[seqfan] Re: A248034 and variants (Was: Digit-counters updating themselves)

Mon Oct 27 23:40:37 CET 2014

On Mon, Oct 27, 2014 at 10:57 PM, Antti Karttunen <antti.karttunen at gmail.com
> wrote:

>
> Dear SeqFans,
>
> Taking Angelini's http://oeis.org/A248034 and Heinz's variant
> http://oeis.org/A249009
> as my starting points, I created some further variants of the same general
> theme:
>
> "Divide natural numbers to their constituent elements by some means, and
> count the number of times such an element selected with some criteria from
> a(n-1) occurs among the terms computed so far, up to and including a(n-1),
> and let that count be the value of a(n)".
>
> With A248034 the "elements" are digits 0-9 in base-10 representation of
> natural numbers, and the element selected from a(n-1) is the least
> significant digit. With A249009 it is otherwise same, but we count the
> occurrences of the most significant digit of a(n-1).
>
> For the latter, an obvious change is to use a factorial base, instead of
> any fixed base, and we get:
>
> A249069 a(n+1) gives the number of occurrences of the first digit of a(n)
> in factorial base (i.e., A099563(a(n))) so far amongst the factorial base
> representations of all the terms up to and including a(n), with a(0)=0.
>
> another variant is to count the occurrences of maximal digit in factorial
> expansion, as in:
>
> A249070 a(n+1) gives the number of occurrences of the maximum digit of
> a(n) in factorial base (i.e., A246359(a(n))) so far amongst the factorial
> base representations of all the terms up to and including a(n), with a(0)=0.
> (which has a slightly more interesting looking graph than the previous
> one).
>
> Another possibility to avoid being limited to a finite number of elements
> which to count is to use the runlengths of the binary expansion, as in:
>
> A249144 a(0) = 0, after which a(n) gives the total number of runs of the
> same length as the rightmost run in the binary representation of a(n-1)
> [i.e., A136480(a(n-1))] among the binary expansions of all previous terms,
> including the runs in a(n-1) itself.
>
> and
>
> A249146    a(0) = 0, after which a(n) gives the total number of runs of
> the same length as the maximal run in the binary representation of a(n-1)
> [i.e., A043276(a(n-1))] among the binary expansions of all previous terms,
> including the runs in a(n-1) itself.
>
> (Note the analogy with A249070).
>
>
> Then of course, we can always count the prime factors, as in:
>
> A249148    a(1) = 1, after which, if a(n-1) = 1, a(n) = 1 + the total
> number of 1's that have occurred in the sequence so far, otherwise a(n) =
> the total number of times the least prime dividing a(n-1) [i.e.,
> A020639(a(n-1))] occurs as a divisor (counted with multiplicity for each
> term) in the previous terms from a(1) up to and including a(n-1).
>
> However, none of these have such a nice graph as Angelini's original
> A248034, except very similar base-8 version A249068.
>
>
> From A249148, I proceeded to
>
> A249336 a(1) = 1; for n>1, a(n) = number of values k in range 1 .. n-1
> such that {sum of prime indices in the prime factorization of a(k)} = {sum
> of prime indices in the prime factorization of a(n-1)}, both counted with
> multiplicity.
>
> (and its minor variant A249337), where the theme is simpler now, to just
> set a(n) as the count the number of terms a(k) from k = 1 to n-1 for which
> f(a(k)) = f(a(n-1)), for some function f. In the above case that function
> is A056239, which involves prime factorization.
>
>
> But the graph is more interesting now:
> http://oeis.org/A249336/graph
> although I swear I have seen many similar ones already in OEIS, as the
> theme certainly is old one.
>

If I use A000010 (phi) as the function f in the above "sequence pattern"
(instead of A056239), and start with a(1) = 1, I will get the following
sequence (first 512 terms):

