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

89, 187, 1058, 529, 1481, 9892, 4946, 2473, 9905, 19855, 118406, 59203, 154523, 708844, 354422, 177211, 949322, 474661, 1241102, 620551, 1275761, 9040972, 4520486, 2260243, 8692463, 18558895, 117444446, 58722223, 146254445, 801698866, 400849433, 1385292733, 11260625954

Offset: 0

Author

M. F. Hasler, Mar 20 2022

Keywords

• nonn
• base

Comments

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.

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.

As a consequence, the sequence tends to a 10-adic limit ...27057751007.

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, ...).

Examples

			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).
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.
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.
From a(37) = 11079547822507 on, all terms appear to have at least two digits 0, and therefore all end in the digits ...07.
From a(46) = 11109941625118561459007 on, all terms appear to have at least three digits 0, and therefore all end in the digits ...007.
From a(52) = 1119999530692487035860091007 on, all terms appear to have at least four digits 0, and therefore all end in the digits ...1007.
a(49) = 9999653161399504894770007 ~ 9.999653e24 appears to be the last term to have:
    (i) not more digits than the preceding term,
   (ii) its leading digit different from 1,
  (iii) a successor a(n+1) ~ 1.999965e25 ~ 2*a(n) and a(51) ~ 1.1999964e26 ~ 6*a(50).
For all n >= 51, a(n) has one more digit than a(n-1), and a(n+1) > 9*a(n).
		

Links

• Eric Angelini, Divide by 2 or add the biggest anagram, math-fun discussion list, Mar 20 2022
• M. F. Hasler, Number of digits 0 in A352542(0..500), PARI/GP plot, Mar 22 2022
• M. F. Hasler, Number of leading digits 1 in A352542(0..500), PARI/GP plot, Mar 22 2022

Crossrefs

Cf. A352544 (the iterated map), A352540 (starting values with infinite orbit), A352541 (number of iterations until a value is repeated).

Programs

PARI

A352542_upto(N)=vector(N+1,i,N=if(i>1,A352544(N),89))

Formula

log_10 a(n) ~ n (asymptotical equivalence, as n -> oo).

a(n+1) > 9*a(n) for all n > 50. (Conjectured.)