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A055392 Number of bracketings of 0#0#0#...#0 giving result 0, where 0#0 = 1, 0#1 = 1#0 = 1#1 = 0. 4
1, 0, 2, 1, 12, 14, 100, 180, 990, 2310, 10920, 30030, 129612, 396576, 1620168, 5318841, 21029580, 72364578, 280735884, 997356360, 3828988020, 13905563100, 53108050320, 195875639310, 746569720572, 2784329809344, 10610782107800 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

Operation # can be interpreted as NOT OR. The ratio a(n)/A000108(n-1) converges to sqrt(3)/3. Thanks to Soren Galatius Smith

Essentially second column of A112519. - Paul Barry, Sep 09 2005

LINKS

Table of n, a(n) for n=1..27.

FORMULA

G.f.: 1/2 + 1/2 (3 - 2 (1 - 4 x)^{1/2})^{1/2}

The g.f. Z is also given by Z(x) = C(x)U(xC(x)), where U(x) = C(-x) and C is the g.f. of the Catalan numbers. - D. G. Rogers, Oct 20 2005

a(n) = sum{j=0..n, (1/n)*(-1)^(j-1)*C(2n-j-1, n-j)*C(2(j-1), j-1)}; - Paul Barry, Sep 09 2005, corrected by Peter Bala, Aug 19 2014

G.f. A(x) satisfies: A(x) = x + 2*A(x)^3 + A(x)^4; thus, A(x - 2*x^3 - x^4) = x. [Paul D. Hanna, Apr 05 2012]

G.f. A(x) satisfies: x = Sum_{n>=1} 1/(1+A(x))^(2*n-1) * Product_{k=1..n} (1 - 1/(1+A(x))^k). [Paul D. Hanna, Apr 05 2012]

Conjecture: 500*n*(n-1)*a(n) +100*(n-1)*(5*n-12)*a(n-1) +20*(25*n^2-463*n+846)*a(n-2) +(-140161*n^2+966559*n-1637508)*a(n-3) +2*(250*n^2-26509*n+105084)*a(n-4) +98036*(4*n-19)*(4*n-21)*a(n-5)=0. - R. J. Mathar, Nov 26 2012

MATHEMATICA

CoefficientList[ Series[1/2 + 1/2(3 - 2(1 - 4x)^(1/2))^(1/2), {x, 0, 27}], x] (* Robert G. Wilson v, May 04 2004 *)

PROG

(PARI) {a(n)=if(n<1, 0, polcoeff(serreverse(x - 2*x^3 - x^4 +x*O(x^n)), n))} /* Paul D. Hanna, Apr 05 2012 */

CROSSREFS

Cf. A055113, A055395, A112521.

Sequence in context: A048854 A151508 A164826 * A045873 A265022 A110060

Adjacent sequences:  A055389 A055390 A055391 * A055393 A055394 A055395

KEYWORD

nonn

AUTHOR

Jeppe Stig Nielsen, Jun 24 2000

STATUS

approved

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Last modified October 21 23:02 EDT 2018. Contains 316431 sequences. (Running on oeis4.)