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A256447
Number of integers in range (prime(n)^2)+1 .. (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n): a(n) = A250477(n) - A250474(n).
7
2, 3, 3, 7, 5, 9, 6, 13, 23, 9, 28, 22, 12, 24, 39, 37, 17, 44, 32, 16, 53, 37, 53, 76, 46, 23, 43, 20, 49, 161, 48, 82, 23, 142, 27, 91, 90, 66, 103, 97, 41, 181, 41, 74, 39, 228, 228, 86, 45, 86, 130, 44, 217, 134, 141, 138, 46, 148, 106, 47, 261, 355, 116, 53, 109, 387, 166, 284, 65, 119, 181, 243, 198, 195, 122, 190, 268, 125, 265, 330, 78
OFFSET
1,1
COMMENTS
a(n) = number of integers in range [(prime(n)^2)+1, (prime(n) * prime(n+1))] whose smallest prime factor is at least prime(n).
All the terms are strictly positive, because at least for the last number in the range we have A020639(prime(n)*prime(n+1)) = prime(n).
See the conjectures in A256448.
FORMULA
a(n) = A250477(n) - A250474(n).
a(n) = A251723(n) - A256448(n).
a(n) = A256448(n) + A256449(n).
a(n) = A256468(n) + 1.
Other identities. For all n >= 1:
a(n+1) = A256446(n) - A256448(n).
EXAMPLE
For n=1, we have in range [(prime(1)^2)+1, (prime(1) * prime(2))], that is, in range [5,6], two numbers, 5 and 6, whose smallest prime factor (A020639) is at least 2, thus a(1) = 2.
For n=2, we have in range [10, 15] three numbers, {11, 13, 15}, whose smallest prime factor is at least 3, thus a(2) = 3.
For n=3, we have in range [26, 35] three numbers, {29, 31, 35}, whose smallest prime factor is at least prime(3) = 5, thus a(3) = 3.
MATHEMATICA
f[n_] := Count[Range[Prime[n]^2 + 1, Prime[n] Prime[n + 1]],
x_ /; Min[First /@ FactorInteger[x]] >=
Prime@n]; Array[f, 81] (* Michael De Vlieger, Mar 30 2015 *)
PROG
(Scheme) (define (A256447 n) (- (A250477 n) (A250474 n)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Mar 29 2015
STATUS
approved