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%I #34 Mar 20 2021 11:04:17
%S 0,0,0,0,0,4,0,6,6,4,6,8,0,8,9,10,0,4,6,8,10,12,6,9,10,12,6,10,12,14,
%T 0,4,8,10,12,14,15,16,0,6,8,12,14,15,16,18,6,9,12,14,15,18,4,6,8,10,
%U 12,14,16,18,20,0,9,10,14,15,16,18,20,21,22,10,15,20,4,6,8,10,12,14,16,18,20,22,24,6,12,15,18,21,24
%N Irregular array read by rows: n-th row contains (in numerical order) the positive integers <= n which are neither divisors of n nor are coprime to n. A 0 is put into row n if there are no such integers.
%C Row n has length A264441(n).
%C The number of nonzero entries in row n is A045763(n).
%C Row n has a 0 if every positive integer <= n is coprime to n or divides n.
%C From _Michael De Vlieger_, Aug 19 2017: (Start)
%C When row n is not empty (and here represented by 0), the terms of row n are composite, since primes p < n must either divide or be coprime to n and the empty product 1 both divides and is coprime to all numbers. For the following, let prime p divide n and prime q be coprime to n.
%C Row n is empty for n < 8 except n = 6.
%C There are two distinct species of term m of row n. The first are nondivisor regular numbers g in A272618(n) that divide some integer power e > 1 of n. In other words, these numbers are products of primes p that also divide n and no primes q that are coprime to n, yet g itself does not divide n. Prime powers n = p^k cannot have numbers g in A272618(n) since they have only one distinct prime divisor p; all regular numbers g = p^e with 0 <= e <= k divide p^k. The smallest n = 6 for which there is a number in A272618. The number 4 is the smallest composite and is equal to n = 4 thus must divide it; 4 is coprime to 5. The number 4 is neither coprime to nor a divisor of 6.
%C The second are numbers h in A272619(n) that are products of at least one prime p that divides n and one prime q that is coprime to n.
%C The smallest n = 8 for which there is a number in A272619 is 8; the number 6 is the product of the smallest two distinct primes. 6 divides 6 and is coprime to 7. The number 6 is neither coprime to nor a divisor of the prime power 8; 4 divides 8 and does not appear in a(8).
%C There can be no other species since primes p <= n divide n and q < n are coprime to n, and products of primes q exclusive of any p are coprime to n.
%C As a consequence of these two species, rows 1 <= n <= 5 and n = 7 are empty and thus have 0 in row n.
%C (End)
%H Robert Israel, <a href="/A133995/b133995.txt">Table of n, a(n) for n = 1..10014</a> (rows 1 to 237, flattened)
%F a(n) = union(A272618(n), A272619(n)). - _Michael De Vlieger_, Aug 19 2017
%e The divisors of 12 are: 1,2,3,4,6,12. The positive integers which are <= 12 and are coprime to 12 are: 1,5,7,11. So row 12 contains the positive integers <= 12 which are in neither of these two lists: 8,9,10.
%e The irregular triangle T(n, k) begins:
%e n\k 1 2 3 4 5 6 7 ...
%e 1: 0
%e 2: 0
%e 3: 0
%e 4: 0
%e 5: 0
%e 6: 4
%e 7: 0
%e 8: 6
%e 9: 6
%e 10: 4 6 8
%e 11: 0
%e 12: 8 9 10
%e 13: 0
%e 14: 4 6 8 10 12
%e 15: 6 9 10 12
%e 16: 6 10 12 14
%e 17: 0
%e 18: 4 8 10 12 14 15 16
%e 19: 0
%e 20: 6 8 12 14 15 16 18
%e ... formatted by _Wolfdieter Lang_, Jan 16 2016
%p row:= proc(n) local r;
%p r:= remove(t -> member(igcd(t, n), [1, t]), [$1..n]):
%p if r = [] then 0 else op(r) fi
%p end proc:
%p A:= [seq](row(n), n=1..30); # _Robert Israel_, Jan 19 2016
%t Table[Select[Range@ n, Nor[Divisible[n, #], CoprimeQ[n, #]] &] /. {} -> {0}, {n, 27}] // Flatten (* _Michael De Vlieger_, Aug 19 2017 *)
%Y Cf. A045763, A133994, A264441, A272618, A272619.
%K nonn,tabf
%O 1,6
%A _Leroy Quet_, Oct 01 2007
%E More terms from _Alvin Hoover Belt_, Jan 21 2008
%E Edited by _Wolfdieter Lang_, Jan 16 2016
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