A289815 The first of a pair of coprime numbers whose factorizations depend on the ternary representation of n (see Comments for precise definition).

1, 2, 1, 3, 6, 3, 1, 2, 1, 4, 10, 5, 12, 30, 15, 4, 10, 5, 1, 2, 1, 3, 6, 3, 1, 2, 1, 5, 14, 7, 15, 42, 21, 5, 14, 7, 20, 70, 35, 60, 210, 105, 20, 70, 35, 5, 14, 7, 15, 42, 21, 5, 14, 7, 1, 2, 1, 3, 6, 3, 1, 2, 1, 4, 10, 5, 12, 30, 15, 4, 10, 5, 1, 2, 1, 3, 6

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=1} "the (e_i+1)-th term of S(k)".

See A289816 for the second coprime number.

See A289838 for the product of this sequence with A289816.

By design, gcd(a(n), A289816(n)) = 1.

Also, the number of distinct prime factors of a(n) equals the number of ones in the ternary representation of n.

We also have a(n) = A289816(A004488(n)) for any n >= 0.

For each pair of coprime numbers, say x and y, there is a unique index, say n, such that a(n) = x and A289816(n) = y; in fact, n = A289905(x,y).

This sequence combines features of A289813 and A289272.

The scatterplot of the first terms of this sequence vs A289816 (both with logarithmic scaling) looks like a triangular cristal.

For any t > 0: we can adapt the algorithm used here and in A289816 in order to uniquely enumerate every tuple of t mutually coprime numbers (see Links section for corresponding program).

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) = 4 * 5 = 20.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Rémy Sigrist, Scatterplot of the first 10000 terms of A289815 vs A289816 (both with logarithmic scaling)
• Rémy Sigrist, PARI program to uniquely enumerate tuples of mutually coprime numbers

Crossrefs

Cf. A000961, A004488, A289272, A289813, A289816, A289838, A289905.

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==1, 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==1:v*=o
            n //= 3
        o+=1
print([a(n) for n in range(101)]) # Indranil Ghosh, Aug 02 2017

Formula

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

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