A335497 - OEIS

%I #37 Jun 14 2022 11:58:14

%S 1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,1,12,2,3,2,4,2,5,2,6,2,7,2,8,

%T 2,9,2,10,2,11,4,3,3,4,4,5,3,5,4,6,3,6,4,7,3,7,4,8,3,8,4,9,3,9,4,10,3,

%U 10,4,11,5,5,6,5,7,5,8,5,9,5,10,5,11,6,6

%N a(1) = 1, and for any n > 0, a(n+1) is the number of times the decimal representation of a(n) appears in the concatenation of the first n terms, possibly with overlap.

%C This sequence is a variant of A276457.

%C This sequence is unbounded.

%C It seems that lim sup a(n)/(n*log(n)) = 0.03 approximately. - _Ya-Ping Lu_, Dec 16 2021

%H Rémy Sigrist, <a href="/A335497/b335497.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A335497/a335497.png">Logarithmic scatterplot of the first 1000000 terms</a>

%H Rémy Sigrist, <a href="/A335497/a335497.pl.txt">Perl program for A335497</a>

%e The first terms, alongside their concatenations with a star in front of each occurrence of a(n), are:

%e   n   a(n)  cat(a(1)...a(n))

%e   --  ----  ---------------------------------

%e    1     1  *1

%e    2     1  *1*1

%e    3     2  11*2

%e    4     1  *1*12*1

%e    5     3  1121*3

%e    6     1  *1*12*13*1

%e    7     4  112131*4

%e    8     1  *1*12*13*14*1

%e    9     5  11213141*5

%e   ...

%e   17     9  1121314151617181*9

%e   18     1  *1*12*13*14*15*16*17*18*19*1

%e   19    10  112131415161718191*10

%e   20     1  *1*12*13*14*15*16*17*18*19*1*10*1

%e   21    12  1*12131415161718191101*12

%e   22     2  11*21314151617181911011*2*2

%o (Perl) See Links section.

%o (Python)

%o a1 = 1; print(a1, end =', '); S = str(a1)

%o for n in range(2, 100): ct = S.count(str(a1)); S += str(ct); print(ct, end = ', '); a1 = ct # _Ya-Ping Lu_, Dec 16 2021

%Y Cf. A179801, A276457.

%K nonn,base

%O 1,3

%A _Rémy Sigrist_, Jun 14 2020