1, 1, 2, 3, 1, 4, 2, 5, 1, 6, 3, 4, 5, 2, 7, 1, 8, 3, 6, 7, 2, 9, 3, 8, 4,
9, 4, 10, 5, 6, 11, 1, 10, 7, 5, 8, 9, 6, 12, 10, 11, 2, 11, 3, 13, 1, 12,
12, 13, 2, 13, 3, 14, 7, 8, 14, 9, 10, 15, 1, 14, 11, 4, 15, 2, 15, 3, 16,
4, 17, 1, 16, 5, 16, 6, 18, 12, 17, 2, 17, 3, 19, 1, 18, 13, 4, 20, 7, 14,
15, 8, 18, 16, 9, 17, 4, 21, 5, 19, 2, 19, 3, 22, 5, 20, 10, 21, 6, 23, 1,
20, 11, 6, 24, 12, 22, 7, 18, 19, 4, 25, 1, 21, 7, 20, 13, 8, 23, 2, 22, 8,
24, 14, 21, 9, 22, 9, 23, 3, 26, 10, 25, 2, 23, 4, 27, 5, 26, 11, 10, 27,
6, 28, 12, 28, 13, 14, 24, 15, 16, 17, 5, 29, 1, 24, 18, 25, 3, 29, 2, 25,
4, 30, 19, 7, 26, 15, 20, 21, 16, 22, 11, 12, 30, 23, 5, 31, 1, 26, 17, 6,
31, 2, 27, 8, 32, 7, 27, 9, 28, 18, 29, 3, 32, 8, 33, 5, 34, 9, 30, 24, 25,
6, 33, 7, 31, 3, 34, 10, 35, 1, 28, 19, 10, 36, 20, 26, 21, 22, 13, 23, 6,
35, 2, 29, 4, 36, 24, 27, 11, 14, 32, 11, 15, 28, 25, 8, 37, 1, 30, 29, 5,
38, 12, 39, 3, 37, 2, 31, 4, 38, 13, 26, 27, 14, 33, 9, 34, 12, 40, 13, 28,
29, 6, 39, 4, 40, 14, 35, 5, 41, 1, 32, 15, 30, 31, 5, 42, 30, 32, 16, 33,
10, 43, 1, 33, 11, 16, 34, 17, 18, 36, 31, 6, 41, 2, 34, 19, 15, 35, 6, 42,
32, 20, 36, 33, 12, 44, 13, 34, 21, 35, 7, 37, 3, 43, 2, 35, 8, 45, 9, 38,
16, 37, 4, 44, 14, 39, 10, 46, 7, 40, 22, 17, 23, 8, 47, 1, 36, 36, 37, 5,
48, 24, 38, 17, 25, 15, 39, 11, 18, 41, 3, 45, 12, 49, 3, 46, 9, 42, 38,
18, 43, 4, 47, 2, 37, 6, 48, 26, 39, 13, 40, 27, 19, 20, 40, 28, 41, 4, 49,
5, 50, 16, 41, 5, 51, 1, 38, 21, 42, 43, 6, 50, 17, 29, 7, 44, 18, 45, 14,
46, 10, 52, 15, 42, 44, 19, 22, 19, 23, 11, 20, 43, 7, 47, 3, 51, 2, 39,
16, 44, 20, 45, 17, 30, 46, 12, 53, 1, 40, 31, 7, 48, 32, 33, 21, 45, 18,
49, 8, 54, 24, 47, 4, 52, 19, 25, 22, 21, 46, 13, 47, 5, 55, 6, 53, 2, 41,
7, 50, 23, 14, 51, 3, 54, 26, 48, 34, 35, 20, 48, 36, 49, 9, 52, 21, 50,
24, 49, 10, 56, 22, 22, 23, 15, 50, 25, 26, 51, 4, 55, 8, 57, 7, 53, 3, 56,
23, 16, 51, 5

which is not yet in OEIS. Should it be, or is just more of the same noise?
(Note how 1's can occur here also after nonprimes, like e.g. after 25).

Antti

>
> Can somebody explain the different slopes of the "streamers" that cross
> each other in that graph?
> There seems to be roughly two different classes of them.
>
>
> Best,
>
> Antti
>
>
>
> On Mon, Oct 20, 2014 at 4:39 AM, <seqfan-request at list.seqfan.eu> wrote:
> >
> >
> > Message: 6
> > Date: Sat, 18 Oct 2014 17:00:39 -0400
> > From: Neil Sloane <njasloane at gmail.com>
> > To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > Subject: [seqfan] Re: Digit-counters updating themselves
> > Message-ID:
> >         <CAAOnSgSrRwqTNSZYs6PhYo_eCLPjTtDU=bx5oSX1piUYUHN=
> iA at mail.gmail.com>
> > Content-Type: text/plain; charset=UTF-8
> >
> > A week ago Eric created a lovely new sequence which Maximilian entered as
> > A248034. It has a spectacular graph and it sounds pretty good too. I
> would
> > like to be able to see more terms and listen to the rest of the music, if
> > someone would create a b-file.
> >
> > I gave it the keywords look and hear.
> >
> >
> > Best regards
> > Neil
> >
> > Neil J. A. Sloane, President, OEIS Foundation.
> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> > Phone: 732 828 6098; home page: http://NeilSloane.com
> > Email: njasloane at gmail.com
> >
> >
> > On Sat, Oct 11, 2014 at 3:28 PM, M. F. Hasler <oeis at hasler.fr> wrote:
> >
> > > Eric,
> > >
> > > my program seems to confirm your data (congrats !),
> > > I submitted a proposal as https://oeis.org/draft/A248034
> > >
> > > -- Maximilian
> > >
> > > (PARI)
> > >
> > >
> c=vector(10);print1(a=0);for(n=1,99,apply(d->c[d+1]++,if(a,digits(a)));print1(","a=c[1+a%10]))
> > >
> > > On Sat, Oct 11, 2014 at 1:59 PM, Eric Angelini <Eric.Angelini at kntv.be>
> > > wrote:
> > > >
> > > >
> > >
> D=0,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,2,2,3,2,4,2,5,2,6,2,7,2,8,2,9,2,10,3,3,4,3,5,3,6,3,7,3,8,3,9,3,10,4,4,5,4,6,4,7,4,8,4,9,4,10,5,5,6,5,7,5,8,5,9,5,10,6,6,7,6,8,6,9,6,10,7,7,8,7,9,7,10,8,8,9,8,10,9,9,10,10,11,20,12,...
> > > >
> > > > Hello SeqFans,
> > > > pick any comma in D.
> > > > Immediately to the left of the comma
> > > > there is a digit 'd'.
> > > > Immediately to the right of the comma
> > > > there is an integer d(n).
> > > > D is such that there are d(n) digit 'd'
> > > > so far in D [from the start of D up to the comma].
> > > >
> > > > In other words, the rightmost digit of d(n) is present d(n+1) times
> in
> > > D, counting from d(1) to d(n).
> > > >
> > > > I'm wondering: do all integers appear
> > > > at least once in D?
> > > >
> > > > P.-S.
> > > > It is possible to compute similar sequences
> > > > for every base. I guess the binary-one is:
> > > >
> > > > B = 0,1,1,10,10,11,110,100,110,111,1110,...
> > > >
> > > > Best,
> > > > É.
> > > >
> > >
> > > _______________________________________________
> > >
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> >
>