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A352542 Trajectory of initial value 89 under iterations of the map A352544: half if even, add largest anagram if odd.

%I A352542 #17 Mar 30 2022 07:03:43
%S A352542 89,187,1058,529,1481,9892,4946,2473,9905,19855,118406,59203,154523,
%T A352542 708844,354422,177211,949322,474661,1241102,620551,1275761,9040972,
%U A352542 4520486,2260243,8692463,18558895,117444446,58722223,146254445,801698866,400849433,1385292733,11260625954
%N A352542 Trajectory of initial value 89 under iterations of the map A352544: half if even, add largest anagram if odd.
%C A352542 89 is the smallest nonnegative integer with an orbit of infinite size under iterations of x -> A352544(x) = {x/2 if x is even, x + A004186(x) if x is odd}. The list of all such numbers is given in A352540, which contains this sequence as subset.
%C A352542 We conjecture that there is a strictly increasing sequence (b(k), k >= 0) = (32, 37, 46, 52, 88, 91, 118, 122, 141, ...) such that all terms a(n) with n >= b(k) have more than k digits 0.
%C A352542 As a consequence, the sequence tends to a 10-adic limit ...27057751007.
%C A352542 Similarly, the number of leading digits 1 appears to grow to infinity; more precisely, a(n) has more than k leading digits 1 for all n > c(k >= 0) = (50, 70, 95, 121, 122, 123, 130, ...).
%H A352542 Eric Angelini, <a href="https://mailman.xmission.com/hyperkitty/list/math-fun@mailman.xmission.com/thread/RI5P2MYI7NKD32LK47XTYTGUAEMQKVVZ/">Divide by 2 or add the biggest anagram</a>, math-fun discussion list, Mar 20 2022
%H A352542 M. F. Hasler, <a href="/A352542/a352542_1.gif">Number of digits 0 in A352542(0..500)</a>, PARI/GP plot, Mar 22 2022
%H A352542 M. F. Hasler, <a href="/A352542/a352542.gif">Number of leading digits 1 in A352542(0..500)</a>, PARI/GP plot, Mar 22 2022
%F A352542 log_10 a(n) ~ n (asymptotical equivalence, as n -> oo).
%F A352542 a(n+1) > 9*a(n) for all n > 50. (Conjectured.)
%e A352542 The initial term a(0) = 89 and its successor a(1) = 187 are odd, so the number with the same digits in decreasing order, 98 resp. 871, are added to find the successor a(n+1).
%e A352542 Then a(2) = 1058 is even (as are a(5..6), a(10), a(13..14), ...), so the successor is obtained dividing it by two.
%e A352542 a(32) = 11260625954 appears to be the last even term. It appears that from this terms on, all terms have at least one digit 0 and therefore all subsequent terms end in the digit 7.
%e A352542 From a(37) = 11079547822507 on, all terms appear to have at least two digits 0, and therefore all end in the digits ...07.
%e A352542 From a(46) = 11109941625118561459007 on, all terms appear to have at least three digits 0, and therefore all end in the digits ...007.
%e A352542 From a(52) = 1119999530692487035860091007 on, all terms appear to have at least four digits 0, and therefore all end in the digits ...1007.
%e A352542 a(49) = 9999653161399504894770007 ~ 9.999653e24 appears to be the last term to have:
%e A352542     (i) not more digits than the preceding term,
%e A352542    (ii) its leading digit different from 1,
%e A352542   (iii) a successor a(n+1) ~ 1.999965e25 ~ 2*a(n) and a(51) ~ 1.1999964e26 ~ 6*a(50).
%e A352542 For all n >= 51, a(n) has one more digit than a(n-1), and a(n+1) > 9*a(n).
%o A352542 (PARI) A352542_upto(N)=vector(N+1,i,N=if(i>1,A352544(N),89))
%Y A352542 Cf. A352544 (the iterated map), A352540 (starting values with infinite orbit), A352541 (number of iterations until a value is repeated).
%K A352542 nonn,base
%O A352542 0,1
%A A352542 _M. F. Hasler_, Mar 20 2022