# [seqfan] Re: degree of integer

Reinhard Zumkeller reinhard.zumkeller at gmail.com
Sun Jun 26 16:06:53 CEST 2011

```concerning arithmetic derivatives (A003315):
insert 18 into 4,8,12,15,16,20,24,26,27,28,32,35,36,39,40,44,45,48,50,51, etc.
and you will get http://oeis.org/A099308, which is complement of
4,8,12,15,16,20,24,26,27,28,32,35,36,39,40,44,45,48,50,51, etc. ==>
http://oeis.org/A099309
In the comment of http://oeis.org/A099307 you might find the desired criterion.
Best
Reinhard

2011/6/25 Vladimir Shevelev <shevelev at bgu.ac.il>
>
> Degree k(P) of a polynomial P(x) one can define by the number m(P) of iterations of derivative up to receiving 0 over the formula k(P)=m(P)-1+delta(P,0), where delta(P,0)=1, if P=0, and 0, otherwise.
> Analogously, using A038554, one can introduce degree a(n) of integer n over the formula a(n)=t(n)-1+delta(n,0), where t(n) is the number of iterations of A038554 up to receiving 0 , delta is the Kronecker symbol. Since, for n>=1, A038554(n)<n, then t(n) always exists. So, we find  a(0)=0,a(1)=0,a(2)=1, a(3)=0, a(4)=2, a(5)=1, a(6)=1, a(7)=1, a(8)=3, etc.
>    In case of derivative A003415, not all integers have a finite degree, and we can consider two sequences: 1) of numbers having A003415-finite degree ("polynomial-like numbers") and 2) of numbers having no A003415-finite degree. But here there is a problem: when a number has no A003415-finite degree? I do not know a criterion, although it seems that the hypothetical sequence of such numbers begins with 4,8,12,15,16,20,24,26,27,28,32,35,36,39,40,44,45,48,50,51, etc.
>   On the other hand, since  the degrees of  the consecutive derivatives of polynomials monotonocally decrease, then  it is natural to define a polynomial-like number as a number which has monotonocally decreasing A003415-iterations. The sequence of such numbers begins with 1,2,3,5,6,7,9,10,11,13,14,17,19,21,22,23,25,29,31,33,34,37,38,41,42,43,46,47,49,53,...
>
> Regards,