A140774 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.

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, 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, 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

Offset: 1

Author

Leroy Quet, May 29 2008

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A140773.

Differs from A099042 for the first time at n=32, where a(32) = 3, while A099042(32) = 2.

Programs

Mathematica

f[n_] := Block[{d = Divisors@ n}, Count[n/((Most@d) (Rest@d)), Integer]]; Array[f, 105] (* _Robert G. Wilson v, May 30 2008 *)

PARI

\\ Two implementations, after the two different interpretations given by the author of the sequence:
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; }
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; }
\\ Antti Karttunen, May 19 2017

Extensions

More terms from Robert G. Wilson v, May 30 2008