# A107736 is duplicate of A067824 ?

Max relf at unn.ac.ru
Mon Jun 13 06:07:48 CEST 2005

```> Ralf Stephan wrote:

>> %S A000001 1,2,2,4,2,6,2,8,4,6,2,16,2,6,6,16,2,16,2,16,
>> %N A000001 Number of polynomials p with coefficients in {0,1} that
>> divide x^n-1 and such that (x^n-1)/{p(x-1)} has all coefficients in
>> {0,1}.

This sequence has been added as A107736 to OEIS.
I've computed more terms:

> 1 2 2 4 2 6 2 8 4 6 2 16 2 6 6 16 2 16 2 16 6 6 2 40 4 6 8 16 2 26 2 32
> 6 6 6 52 2 6 6 40 2 26 2 16 16 6 2 96 4 16 6 16 2 40 6 40 6 6 2 88 2 6
> 16 64 6 26 2 16 6 26 2 152 2 6 16 16 6 26 2 96 16 6 2 88 6 6 6 40 2 88 6
> 16 6 6 6 224 2 16 16 52

And it looks like this sequence coincide with A067824.
I've tested their equality for n=1..300.

%I A067824
%S A067824 1,2,2,4,2,6,2,8,4,6,2,16,2,6,6,16,2,16,2,16,6,6,2,40,4,6,8,16,2,26,2,
%T A067824 32,6,6,6,52,2,6,6,40,2,26,2,16,16,6,2,96,4,16,6,16,2,40,6,40,6,6,2,88,
%U A067824 2,6,16,64,6,26,2,16,6,26,2,152,2,6,16,16
%N A067824 a(n) = if n=1 then 1 else 1 + sum{a(d): 0<d<n and d divides n }.
%e A067824 a(12) = 1 + a(6) + a(4) + a(3) + a(2) + a(1)
%e A067824 = 1+[1+a(3)+a(2)+a(1)]+[1+a(2)+a(1)]+[1+a(1)]+[1+a(1)]+[1]
%e A067824 = 1+[1+(1+a(1))+(1+a(1))+1]+[1+(1+a(1))+1]+[1+1]+[1+1]+[1]
%e A067824 = 1+[1+(1+1)+(1+1)+1]+[1+(1+1)+1]+[1+1]+[1+1]+[1]
%e A067824 = 1 + 6 + 4 + 2 + 2 + 1 = 16.
%Y A067824 Cf. A000005.
%Y A067824 Sequence in context: A083260 A046523 A071364 this_sequence A046801 A090624 A099735
%Y A067824 Adjacent sequences: A067821 A067822 A067823 this_sequence A067825 A067826 A067827
%K A067824 nonn
%O A067824 1,2
%A A067824 Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Feb 08 2002

I believe there exists a simple proof for that but I don't see it. Do you?

Max

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