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A295884
Number of exponents larger than 3 in the prime factorization of n.
5
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
1,1296
COMMENTS
a(1296) is the first term greater than 1, and a(810000) is the first term greater than 2. - Harvey P. Dale, Dec 22 2017
FORMULA
Additive with a(p^e) = 1 when e > 3, 0 otherwise.
a(n) = A295659(n) - A295883(n).
a(n) = A056170(A062378(n)) = A056170(A003557(A003557(n))) = A001221(A003557^3(n)).
a(n) = A001221(A053164(n)) = A001221(A008835(n)).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^4 = 0.076993... (A085964). - Amiram Eldar, Nov 01 2020
EXAMPLE
For n = 16 = 2^4, there is one exponent and it is larger than 3, thus a(16) = 1.
For n = 96 = 2^5 * 3^1, there are two exponents, and the other one is larger than 3, thus a(96) = 1.
For n = 1296 = 2^4 * 3^4, there are two exponents larger than 3, thus a(1296) = 2.
MATHEMATICA
Array[Total@ Map[Boole[# > 3] &, FactorInteger[#][[All, -1]]] &, 120] (* Michael De Vlieger, Nov 29 2017 *)
Table[Count[FactorInteger[n][[All, 2]], _?(#>3&)], {n, 130}] (* Harvey P. Dale, Dec 22 2017 *)
PROG
(Scheme, with memoization-macro definec)
(definec (A295884 n) (if (= 1 n) 0 (+ (if (> (A067029 n) 3) 1 0) (A295884 (A028234 n)))))
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 29 2017
STATUS
approved