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%I #11 Dec 05 2018 03:02:22
%S 1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,1,10,11,1,12,13,14,15,2,16,2,17,2,
%T 18,2,3,19,2,4,20,2,5,21,2,6,3,22,2,7,3,8,2,9,3,10,23,11,3,4,12,13,14,
%U 3,5,3,15,3,6,24,3,16,7,25,26,8,4,27,17,28,9,29
%N a(1) = 1, and for any n > 0, a(n+1) is the number of occurrences of the n-th digit of the sequence among the first n digits of the sequence.
%C In other words, if we take the ordinal transform of the digits of the sequence and prepend the number 1, then we obtain the sequence again.
%C The number 1 appears 11 times.
%C Any number > 1 appears 10 times.
%C The sequence contains arbitrarily large runs of consecutive numbers.
%H Rémy Sigrist, <a href="/A322182/b322182.txt">Table of n, a(n) for n = 1..10000</a>
%H Rémy Sigrist, <a href="/A322182/a322182.png">Colored scatterplot of (n, a(n)) for n = 1..1000000</a> (where the color is function of the n-th digit of the sequence)
%e The first terms of the sequence, alongside the (n-1)-th digit of the sequence, are:
%e n a(n) (n-1)-th digit
%e --- ---- --------------
%e 1 1 N/A
%e 2 1 1
%e 3 2 1
%e 4 1 2
%e 5 3 1
%e 6 1 3
%e 7 4 1
%e 8 1 4
%e 9 5 1
%e 10 1 5
%e 11 6 1
%e 12 1 6
%e 13 7 1
%e 14 1 7
%e 15 8 1
%e 16 1 8
%e 17 9 1
%e 18 1 9
%e 19 10 1
%e 20 11 1
%e 21 1 0
%o (PARI) a = [1]; ord = vector(base = 10); for (k=1, 59, a = concat(a, apply(d -> ord[1+d]++, digits(a[k], #ord)))); print (a)
%Y Cf. A248034.
%K nonn,base
%O 1,3
%A _Rémy Sigrist_, Nov 30 2018