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A245595
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Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that S(n) == r (mod prime(n)), where S(n) is the large Schroeder number A006318(n).
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1
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0, 0, 2, -1, -2, -1, 7, -5, -5, 11, 10, -11, 11, 12, 2, 17, -2, 19, -15, -26, 33, 17, -22, -11, 18, 8, 18, -27, 17, 51, -37, -34, 28, -4, 66, -37, -69, -58, 45, -81, -20, -86, -19, 17, -12, -30, 35, -32, 5, -11, -8, -45, 12, -111, -28, -71, 76, 59, 102, -25
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OFFSET
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1,3
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COMMENTS
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Conjecture: (i) For any integer n > 2, the term a(n) is nonzero, i.e., prime(n) does not divide the large Schroeder number S(n).
(ii) For any integer n > 2, prime(n) does not divide the Bell number B(2*n) = A000110(2*n).
We have verified parts (i) and (ii) for n up to 440000 and 66000 respectively.
Conjecture (i) fails for the first time for n=20239789. In particular, a(20239789)=0. - Max Alekseyev, Oct 05 2015
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LINKS
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EXAMPLE
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a(5) = -2 since S(5) = 394 == -2 (mod prime(5)=11).
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MATHEMATICA
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rMod[m_, n_]:=Mod[m, n, -(n-1)/2]
S[n_]:=Sum[Binomial[n+k, 2k]*Binomial[2k, k]/(k+1), {k, 0, n}]
a[n_]:=rMod[S[n], Prime[n]]
Table[a[n], {n, 1, 60}]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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