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

0,1

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, ...).

Links

Table of n, a(n) for n=0..32.

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

Formula

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

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

Example

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).

Prog

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

Crossrefs

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

Sequence in context: A230168 A325082 A033670 * A044421 A044802 A142499

Adjacent sequences: A352539 A352540 A352541 * A352543 A352544 A352545

Keyword

nonn,base

Author

M. F. Hasler, Mar 20 2022

Status

approved