A308170 - OEIS

%I #37 May 17 2019 20:11:09

%S 1,6,3,9,1,8,1,4,6,1,6,1,6,3,1,6,3,1,6,3,9,1,6,3,9,1,6,3,9,1,8,1,6,3,

%T 9,1,8,1,6,3,9,1,8,1,4,6,1,6,3,9,1,8,1,4,6,1,6,3,9,1,8,1,4,6,1,6,1,6,

%U 3,1,6,3,9,1,8,1,4,6,1,6,1,6,3,1,6,3,9

%N Start with the number 7, repeatedly square every digit in place to get a new number; in the limit this process converges (reading from right to left) to the string shown here.

%C If we start with 3 or 9, we get the same sequence. If instead we start with 2, 4, 6, or 8 we get the same sequence but without the initial 1. If we start with 5 we get A308171. [Corrected by _M. F. Hasler_, May 15 2019]

%C Comment from _Jean-Paul Allouche_, May 15 2019: These sequences can be obtained in the order shown (that is, in the right-to-left order) by starting with 7 (say), and repeatedly applying the morphism on the alphabet {1,2,3,...,9} defined by 1 -> 1, 3 -> 9, 4 -> 61, 5 -> 52, 6 -> 63, 7 -> 94, 8 -> 46, 9 -> 18.

%C If we start with 1, application of the morphism will never change that initial sequence of length 1: this is the third fixed point of the morphism. - _M. F. Hasler_, May 15 2019

%H Rémy Sigrist, <a href="/A308170/b308170.txt">Table of n, a(n) for n = 1..25000</a>

%e The successive numbers that arise are

%e 7

%e 49

%e 1681

%e 136641

%e 193636161

%e 1819369361361

%e 1641819368193619361

%e ...

%e and reading from the right we see 1,6,3,9, ...

%o (PARI) { wanted = 87; a = [7]; while (1, b = concat(apply(d -> if (d,digits(d^2),[0]),a)); if (#b > wanted, b = b[#b-wanted+1..#b]); if (a==b, break, a = b)); print (Vecrev(a)) } \\ _Rémy Sigrist_, May 15 2019

%o (PARI) A308170_vec(N,a=[9])={while(a!=a=concat(apply(t->digits(t^2),if(#a>N,a[-N..-1],a))),); Vecrev(a[-N..-1])} \\ _M. F. Hasler_, May 15 2019

%Y Cf. A048385, A061588, A082027, A308171.

%K nonn,base

%O 1,2

%A _N. J. A. Sloane_, May 15 2019, following a suggestion from _Jeremy Gardiner_.

%E More terms from _Rémy Sigrist_, May 15 2019