# [seqfan] Re: Mini-max numbers

Vladimir Shevelev shevelev at bgu.ac.il
Mon May 13 21:57:43 CEST 2013

```Thank you, Sven, for your question!

I used only tables of primes.
Indeed, my algorithm of search is the following. I take an arbitrary n-digit prime and calculate iterations of Kaprekar-like map which I described and wathed an appearance of a loop. Surprisingly I rather quickly find it!
For example, in case n=4  I took a prime 4021. I got iterations: 4210-0124=4096, P(4096)=4091;  9410-0149=9261,
P(9261)=9277; 9772-2779=6993 , P(6993)=6997; 9976-6799=3177, P(3177)=3181; 8311-1138=7173, P(7173)=7177; 7771-1777=5994, P(5994)=6007; 7600-0067=7533, P(7533)=7537;  7753-3577=4176, P(4176)=4177; 7741-1477=6264, P(6264)=6269; 9662-2669=6993, P(6993)=6997. We see a loop of lendth 7. Using a computer realization of this algorithm, it seems that it is not difficult to find all trajactories for n-digits primes and thus to find all mini-max primes with their cycles.

Best regards,

----- Original Message -----
From: Sven H Simon <sven-h.simon at t-online.de>
Date: Sunday, May 12, 2013 9:18
Subject: [seqfan] Re: Mini-max numbers
To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>

> and you really calculated those values with handy in these
> modern times ?    I just ask, because I used a HP
> pocket calculator to proof a 21digit prime a lot of years ago.
> Using Chinese remainders it was not that difficult. And
> calculated some numbers for Kaprekar with it too.
>
> Best regards
> Sven.
>
> -----Ursprüngliche Nachricht-----
> Von: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] Im Auftrag
> von Vladimir Shevelev
> Gesendet: Sonntag, 12. Mai 2013 16:50
> An: Sequence Fanatics Discussion list
> Betreff: [seqfan] Re: Mini-max numbers
>
> 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,
>
>
> ----- 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,
> >
> >
> >
> > ----- 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
> > >
> > > --
> > > Joseph S. Myers
> > > jsm at polyomino.org.uk
> > >
> > > _______________________________________________
> > >
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> >  Shevelev Vladimir‎
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
>
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