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A210255
a(n) is the number of numbers m for which n is in interval (A007814(m!), A007814(m!) + A007814(m)].
4
0, 1, 0, 1, 2, 0, 0, 1, 2, 1, 1, 2, 0, 0, 0, 1, 2, 1, 2, 2, 0, 0, 1, 2, 1, 1, 2, 0, 0, 0, 0, 1, 2, 1, 2, 3, 0, 0, 1, 2, 1, 1, 2, 0, 0, 0, 1, 2, 1, 2, 2, 0, 0, 1, 2, 1, 1, 2, 0, 0, 0, 0, 0, 1, 2, 1, 2, 3, 1, 0, 1, 2, 1, 1, 2, 0, 0, 0, 1, 2, 1, 2, 2, 0, 0, 1, 2, 1, 1, 2, 0, 0, 0, 0, 1, 2, 1, 2, 3, 0
OFFSET
1,5
COMMENTS
Using induction, one can prove that the sequence takes only the values 0, 1, 2, and 3. - Vladimir Shevelev, Mar 28 2012
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
FORMULA
For k >= 1, a(2^k) = 1 and a(2^k-1) = 0; for k >= 2, a(2^k+1) = 2; for k >= 5, a(2^k+4) = 3.
One can prove many formulas of the following type: a(h - A000120(h) + 5) = 2, if A007814(h) = 3 or 4, and a(h - A000120(h) + 5) = 3, if A007814(h) >= 5. - Vladimir Shevelev, Mar 28 2012
EXAMPLE
Let n = 36. Up to m = 30, the maximal n contained in the interval (A007814(m!), A007814(m!) + A007814(m)] is 27. Evidently, it is sufficient to consider even numbers m. For m = 32, 34, 36, 38, and 40, we have the intervals (31, 36], (32, 33], (34, 36], (35, 36], and (38, 41], respectively. Thus, 36 occurs 3 times, and a(36) = 3.
MATHEMATICA
Map[Count[Flatten[Map[Rest[Apply[Range, #]]&, Map[{IntegerExponent[#!, 2], IntegerExponent[#!, 2] + IntegerExponent[#, 2]}&, Range[2, 110, 2]]]], #]&, Range[100]] (* Peter J. C. Moses, Mar 27 2012 *)
PROG
(PARI) list(N)=my(v=vector(N), t, n, s); while((s+=t=valuation(n++, 2))<=N, for(i=s+1, min(s+t, N), v[i]++)); v \\ Charles R Greathouse IV, Mar 28 2012
CROSSREFS
Sequence in context: A237123 A024942 A372809 * A283319 A376081 A049321
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Mar 19 2012
STATUS
approved