A210255 a(n) is the number of numbers m for which n is in interval (A007814(m!), A007814(m!) + A007814(m)].

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

Cf. A000142, A007814.

Sequence in context: A237123 A024942 A372809 * A283319 A049321 A204425

Adjacent sequences: A210252 A210253 A210254 * A210256 A210257 A210258

Keyword

nonn

Author

Vladimir Shevelev, Mar 19 2012

Status

approved