced with itself (from the second term), i.e. c = 2*(0, 1, 1, 5, 5, 21, 21, 85, 85, 341, 341, ). Let d be powers of 4 interlace with the zero-sequence: d = (1, 0, 4, 0, 16, 0, 64, 0, 256, 0,). Then a(n+1) = c(n) + d(n). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), May 09 2005

Inverse binomial transform of A094705 (0, 1, 4, 15). - Paul Curtz (bpcrtz(AT)free.fr), Jun 15 2008

J. Brillhart and P. Morton, A case study in mathematical research: the Golay- Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869 (contains the sequence of the odd-subscripted terms and that of the even-subscripted terms).

E. Deutsch, Problem 10902, Amer. Math. Monthly, 110 (2003), 639.

T. Mansour and A. Robertson, Refined restricted permutations....

a(n) = (3/8)*2^n +(1/24)*(-2)^n - 2/3; recurrence relation: a(n) = 4a(n-2) + 2, a(1)=0, a(2)=1; generating function = z^2*(1+z)/((1-z)(1-4z^2); a(n)=A020989((n-2)/2) for n=2, 4, 6, ... and A020988((n-3)/2) for n=3, 5, 7, ...

a(n+1)-2a(n) = A078008 signed. Differences: doubled A000302. - Paul Curtz (bpcrtz(AT)free.fr), Jun 15 2008

a(4)=6 because the only 132- and 213-avoiding permutations of {1,2,3,4} without fixed points are: 2341, 3412, 3421, 4123, 4312 and 4321.

Floretion Algebra Multiplication Program, FAMP Code: jesseq[ + 'i - .5'j + i' - .5j' + 'kk' + 'ik' + 'jk' + 'ki' + 'kj']

Cf. A020988, A020989.

Sequence in context: A055237 A057434 A162581 this_sequence A119459 A102581 A152000

Adjacent sequences: A061544 A061545 A061546 this_sequence A061548 A061549 A061550

nonn

Emeric Deutsch (deutsch(AT)duke.poly.edu), May 16 2001