

A295884


Number of exponents larger than 3 in the prime factorization of n.


4



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
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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


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n


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)).


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 memoizationmacro definec)
(definec (A295884 n) (if (= 1 n) 0 (+ (if (> (A067029 n) 3) 1 0) (A295884 (A028234 n)))))


CROSSREFS

Cf. A000188, A001221, A003557, A008835, A053164, A056170, A062378, A295659, A295883.
Sequence in context: A023974 A277164 A011730 * A101637 A337380 A011729
Adjacent sequences: A295881 A295882 A295883 * A295885 A295886 A295887


KEYWORD

nonn


AUTHOR

Antti Karttunen, Nov 29 2017


STATUS

approved



