A285110 a(n) = A001222(A285323(n)).

0, 2, 2, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1

Offset

0,2

Comments

The sequence is completely determined by the positions of two least significant 1-bits of n: After initial zero, if n is a power of two (only one 1-bit present) or if prime(1+A285099(n)) > prime(1+A007814(n))^2, a(n) = 2, otherwise a(n) = 1.

Links

Antti Karttunen, Table of n, a(n) for n = 0..8192

Formula

a(n) = A001222(A285323(n)).

Prog

(Scheme)
(define (A285110 n) (A001222 (A285323 n)))
(define (A285110 n) (cond ((zero? n) n) ((or (= 1 (A000120 n)) (> (A000040 (+ 1 (A285099 n))) (A000290 (A000040 (+ 1 (A007814 n)))))) 2) (else 1)))
(Python)
from operator import mul
from sympy import prime, primefactors
from functools import reduce
def a001222(n): return 0 if n<2 else a001222(n//min(primefactors(n))) + 1
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n += r
    while a007947(n)!=r:
        n+=r
    return n
def a285323(n): return a065642(a065642(a019565(n)))//a019565(n)
def a(n): return a001222(a285323(n))
print([a(n) for n in range(121)]) # Indranil Ghosh, Apr 20 2017

Crossrefs

Cf. A000040, A000290, A001222, A007814, A285099, A285323, A285324.

Sequence in context: A131837 A232628 A087889 * A323018 A192064 A225182

Adjacent sequences: A285107 A285108 A285109 * A285111 A285112 A285113

Keyword

nonn

Author

Antti Karttunen, Apr 19 2017

Status

approved