[seqfan] degree of integer
Vladimir Shevelev
shevelev at bgu.ac.il
Sat Jun 25 13:49:40 CEST 2011
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,
Vladimir
Shevelev Vladimir
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