a propos divisors...

[Peter Pein]

Peter Pein schrieb:
> 
> Although the interst seems to be bounded by a fairly small n (0 until
> now), I would like to ask you if you could please help me.
>
....

> 
> Thank you for your attention and in advance for CPU-time,
> 
> Peter
> 

Well, there are other groups on the net...

THX for ignoring me completely
Peter



Peter,   I at least did not ignore you!   I saved
all your messages to be read later. 

(I just didn't reply yet!)

Neil



>%I A000001
>%S A000001 3, 5, 7, 8, 9, 10, 11, 13, 14, 15, 17, 18, 19, 20, 21, 22,
>23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42,
>43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63
>%N A000001 Those integers for which a smaller positive integer exists
>which has the same number of divisors
>%F A000001 a(n)= n-th element of {x>0, there exists a k with
>0<k<x and the same number of divisors as x)
>%e A000001 a(4)=8 because it is the fourth integer for which a smaller
>integer with the same number of divisors exists (after 3, 5 and 7).
>divisors of 8 are 1,2,4,8 which are four and the divisors of 6 which is
>less than 8 are (1, 2, 3, 6) which are four.
>%t A000001 Clear[tmp];
>Function[n, If[Head[#1] === tmp, #1 = n; Unevaluated[Sequence[]], n] &
>   [tmp[DivisorSigma[0, n]]]] /@ Range[64]
>%Y A000001 Cf. A069822, A131902-A131908
>%O A000001 1
>%K A000001 ,easy,nonn,
>%A A000001 Peter Pein (petsie at dordos.net), Jul 26 2007
>RH
>RA 192.20.225.32
>RU
>RI

Appears to be the complement of A007416.

Tony