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%I #17 Mar 06 2024 04:47:33
%S 0,1,1,1,1,2,1,2,1,2,1,3,1,2,2,2,1,3,1,2,2,2,1,4,1,2,2,2,1,4,1,3,2,2,
%T 2,3,1,2,2,4,1,3,1,2,3,2,1,5,1,3,2,2,1,4,2,4,2,2,1,6,1,2,3,3,2,3,1,2,
%U 2,4,1,5,1,2,3,2,2,3,1,5,2,2,1,5,2,2,2,3,1,5,2,2,2,2,2,6,1,3,2,3,1,3,1,3,4
%N Consider the products of all pairs of consecutive (when ordered by size) positive divisors of n. a(n) = the number of these products that divide n. a(n) also = the number of the products that are divisible by n.
%C Least number k whose value, a(k)=j beginning with j=0, is: 1, 2, 6, 12, 24, 48, 60, 168, 336, 240, 360, 672, 840, 720, 1512, 1680, 1440, 4320, 2520, 4200, 5040, 6720, 7560, 12480, 13440, 15840, ..., . - _Robert G. Wilson v_, May 30 2008
%H Antti Karttunen, <a href="/A140774/b140774.txt">Table of n, a(n) for n = 1..10000</a>
%e The divisors of 20 are 1,2,4,5,10,20. There are 2 pairs of consecutive divisors whose product divides 20: 1*2=2, 4*5 = 20. Likewise, there are 2 such products that are divisible by 20: 4*5=20, 10*20=200. So a(20) = 2.
%t f[n_] := Block[{d = Divisors@ n}, Count[n/((Most@d) (Rest@d)), _Integer]]; Array[f, 105] (* _Robert G. Wilson v_, May 30 2008 *)
%o (PARI)
%o \\ Two implementations, after the two different interpretations given by the author of the sequence:
%o A140774v1(n) = { my(ds = divisors(n),s=0); if(1==n,0,for(i=1,(#ds)-1,if(!(n%(ds[i]*ds[1+i])),s=s+1))); s; }
%o A140774v2(n) = { my(ds = divisors(n),s=0); if(1==n,0,for(i=1,(#ds)-1,if(!((ds[i]*ds[1+i])%n),s=s+1))); s; }
%o \\ _Antti Karttunen_, May 19 2017
%Y Cf. A140773.
%Y Differs from A099042 for the first time at n=32, where a(32) = 3, while A099042(32) = 2.
%K nonn
%O 1,6
%A _Leroy Quet_, May 29 2008
%E More terms from _Robert G. Wilson v_, May 30 2008
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