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A189202
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Let s_k(n) denote the sum of digits of n in base k. Then a(n) is the smallest m>0 such that both s_2(m*(n-1)) and s_n(2*m*(n-1))/(n-1) are even, or a(n)=0, if such m does not exist.
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1
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3, 5, 5, 3, 13, 4, 9, 5, 11, 6, 19, 20, 15, 47, 17, 9, 19, 10, 21, 32, 23, 12, 37, 13, 40, 41, 29, 15, 46, 16, 33, 17, 35, 18, 37, 56, 39, 20, 41, 21, 85, 22, 45, 68, 47, 72, 73, 25, 51, 26, 79, 80, 109, 28, 57, 87, 59, 30, 91, 153, 63, 191
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OFFSET
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2,1
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COMMENTS
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Conjecture: For all n>=2, a(n)>0.
For a general problem, see SeqFan link.
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LINKS
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MAPLE
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s:= proc(n, b) local m, t;
t:= 0; m:= n;
while m<>0 do t:= t+ irem(m, b, 'm') od; t
end:
a:= proc(n) local m;
for m while irem(s(m*(n-1), 2), 2)<>0 or
irem(s(2*m*(n-1), n)/(n-1), 2)<>0 do od; m
end:
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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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