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A308170
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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.
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2
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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, 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, 3, 1, 6, 3, 9, 1, 8, 1, 4, 6, 1, 6, 1, 6, 3, 1, 6, 3, 9
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OFFSET
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1,2
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COMMENTS
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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]
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.
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
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LINKS
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EXAMPLE
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The successive numbers that arise are
7
49
1681
136641
193636161
1819369361361
1641819368193619361
...
and reading from the right we see 1,6,3,9, ...
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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