%I #17 Dec 19 2017 18:36:47
%S 22,10213223,10311233,10313314,10313315,10313316,10313317,10313318,
%T 10313319,21322314,21322315,21322316,21322317,21322318,21322319,
%U 31123314,31123315,31123316,31123317,31123318,31123319,31331415,31331416,31331417,31331418,31331419,31331516,31331517,31331518
%N Autobiographical numbers in base 10: numbers which are fixed or belong to a cycle under the operator T.
%C The T operator numerically summarizes the frequency of digits 0 through 9 in that order when they occur in a number. The numbers and the frequency are written in base 10.
%C These are all autobiographical numbers in base 10 which lead to a fixed-point or belong to a cycle.
%C 109 numbers are fixed-points. There are 31 cycles with length 2 (62 numbers) and 10 cycles with length 3 (30 numbers).
%D Antonia Münchenbach and Nicole Anton George, "Eine Abwandlung der Conway-Folge", contribution to "Jugend forscht" 2016, 2016
%H Antonia Münchenbach, <a href="/A267498/b267498.txt">Table of n, a(n) for n = 1..201</a>
%H Andre Kowacs, <a href="https://arxiv.org/abs/1708.06452">Studies on the Pea Pattern Sequence</a>, arXiv:1708.06452 [math.HO], 2017.
%H Antonia Münchenbach, <a href="/A267498/a267498.txt">Numbers sorted as fixed-points, cycles with length 2 and 3</a>
%e 10213223 contains one 0, two 1's, three 2's and two 3's, so T(10213223) = 1 0 2 1 3 2 2 3, and this is fixed under T.
%e 103142132415 and 104122232415 belong to the cycle of length 2, so T(T(103142132415)) = T(1 0 4 1 2 2 2 3 2 4 1 5) = 1 0 3 1 4 2 1 3 2 4 1 5.
%Y Cf. A047841, A267491, A267492, A267493, A267494, A267495, A267496, A267497, A267498, A267499, A267500, A267502.
%K nonn,base,fini,full
%O 1,1
%A _Antonia Münchenbach_, Jan 25 2016