# [seqfan] Re: d(a(n)) = d(a(n)-a(n-1))

Farideh Firoozbakht f.firoozbakht at sci.ui.ac.ir
Thu May 28 23:16:07 CEST 2009

```Dear Seqfans,

Sorry a few hours ago four empty emails were sent to you. This was due
to problem with our mailing system.

> Does every positive integer occur in A160691?

It seems that no, all the first 200000 terms of the sequence are in the set
{1,2,4,6,8,12}.
In fact for the first 200000 natural numbers n, we have:

1. For one number n, A160691(n)=1 (A160691(1)=1).
2. For 13 numbers n,  A160691(n)=12.
3. For 4785 numbers n, A160691(n)=6.
4. For 6706 numbers n, A160691(n)=8.
5. For 26790 numbers n, A160691(n)=2.
6. For 161705 numbers n, A160691(n)=4.

Also n = 2 is the only number n (less than 200000) that
A160691(n)=A160691(n+1) =A160691(n+2)=2.

For the 53 consecutive numbers 64833, 64834, ... , 64885 we have A160691(n)=4.

There is no number n (up to 200000) such that A160691(n)=A160691(n+1)=
A160691(n+2)=6.

For n in the set {107486, 126785, 160804, 162574} (up to 200000), A160691(n)=
A160691(n+1)=A160691(n+2)=8.

For n in the set {2993, 63254, 63792, 82213, 104728, 130811, 135767, 137994,
149813, 156335, 193041, 193776, 199879}, A160691(n)=12.

What is the first number n such that A160691(n) isn't in the set {1,2,4,6,8,
12}?

The first 100 terms of the sequences A160689, A160690 & A160691
respectively are:

1,2,2,2,8,2,2,8,2,2,8,2,2,8,2,21,5,6,6,15,3,6,8,6,2,10,12,6,12,2,10,22,8,
6,34,6,6,22,8,6,8,2,2,6,8,8,2,6,15,31,6,2,6,8,6,2,2,6,10,2,6,6,15,13,6,2,
6,8,2,8,6,10,6,10,8,8,6,8,6,10,8,2,2,10,2,10,6,2,38,10,6,10,8,10,8,10,2,
22,2,6

1,3,5,7,15,17,19,27,29,31,39,41,43,51,53,74,79,85,91,106,109,115,123,
129,131,141,153,159,171,173,183,205,213,219,253,259,265,287,295,301,
309,311,313,319,327,335,337,343,358,389,395,397,403,411,417,419,421,
427,437,439,445,451,466,479,485,487,493,501,503,511,517,527,533,543,
551,559,565,573,579,589,597,599,601,611,613,623,629,631,669,679,685,
695,703,713,721,731,733,755,757,763

1,2,2,2,4,2,2,4,2,2,4,2,2,4,2,4,2,4,4,4,2,4,4,4,2,4,6,4,6,2,4,4,4,4,4,
4,4,4,4,4,4,2,2,4,4,4,2,4,4,2,4,2,4,4,4,2,2,4,4,2,4,4,4,2,4,2,4,4,2,4,
4,4,4,4,4,4,4,4,4,4,4,2,2,4,2,4,4,2,4,4,4,4,4,4,4,4,2,4,2,4

Regards,

Farideh

Quoting Leroy Quet <q1qq2qqq3qqqq at yahoo.com>:

>
> Consider these three sequences I just submitted:
>
> %I A160689
> %S A160689 1,2,2,2,8,2,2,8,2,2,8,2,2,8,2,21,5
> %N A160689 a(1)=1. a(n) = the smallest positive integer such that
> d(a(n)) = d(sum{k=1 to n} a(k)), where d(m) = the number of divisors
>  of m.
> %C A160689 sum{k=1 to n} a(k) = A160690(n). d(A160689(n)) =
> d(A160690(n)) = A160691(n).
> %Y A160689 A160690,A160691
> %K A160689 more,nonn
> %O A160689 1,2
>
> %I A160690
> %S A160690 1,3,5,7,15,17,19,27,29,31,39,41,43,51,53,74,79
> %N A160690 a(1)=1. a(n) = the smallest integer > a(n-1) such that
>  d(a(n)) = d(a(n)-a(n-1)), where d(m) = the number of divisors of m.
> %C A160690 A160690(n)-A160690(n-1) = A160689(n), for n >= 2.
> d(A160689(n)) = d(A160690(n)) = A160691(n).
> %Y A160690 A160689,A160691
> %K A160690 more,nonn
> %O A160690 1,2
>
> %I A160691
> %S A160691 1,2,2,2,4,2,2,4,2,2,4,2,2,4,2,4,2
> %N A160691 a(n) = the number of divisors of A160689(n) = the number
> of divisors of A160690(n).
> %Y A160691 A160689,A160690
> %K A160691 more,nonn
> %O A160691 1,2
>
> Does every positive integer occur in A160691?
>
> If so, could someone calculate a few of the terms of the sequence
> {b(k)},T where b(n) = the smallest positive integer such that
> A160691(b(n)) = n?
>
> Thanks,
> Leroy Quet
>
>

----------------------------------------------------------------
University of Isfahan (http://www.ui.ac.ir)

```