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A013585
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Smallest m such that 1!+...+m! is divisible by 2n+1, or 0 if no such m exists.
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2
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1, 2, 0, 0, 3, 4, 0, 0, 5, 0, 0, 12, 0, 7, 19, 0, 4, 0, 24, 0, 32, 19, 0, 0, 0, 5, 20, 0, 0, 0, 0, 0, 0, 20, 12, 0, 7, 0, 0, 57, 7, 0, 0, 19, 0, 0, 0, 0, 6, 8, 83, 0, 0, 15, 33, 24, 0, 0, 0, 0, 12, 32, 0, 38, 19, 9, 0, 0, 0, 23, 0, 0, 0, 0, 70, 71, 5, 0, 57, 20, 0, 17, 0, 0, 0, 0, 26, 0, 0, 0, 0, 0, 0, 0, 0, 28
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OFFSET
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0,2
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COMMENTS
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a(n) < 2*n for n > 1.
If a(n) = 0, then a((2*k+1)*n + k) = 0 for all k >= 0.
(End)
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REFERENCES
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M. R. Mudge, Smarandache Notions Journal, University of Craiova, Vol. VII, No. 1, 1996.
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LINKS
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MAPLE
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f:= proc(n) local t, r, m;
r:= 1; t:= 0;
for m from 1 do
r:= r*m mod (2*n+1);
if r = 0 then return 0 fi;
t:= t + r mod (2*n+1);
if t = 0 then return m fi;
od;
end proc:
f(0):= 1:
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MATHEMATICA
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a[n_] := Module[{t, r, m}, r = 1; t = 0; For[m = 1, True, m++, r = Mod[r m, 2 n + 1]; If[r == 0, Return[0]]; t = Mod[t + r, 2 n + 1]; If[t == 0, Return[m]]]];
a[0] = 1;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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