A325273 Prime omicron of n!.

0, 0, 1, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Author

Gus Wiseman, Apr 18 2019

Keywords

• nonn

Comments

We define the omega-sequence of n (row n of A323023) to have length A323014(n) = adjusted frequency depth of n, and the k-th term is Omega(red^{k-1}(n)), where Omega = A001222 and red^{k} is the k-th functional iteration of red = A181819, defined by red(n = p^i*...*q^j) = prime(i)*...*prime(j) = product of primes indexed by the prime exponents of n. For example, we have 180 -> 18 -> 6 -> 4 -> 3, so the omega-sequence of 180 is (5,3,2,2,1).

The prime omicron of n (A304465) is 0 if n is 1, 1 if n is prime, and otherwise the second-to-last part of the omega-sequence of n. For example, the prime omicron of 180 is 2.

Conjecture: all terms after a(10) = 4 are less than 4.

From James Rayman, Apr 17 2021: (Start)

The conjecture is false. a(3804) = 4. In fact, there are 91 values of n < 10000 such that a(n) = 4.

The first value of n such that a(n) = 5 is 37934. For any other n < 5*10^5, a(n) < 5. (End)

Links

• James Rayman, Table of n, a(n) for n = 0..10000

Crossrefs

a(n) = A055396(A325275(n)/2).

Cf. A000142, A006939, A303555, A323023, A325238, A325272, A325274, A325275, A325276, A325277.

Omega-sequence statistics: A001222 (first omega), A001221 (second omega), A071625 (third omega), A323022 (fourth omega), A304465 (second-to-last omega), A182850 or A323014 (length/frequency depth), A325248 (Heinz number), A325249 (sum).

Programs

Mathematica

omseq[n_Integer]:=If[n<=1,{},Total/@NestWhileList[Sort[Length/@Split[#]]&,Sort[Last/@FactorInteger[n]],Total[#]>1&]];
omicron[n_]:=Switch[n,1,0,?PrimeQ,1,,omseq[n][[-2]]];
Table[omicron[n!],{n,0,100}]

Python

from sympy.ntheory import *
def red(v):
    r = {}
    for i in v: r[i] = r.get(i, 0) + 1
    return r
def omicron(v):
    if len(v) == 0: return 0
    if len(v) == 1: return v[0]
    else: return omicron(list(red(v).values()))
f, a_list = {}, []
for i in range(101):
    a_list.append(omicron(list(f.values())))
    g = factorint(i+1)
    for k in g: f[k] = f.get(k, 0) + g[k]
print(a_list) # James Rayman, Apr 17 2021

Extensions

More terms from James Rayman, Apr 17 2021