A060826 a(n) is the largest number such that 3^a(n) [also 6^a(n)] divides A025487(n) (least prime signatures).

0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 0, 2, 2, 1, 1, 3, 1, 0, 2, 2, 1, 1, 3, 1, 0, 2, 2, 1, 1, 3, 2, 1, 0, 3, 2, 2, 4, 2, 1, 1, 3, 2, 1, 0, 3, 2, 1, 2, 4, 2, 1, 1, 3, 2, 1, 0, 3, 2, 1, 2, 4, 3, 2, 1, 2, 4, 1, 3, 2, 3, 1, 5, 0, 3, 2, 1, 2, 4, 3, 2, 1, 2, 4, 1, 3, 2, 2, 3, 1, 5, 0, 3, 2, 1, 2, 4, 3, 2

Offset

1,11

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = A007949(A025487(n)) = A122841(A025487(n)). - Amiram Eldar, Dec 29 2020

Prog

(Python)
from functools import lru_cache
from itertools import count
from sympy import prime, integer_log, prime, multiplicity
from oeis_sequences.OEISsequences import bisection
def A060826(n):
    @lru_cache(maxsize=None)
    def g(x, m, j): return sum(g(x//(prime(m)**i), m-1, i) for i in range(j, integer_log(x, prime(m))[0]+1)) if m-1 else max(0, x.bit_length()-j)
    def f(x):
        c, p = n-1+x, 1
        for k in count(1):
            p *= prime(k)
            if p>x:
                break
            c -= g(x, k, 1)
        return c
    return multiplicity(3, bisection(f, n, n)) # Chai Wah Wu, Apr 09 2026

Crossrefs

Cf. A007949, A025487, A036041, A051282, A122841.

Sequence in context: A115211 A097516 A219052 * A078134 A282380 A083661

Adjacent sequences: A060823 A060824 A060825 * A060827 A060828 A060829

Keyword

nonn

Author

Henry Bottomley, Apr 30 2001

Extensions

Name edited and offset corrected by Amiram Eldar, Dec 29 2020

Status

approved