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A308171
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Limiting sequence of digits, read from the right, when starting with 5 we repeatedly replace each digit with its square (as in A308170).
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2
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5, 2, 4, 6, 1, 6, 3, 1, 6, 3, 9, 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, 1, 6, 3, 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, 8, 1, 4, 6, 1, 6, 1, 6, 3, 1, 6, 3, 1, 6, 3, 9, 1, 6, 3, 9, 1
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OFFSET
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1,1
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COMMENTS
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As Jean-Paul Allouche remarks on the SeqFan list, also the limiting sequence of the morphism 5 -> 52, 2 -> 4, 4 -> 61, 6 -> 63, 1 -> 1, 3 -> 9, 9 -> 18, 8 -> 46 over the alphabet {1..9} \ {7}, iterated on an initial value of 5. The digit 7 never occurs, and digits 2 and 5 only occur as a(1) and a(2). - M. F. Hasler, May 15 2019 [Corrected by N. J. A. Sloane, May 16 2019]
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LINKS
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EXAMPLE
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Replacing each digit with its square, we get 5 -> 25 -> 425 -> 16425 -> 13616425 -> .... The final digits converge to ...16425, or read from the right, to this sequence: 5, 2, 4, 6, 1, ... - M. F. Hasler, May 15 2019
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MATHEMATICA
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s = {5}; Do[s = Flatten[ Reverse@ IntegerDigits[#^2] & /@ s]; If[Length[s] > 100, s = Take[s, 100]], {100}]; s (* Giovanni Resta, Jul 03 2019 *)
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PROG
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(PARI) { wanted = 87; a = [5]; 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) A308171_vec(N, a=[5])={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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