A289816 The second of a pair of coprime numbers whose factorizations depend on the ternary representation of n (See Comments for precise definition).

1, 1, 2, 1, 1, 2, 3, 3, 6, 1, 1, 2, 1, 1, 2, 3, 3, 6, 4, 5, 10, 4, 5, 10, 12, 15, 30, 1, 1, 2, 1, 1, 2, 3, 3, 6, 1, 1, 2, 1, 1, 2, 3, 3, 6, 4, 5, 10, 4, 5, 10, 12, 15, 30, 5, 7, 14, 5, 7, 14, 15, 21, 42, 5, 7, 14, 5, 7, 14, 15, 21, 42, 20, 35, 70, 20, 35, 70

Offset: 0

Author

Rémy Sigrist, Jul 12 2017

Keywords

• nonn
• base
• look

Comments

For n >= 0, with ternary representation Sum_{i=1..k} t_i * 3^e_i (all t_i in {1, 2} and all e_i distinct and in increasing order):

- let S(0) = A000961 \ { 1 },

- and S(i) = S(i-1) \ { p^(f + j), with p^f = the (e_i+1)-th term of S(i-1) and j > 0 } for any i=1..k,

- then a(n) = Product_{i=1..k such that t_i=2} "the (e_i+1)-th term of S(k)".

See A289815 for the first coprime number and additional comments.

The number of distinct prime factors of a(n) equals the number of twos in the ternary representation of n.

Examples

			For n=42:
- 42 = 2*3^1 + 1*3^2 + 1*3^3,
- S(0) = { 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, ... },
- S(1) = S(0) \ { 3^(1+j) with j > 0 }
       = { 2, 3, 4, 5, 7, 8,    11, 13, 16, 17, 19, 23, 25,     29, ... },
- S(2) = S(1) \ { 2^(2+j) with j > 0 }
       = { 2, 3, 4, 5, 7,       11, 13,     17, 19, 23, 25,     29, ... },
- S(3) = S(2) \ { 5^(1+j) with j > 0 }
       = { 2, 3, 4, 5, 7,       11, 13,     17, 19, 23,         29, ... },
- a(42) = 3.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000961, A004488, A289815.

Programs

PARI

a(n) = my (v=1, x=1);                   \
       for (o=2, oo,                           \
           if (n==0, return (v));              \
           if (gcd(x,o)==1 && omega(o)==1,     \
               if (n % 3,    x *= o);          \
               if (n % 3==2, v *= o);          \
               n \= 3;                         \
           );                                  \
       );

Python

from sympy import gcd, primefactors
def omega(n): return 0 if n==1 else len(primefactors(n))
def a(n):
    v, x, o = 1, 1, 2
    while True:
        if n==0: return v
        if gcd(x, o)==1 and omega(o)==1:
            if n%3: x*=o
            if n%3==2:v*=o
            n //= 3
        o+=1
print([a(n) for n in range(101)]) # Indranil Ghosh, Aug 02 2017

Formula

a(n) = A289815(A004488(n)) for any n >= 0.

a(A005836(n)) = 1 for any n > 0.

a(2 * A005836(n)) = A289272(n-1) for any n > 0.