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A033093
Number of 0's when n is written in base b for 2<=b<=n+1.
15
0, 1, 1, 3, 2, 3, 1, 5, 5, 5, 3, 6, 2, 3, 3, 8, 5, 9, 5, 8, 5, 4, 2, 9, 5, 5, 7, 9, 5, 8, 2, 11, 9, 8, 8, 13, 6, 7, 6, 11, 5, 9, 3, 7, 8, 5, 3, 13, 7, 10, 8, 9, 5, 12, 7, 11, 6, 5, 3, 13, 3, 4, 6, 15, 12, 14, 8, 11, 9, 12, 6, 18, 8, 9, 11, 11, 9, 11, 5, 14, 13
OFFSET
1,4
LINKS
FORMULA
Conjecture: lim inf a(n) = infinity. The lim inf grows quite slowly: e.g. a(2047)=7. Weaker conjecture: 2, 3 and 7 are the only n for which a(n) = 1. Note that a(n)=1 implies n=2 or n=2^k-1; probabilistically, the chance that 2^k-1 has no zeros just in base 3 is C^k, where C = (2/3)^(log(2)/log(3)) ~ .774, so the number of exceptions should be finite. It appears that 2^15-1 is the last n with no zeros in either base 2 or 3 (and it does have zeros in base 5). - Franklin T. Adams-Watters, Nov 07 2005
G.f.: (Sum_{b>=2} (Sum_{k>=0} x^(b^(k+1))/(Sum_{0<=i<b} x^(i*b^k)))/(1-x)) - Franklin T. Adams-Watters, Nov 07 2005
MATHEMATICA
f[n_] := Count[Flatten@ Table[ IntegerDigits[n, b], {b, 2, n + 1}], 0]; Array[f, 90] (* Robert G. Wilson v, Nov 14 2012 *)
KEYWORD
nonn,base
STATUS
approved