[seqfan] Re: Mini-max numbers

Sun May 12 16:50:09 CEST 2013

Consider now Kaprekar-like primes. In this case denote P=P(N) the nearest to N prime >=N. 
If (P(N))_max - (P(N))_min=N_1 and P(N_1)=P, then we call P a mini-max prime of cycle 1, otherwise,
if (P(N_1))_max - (P(N_1))_min=N_2 and P(N_2)=P, then we call P  a mini-max prime of cycle 2, etc.
Without a computer I found for n=2, P=37 of cycle 1; for n=3, P=199 of cycle 2; for n=4, P=6997 of cycle 7; for  n=5, P=61001 of cycle 1; for n=6, P=640859 of cycle 7, for n=7, P=8309869 of cycle 5.
For example, in case N=199, P(N)=199, 991-199=792, P(792)=797, 977-779=198, P(198)=199=P(N).
Formally, for the same n could be more than one mini-max prime. It is interesting to find all of them for a few values of n>=2.

Best regards,
Vladimir

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Saturday, May 11, 2013 3:19
Subject: [seqfan] Re: Mini-max numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Thank you, Joseph, for your theorems and conjectures and for key 
> words "Kaprekar map". Using them, I found paper 
> http://mathworld.wolfram.com/KaprekarRoutine.html and several 
> interesting references there.
>  
> Consider also the following Kaprekar-like map. To every even 
> digit of N we add 1. Denote new number by [N]. 
> Let N_1=[N]_max-[N]_min. If [N_1]=[N], we say that [N] is 
> Kaprekar-like number of cycle 1, otherwise, let N_2 =
> [N_1]_max-[N_1]_min. If [N_2]=[N], we say that [N] is Kaprekar-
> like number of cycle 2, etc.
>  
> By handy, I found the following  Kaprekar-like numbers with 
> odd digits for n=2,3,4,5,6:  37 of cycle 1, 595 of cycle 2, 
> 7175 of cycle 1, 75935 of cycle 4, 539957 of cycle 5.
>  
> Best regards,
> Vladimir
> 
> 
> 
> ----- Original Message -----
> From: "Joseph S. Myers" <jsm at polyomino.org.uk>
> Date: Friday, May 10, 2013 10:16
> Subject: [seqfan] Re: Mini-max numbers
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> 
> > On Fri, 10 May 2013, Vladimir Shevelev wrote:
> > 
> > > Let, for n>=3, N be n-digit number such that the first zeros 
> > are allowed 
> > > but not all digits are the same. Let consider digits of N in 
> > > non-increasing order and the corresponding number denote by 
> > N_max; after 
> > > that we consider digits of N in non-decreasing order and the 
> > > corresponding number denote by N_min. If N_max-N_min=N, then 
> > we call N a 
> > > mini-max number of cycle 1; if N_max-N_min=N_1 which differs 
> > of N, but 
> > > (N_1)_max-(N_1)_min=N, then we call N a mini-max number of 
> > cycle 2, etc. 
> > > By handy, I found the following numbers:
> > > If n=3, N=495 is a mini-max number of cycle 1. Indeed, 954-
> 459=495;> > If n=4, N=6174 is a mini-max number of cycle 1. 
> Indeed, 7641-
> > 1467=6174;> If n=5, N=53955 is a mini-max number of cycle 2, 
> > since 
> > > N_1=95553-35559=59994 and N_2=99954-45999=53955=N;
> > > If n=6, N=840852 is a mini-max number of cycle 7, since 
> > > N_1=885420-024588=860832;  N_2=886320-023688=862632;
> > > N_3=866322-223668=642654; N_4=665442-244566=420876; 
> > > N_5=876420-024678=851742; N_6=875421-124578=750843; 
> > > N_7=875430-034578=840852=N.
> > > It is interesting to find for every n the set of mini-max 
> > numbers with 
> > > their cycles. Are mini-max numbers known, maybe, under 
> another name?
> > 
> > Look up sequences related to the "Kaprekar map".  (That's 
> > slightly 
> > different in that the number of digits goes down if the 
> > difference would 
> > have a leading zero when interpreted as an n-digit 
> number.  
> > But you don't 
> > get anything interesting by allowing leading zeros.)
> > 
> > These were discussed a bit on seqfan in August 2009, but then 
> > the 
> > discussion moved off-list.  The following are 
> observations 
> > of mine from 
> > that time:
> > 
> > Theorems:
> > 
> > * In base 2, all cycles are fixed points and a complete 
> classification> of fixed points is easy; they are numbers of the 
> form 2^n - 2^(n-
> > k) -
> > 2^k + 1 for 1 < k < n-1 (note that k and n-k yield the 
> > same fixed
> > point, which results from any n-digit number with k or n-k 1s 
> in it).
> > 
> > * In any base b, there are finitely many "types" of c-cycle 
> for any
> > fixed constant c: at most O((3 * 2^(b+1))^c) such types if I've
> > calculated correctly.  A type is given by O(bc) linear 
> > equations and
> > inequalities in 1+ceil(b/2) variables, whose coefficients are 
> O(2^c);> to go from types to the number of n-digit c-cycles you 
> need to count
> > (as a function of n probably given by a generating function) the
> > number of lattice points with each n-coordinate inside a possibly
> > empty convex body described by those equations and 
> inequalities (which
> > does always give a rational generating function; the question 
> is just
> > the complexity of calculating it).  This method would 
> > actually find
> > and count the number of n-digit numbers that are in a cycle of 
> length> dividing c, so to count c-cycles the numbers in shorter 
> cycles need
> > removing and the result needs dividing by c.  Certainly 
> > this seems
> > infeasible for determining whether there are 6-cycles in base 10;
> > since the calculations needed for each case are quite complicated,
> > even using it for 4-cycles in base 4 seems of borderline 
> feasibility.> 
> > * In any even base b > 2, consider the set of n-digit numbers 
> > that are
> > members of cycles and satisfy "number of digits (b-1) - number of
> > digits 0 = o(n)".  Almost all numbers in that set are in 
> a 
> > cycle of
> > length exactly A003558(b/2 - 1), and so almost all cycles containing
> > at least one number with that property have that length.
> > 
> > Conjectures:
> > 
> > * In any odd base, there are arbitrarily long cycles.  A 
> stronger> conjecture would be that all cycle lengths are present.
> > 
> > * In any even base, there are only finitely many cycle lengths and
> > there exists an algorithm to determine them (or equivalently to
> > produce an upper bound on cycle length).
> > 
> > * In any even base b > 2, almost all n-digit numbers in cycles 
> > are in
> > a cycle of length exactly A003558(n/2 - 1) (i.e., the 
> condition above
> > about not having many more b-1 digits than 0 digits is not 
> necessary)> and so almost all cycles of n-digit numbers have 
> that length.
> > 
> > I hope there are much better ways of classifying and counting fixed
> > points in any base than considering O(2^b) cases, but don't 
> have a
> > specific conjecture about this.
> > 
> > -- 
> > Joseph S. Myers
> > jsm at polyomino.org.uk
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
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> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