%I #21 Dec 19 2017 18:36:27

%S 22,10213223,10311233,10313314,10313315,10313316,21322314,21322315,

%T 21322316,31123314,31123315,31123316,31331415,31331416,31331516,

%U 1031223314,1031223315,1031223316,3122331415,3122331416,3122331516,103142132415,104122232415,103142132416,104122232416,314213241516,412223241516,1011112131415,1011112131416,1011112131516,1011112141516,1011113141516,1111213141516,10414213142516,10413223241516,10512223142516,10512213341516,101112213141516

%N Autobiographical numbers in base 7: 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 6 in that order when they occur in a number. The numbers and the frequency are written in base 7.

%C These are all autobiographical numbers in base 7 which lead to a fixed-point or belong to a cycle.

%C There are three cycles with length 2 (103142132415 /104122232415, 103142132416/104122232416, 314213241516/412223241516), one cycle with length 3 (10512213341516/10512223142516/10414213142516). 29 numbers are fixed-points.

%D Antonia Münchenbach and Nicole Anton George, "Eine Abwandlung der Conway-Folge", contribution to "Jugend forscht" 2016, 2016

%H Andre Kowacs, <a href="https://arxiv.org/abs/1708.06452">Studies on the Pea Pattern Sequence</a>, arXiv:1708.06452 [math.HO], 2017.

%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

%e 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 16 2016