

A166133


After initial 1,2,4, a(n+1) is the smallest divisor of a(n)^21 that has not yet appeared in the sequence.


26



1, 2, 4, 3, 8, 7, 6, 5, 12, 11, 10, 9, 16, 15, 14, 13, 21, 20, 19, 18, 17, 24, 23, 22, 69, 28, 27, 26, 25, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 201, 80, 79
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

The initial 1,2,4 provides the smallest example with this rule that is not simply the integers in order, nor (apparently) ends with all divisors of a(n)^21 already present.
Apparently the sequence is infinite, and apparently includes every positive integer.
Apr 05 2015: John Mason has computed the first ten million terms. See link to zipped file.  N. J. A. Sloane, Apr 06 2015
The sequence contains many strings of incrementing and decrementing values. In the 1200 steps following the 4, there are 136 increments, 706 decrements, and 358 larger steps. What is the limiting distribution for these steps? {Click the "listen" button to appreciate these strings.  N. J. A. Sloane, Apr 03 2015]
After 3, 198, 270, 570, 522, 600, 822, and 882, we have a(n+1) = a(n)^21. Does this happen infinitely often? Cf. A256406, A256407.
A256543 gives numbers m such that a(m+1) = a(m)1 or a(m+1) = a(m)+1.  Reinhard Zumkeller, Apr 01 2015
If this is a permutation, then A255833 is the inverse permutation.  M. F. Hasler, Apr 01 2015
a(A256703(n)+1) = a(A256703(n))^2  1.  Reinhard Zumkeller, Apr 08 2015
For n > 3: a(n) = A027750(a(n1)^21,A256751(n)).  Reinhard Zumkeller, Apr 09 2015


LINKS

Franklin T. AdamsWatters and N. J. A. Sloane, Table of n, a(n) for n = 1..20000 [First 1203 terms from Franklin T. AdamsWatters]
H. Havermann, Log plot of 450000+ terms [Produced by Mathematica's ListLogPlot command]
H. Havermann, Oversized pointjoined (over 250000 terms) graph of the sequence [Heavily clipped, which explains the strange appearance.  N. J. A. Sloane, Apr 01 2015]
John Mason, Table of n, a(n) for n = 1..711888 [10 megabytes]
John Mason, Table of n, a(n) for n = 1..2000000 [32 megabytes]
John Mason, Ten million terms (zipped file)
N. J. A. Sloane and others, "Blog" about A166133
John Mason, Java program to generate this sequence, used to generate 10M terms, and some other associated sequences; it requires splitting into single classes for use.


EXAMPLE

After a(24) = 22, the divisors of 22^21 = 483 are 1, 3, 7, 21, 23, 69, 161, and 483; 1, 3, 7, 21, and 23 have already occurred, so a(25) = 69.


MATHEMATICA

s = {1, 2, 4}; e = 4; Do[d = Divisors[e^2  1]; i = 1;
While[MemberQ[s, d[[i]]], i++]; e = d[[i]]; AppendTo[s, e], {19997}]; s (* Hans Havermann, Apr 03 2015 *)


PROG

(PARI) al(n, m=4, u=6)={local(ds, db);
u=bitor(u, 1<<m); print1(m);
for(i=1, n,
ds=divisors(m^21);
for(k=2, #ds, m=ds[k]; db=1<<m; if(!bitand(u, db), break));
u=bitor(u, db); print1(", "m))}
/* This prints the sequence without the initial 1, 2. */
(Haskell)
import Data.List (delete); import Data.List.Ordered (isect)
a166133 n = a166133_list !! (n1)
a166133_list = 1 : 2 : 4 : f (3:[5..]) 4 where
f zs x = y : f (delete y zs) y where
y = head $ isect (a027750_row' (x ^ 2  1)) zs
 Reinhard Zumkeller, Apr 01 2015


CROSSREFS

Cf. A166134, A000027, A122280, A005563, A256406, A256407, A027750, A005563, A256557, A256559.
For records see A256403, A256404.
Smallest missing numbers: A256405, A256408, A256409.
Cf. A256541 (first differences), A256543.
Inverse (conjectured): A255833.
Cf. A256564 (smallest prime factors), A244080 (largest prime factors), A256578 (largest proper divisors), A256542 (number of divisors).
Upper envelope: the sequence of pairs (A256422(n),A256423(n)).
Cf. A256703.
Cf. A256751.
Sequence in context: A243496 A125566 A255833 * A237739 A111699 A067179
Adjacent sequences: A166130 A166131 A166132 * A166134 A166135 A166136


KEYWORD

nonn,nice,hear


AUTHOR

Franklin T. AdamsWatters, Oct 07 2009


STATUS

approved



