Composite Integer Sequences consisting of multiple interleaved primitive sequences

[Max Alekseyev]

On 9/22/07, Alexander Povolotsky <apovolot at gmail.com> wrote:

> Below is my (unpublished so far in OEIS) analysis of cases for "failed"
> "n"'s for the following description of my sequence:
> a(n) = (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^k + k)/k!

[...]

> i=||  {n }Failures
> ------------------------------------------------------
> 14|| {2,9,16,23,30,37,51,58,65,72,79,86,100,...}
> 15|| {2,9,16,23,30,37,51,58,65,72,79,86,100,...}
> 16|| {2,9,16,23,30,37,51,58,65,72,79,86,100,...}
> 22|| {7,18,29,40,51,62,73,84,95,...}
> 23|| {7,18,29,40,51,62,73,84,95,...}
> 24|| {7,18,29,40,51,62,73,84,95,...}
> 25||
> {2,7,12,18,22,27,29,37,40,47,51,52,62,72,73,77,84,87,95,97,...}
> 26|| {7,11,18,24,29,37,40,50,51,62,73,76,84,89,95,...}
>
> My (AP) analysis of above computational results comes to the following
> observations:
>
> >Failures
> >i =
> >14 {2,9,16,23,30,37,51,58,65,72,79,86,100,...}
> >15  ....
> >16  ....
>
> Above sequence of "n" failures could be described as:
> n = 7k + 2;  for k=0,1,3,...,12, !!14, (16?)

To check whether 14! divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 +
14) it is enough to check whether every prime power q from prime
factorization of 14! divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 +
14) for n=0,1,...,q-1.

Note that 14! = 2^11 * 3^5 * 5^2 * 7^2 * 11 * 13.

It is easy to verify that:
i) 2^11 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for
every n=0,1,...,2^11-1;
ii) 3^5 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for
every n=0,1,...,3^5-1;
iii) 7^2 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for
every n=0,1,...,7^2-1, except for n=2, 9, 16, 23, 30, 37 (i.e., n of
the form 7m+2 but not 49m+44);
iv) 11 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every
n=0,1,...,11-1;
v) 13 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every
n=0,1,...,13-1.
This proves that for k=14, a(n) is not integer only for n in the set
difference { 7m+2} \ { 49m+44 } .
For k=15 and k=16, proof is similar.

> Note that Multiplier K for above is 7
>
> > ...
> >22 {7,18,29,40,51,62,73,84,95,...}
> >23
> >24
>
> n = 11k + 7;  k=0,1,...,?

22! = 2^19 * 3^9 * 5^4 * 7^3 * 11^2 * 13 * 17 * 19

Here we find that all prime powers divide (n^1 + 1)*(n^2 + 2)*(n^3 +
3)*...*(n^22 + 22), except for 11^2.
11^2 does not divide (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^22 + 22) for
n = 7, 18, 29, 40, 51, 62, 73, 84, 95, 106, i.e., of the form 11m+7
but not 121m+117.

> >25
> {2,7,12,18,22,27,29,37,40,47,51,52,62,72,73,77,84,87,95,97,...}

25! = 2^22 * 3^10 * 5^6 * 7^3 * 11^2 * 13 * 17 * 19 * 23

There is no divisibility for 5^6 and n in { 50m+2, 50m+12, 50m+22,
50m+27, 50m+37, 50m+47 } and for 11^2 and n in {11m+7} \ {121m+117}.
Therefore, a(n) is not integer for n in {50m+2} U {50m+12} U {50m+22}
U {50m+27} U {50m+37} U {50m+47} U ( {11m+7} \ {121m+117} ).

> >26 {7,11,18,24,29,37,40,50,51,62,73,76,84,89,95,...}

26! = 2^23 * 3^10 * 5^6 * 7^3 * 11^2 * 13^2 * 17 * 19 * 23

There is no divisibility for 11^2 and n in {11m+7} \ {121m+117} and
for 13^2 and n in {13m+11} \ {169m+63}. Therefore, a(n) is not integer
for n in the union ( {11m+7} \ {121m+117} ) U ( {13m+11} \ {169m+63}
).

Regards,
Max



Let an "isolated divisor" of n be a positive divisor, k, of n where 
neither (k-1) nor (k+1) divides n.

Let a "non-isolated divisor" of n be a positive divisor, k, of n where 
either (k-1) and/or (k+1) divides n. In other words, a non-isolated 
divisor of n is a positive integer, k, where k(k-1) and/or k(k+1) divides 
n.

For example, the positive divisors of 20 are 1,2,4,5,10,20. Of these, 1 
and 2 are adjacent, and 4 and 5 are adjacent. So the isolated divisors of 
20 are 10 and 20. While the non-isolated divisors of 20 are 1,2,4,5.

(And of course, every positive divisor of n is either isolated or 
non-isolated and is not both.)

I have submitted several sequences regarding these divisors. But that 
brings me to my question. I have a hard time believing that this concept 
is original. (I have come up with the terms "isolated divisor" and 
"non-isolated divisor".) What I wonder is, are these types of divisors 
already named something else, and I have made a mistake by submitting a 
number of sequences with my terminology?
I have tried to find alternative terms for these divisors on-line, but so 
far have had no luck.

Thanks,
Leroy Quet