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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 (list; graph; refs; listen; history; text; internal format)
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 memoization-macro 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

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Last modified September 21 03:13 EDT 2020. Contains 337266 sequences. (Running on oeis4.)