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A055113 Number of bracketings of 0^0^0^...^0 giving result 0, with conventions that 0^0=1^0=1^1=1, 0^1=0. 7
1, 0, 1, 1, 5, 11, 41, 120, 421, 1381, 4840, 16721, 59357, 210861, 759071, 2744393, 10000437, 36609977, 134750450, 498016753, 1848174708, 6882643032, 25715836734, 96365606679, 362102430069, 1364028272451, 5150156201026 (list; graph; refs; listen; history; internal format)
OFFSET

1,5

COMMENTS

Total number of bracketings of 0^0^...^0 is A000108(n-1) (this is Catalan's problem). So the number of bracketings giving 1 is A000108(n-1) - a(n).

Also bracketings of f => f => ... => f where f is "false" and "=>" is implication.

REFERENCES

Thanks to Soren Galatius Smith, Jesper Torp Kristensen et al.

LINKS

T. D. Noe, Table of n, a(n) for n=1..200

E. A. Bender and S. G. Williamson, Foundations of Combinatorics with Applications (see Chap. 11, Example 11.3, pp. 312-313 and Example 11.31, pp. 351-352).

Douglas Rogers, Comments on A111160, A055113 and A006013

FORMULA

G.f.: - 1/4 - (1/4)*(1 - 4*x)^(1/2) + (1/4)*(2 + 2*(1 - 4*x)^(1/2) + 12*x)^(1/2).

The ratio a(n)/A000108(n-1) converges to (5-sqrt(5))/10 as n->oo.

a(n)=sum(j=0..n-1, binomial(2*j+n-1,j+n-1)*(-1)^(n-j-1)*binomial(2*n-1,j+n))/(2*n-1); [From Vladimir Kruchinin kru(AT)ie.tusur.ru, May 10 2011]

EXAMPLE

Number of bracketings of 0^0^0^0^0^0 giving 0 is 11; so a(6)=11

MATHEMATICA

Rest[ CoefficientList[ Series[(-1 - Sqrt[1 - 4x] + Sqrt[2]Sqrt[1 + Sqrt[1 - 4x] + 6x])/4, {x, 0, 28}], x]] (from Robert G. Wilson v (rgwv(at)rgwv.com), Oct 28 2005)

PROG

(Maxima)

a(n):=sum(binomial(2*j+n-1, j+n-1)*(-1)^(n-j-1)*binomial(2*n-1, j+n), j, 0, n-1)/(2*n-1); [From Vladimir Kruchinin kru(AT)ie.tusur.ru, May 10 2011]

CROSSREFS

Cf. A055392, A055395, A006632.

Sequence in context: A159049 A047976 A006382 * A129015 A141355 A097056

Adjacent sequences:  A055110 A055111 A055112 * A055114 A055115 A055116

KEYWORD

nonn,nice,easy

AUTHOR

Jeppe Stig Nielsen (sequence(AT)jeppesn.dk), Jun 15 2000

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Last modified February 16 20:14 EST 2012. Contains 205962 sequences.