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%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
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