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A087698
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Triangle read by rows, giving T(n,k) = maximum number of examples (Boolean inputs) at Hamming distance 2 for symmetric Boolean functions that can have different outputs.
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2
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1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 4, 3, 1, 1, 4, 7, 8, 7, 4, 1, 1, 5, 11, 15, 15, 11, 5, 1, 1, 6, 16, 26, 30, 26, 16, 6, 1, 1, 7, 22, 42, 56, 56, 42, 22, 7, 1, 1, 8, 29, 64, 98, 112, 98, 64, 29, 8, 1, 1, 9, 37, 93, 162, 210, 210, 162, 93, 37, 9, 1, 1, 10, 46, 130, 255, 372, 420
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OFFSET
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0,12
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COMMENTS
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This sets an upper bound on the second order term of the complexity measure introduced by Franco, 2001 for symmetric Boolean functions. The sum of the terms for a given N is equal to 2^(N-1).
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LINKS
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FORMULA
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T(n, N) = ((N-n)^2 + n^2 - N) * C(N, n) / (N^2 - N) n is the term for the series containing N+1 terms
Except for (n,k) = (1,0) the formula T(n,k) = C(n,k) - 2*C(n-1,n-k-1) + 2*C(n-2,n-k-2), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0, appears to give the correct table entries.
Appears to equal A159853, the Riordan array ((1-2*x+2*x^2)/(1-x), x/(1-x)), except for the entry T(1,0). If this is correct then provided n =! 1 we have exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + x + x^2/2! + x^3/3!) = 1 + 2*x + 2*x^2/2! + 4*x^3/3! + 8*x^4/4! + 15*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). (End)
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EXAMPLE
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Triangle begins:
1 N=0
1 1 N=1
1 0 1 N=2
1 1 1 1 N=3
1 2 2 2 1 N=4
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CROSSREFS
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KEYWORD
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AUTHOR
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Leonardo Franco (Leonardo.Franco(AT)psy.ox.ac.uk), Sep 24 2003
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STATUS
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approved
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