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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)].

%I A210255 #40 Dec 15 2019 16:37:09
%S A210255 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,
%T A210255 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,
%U A210255 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
%N A210255 a(n) is the number of numbers m for which n is in interval (A007814(m!),  A007814(m!) + A007814(m)].
%C A210255 Using induction, one can prove that the sequence takes only the values 0, 1, 2, and 3. - _Vladimir Shevelev_, Mar 28 2012
%H A210255 Charles R Greathouse IV, <a href="/A210255/b210255.txt">Table of n, a(n) for n = 1..10000</a>
%F A210255 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.
%F A210255 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
%e A210255 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.
%t A210255 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 *)
%o A210255 (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
%Y A210255 Cf. A000142, A007814.
%K A210255 nonn
%O A210255 1,5
%A A210255 _Vladimir Shevelev_, Mar 19 2012