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Search: id:A055113
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| A055113 |
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Number of bracketings of 0^0^0^...^0 giving result 0, with conventions that 0^0=1^0=1^1=1, 0^1=0. |
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+0
6
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| 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; listen)
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OFFSET
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1,5
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COMMENT
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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.
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REFERENCES
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Thanks to Soren Galatius Smith, Jesper Torp Kristensen et al.
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LINKS
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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 A1111160, A055113 and A006013
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FORMULA
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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.
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EXAMPLE
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Number of bracketings of 0^0^0^0^0^0 giving 0 is 11; so a(6)=11
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MATHEMATICA
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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)
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CROSSREFS
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Cf. A055392, A055395, A006632.
Sequence in context: A023271 A047976 A006382 this_sequence A129015 A097056 A092358
Adjacent sequences: A055110 A055111 A055112 this_sequence A055114 A055115 A055116
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Jeppe Stig Nielsen (sequence(AT)jeppesn.dk), Jun 15 2000
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