A272618 Irregular array read by rows: n-th row contains (in ascending order) the nondivisors 1 <= k < n such that all the prime divisors p of k also divide n.

0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 8, 0, 8, 9, 0, 4, 8, 9, 0, 0, 4, 8, 12, 16, 0, 8, 16, 9, 4, 8, 16, 0, 9, 16, 18, 0, 4, 8, 16, 0, 8, 16, 0, 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 0, 0, 9, 27, 4, 8, 16, 32, 25, 8, 16, 24, 27, 32, 0, 4, 8, 16, 32, 9, 27, 16, 25, 32, 0, 4, 8, 9, 12, 16, 18, 24, 27, 28, 32

Offset: 1

Author

Michael De Vlieger, May 03 2016

Keywords

• nonn
• tabf

Comments

The k are the "semidivisors" or nondivisor regular numbers of n as counted by A243822(n).

All nonzero terms k are composite and pertain to composite rows n. This is because prime k must either divide or be coprime to n, and k = 1 is both a divisor of and coprime to n.

Row n for prime p contains zero, since numbers 1 <= k < p must either divide or be coprime to prime p.

Row n for prime powers p^e contains zero, since there is only one prime divisor p of p^e and every power 1 <= m <= e of p divides p^e.

Row n = 4 is a special case of composite n that contains zero. This is because 4 is the smallest composite number; there are no composites k < n.

Thus rows n for composite n > 4 contain at least 1 nonzero value.

In base n, 1/a(n) has a terminating expansion with at least 2 places.

Examples

			For n = 12, the numbers 1 <= k < n such that the prime divisors p of k also divide n are {2, 3, 4, 6, 8, 9}; {2, 3, 4, 6} divide n = 12, thus row n = 12 is {8, 9}.
n: k
1: 0
2: 0
3: 0
4: 0
5: 0
6: 4
7: 0
8: 0
9: 0
10: 4 8
11: 0
12: 8 9
13: 0
14: 4 8
15: 9
16: 0
17: 0
18: 4 8 12 16
19: 0
20: 8 16
		

References

• G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, pp. 144-145, Theorem 136.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10814 (rows 1 to 1000, flattened).
• M. De Vlieger, Exploring Number Bases as Tools, ACM Inroads, March 2012, Vol. 3, No. 1, pp. 4-12.
• M. De Vlieger, Neutral Numbers.
• M. De Vlieger, Sequence page.

Crossrefs

Union of A027750 and nonzero terms of a(n) = A162306, thus A000005(n) + A243822(n) = A010846(n).

The union of nonzero terms of a(n) and A272619 = A133995, thus A243822(n) + A243823(n) = A045763(n).

Programs

Mathematica

Table[With[{r = First /@ FactorInteger@ n}, Select[Range@ n,
And[SubsetQ[r, Map[First, FactorInteger@ #]], ! Divisible[n, #]] &]], {n, 30}] /. {} -> 0 // Flatten (* Michael De Vlieger, May 03 2016 *)