|
|
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
(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
|
|
|
FORMULA
|
Additive with a(p^e) = 1 when e > 3, 0 otherwise.
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)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|