%I #42 Mar 03 2025 13:27:25

%S 1,3,6,5,20,18,8,56,10,90,12,132,14,182,16,240,19,108,323,100,22,462,

%T 24,552,26,650,28,756,30,870,32,992,34,1122,36,1260,38,1406,40,1560,

%U 42,1722,44,1892,46,2070,48,2256,50,2450,52,2652,54,2862,57,448,3135,59,3422,61,3660,63,3906,65,4160

%N Lexicographically earliest positive integer sequence satisfying a(n) = a(a(n))/n.

%C While extending the sequence at a(k) we will check if k equals a previous term in the sequence. If such a term a(m) = k is found a(k) is determined as a(k) = a(m)*m. If no previous term matches k we may choose a(k) = k+c with the least c such that c > 0 and k+c does not equal any previous term in the sequence. It is conjectured that this check is sufficient. Reasoning behind this conjecture:

%C The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist, because this cannot be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3.

%H Thomas Scheuerle, <a href="/A381464/b381464.txt">Table of n, a(n) for n = 1..5000</a>

%H Thomas Scheuerle, <a href="/A381464/a381464.png">Scatter plot log(a(n)/n^2) for n = 1..800</a>.

%F Let b(n, m) be m times recursion into a(n), for example b(3, 2) = a(a(3)).

%F b(3, m) = A000304(m+1) for m > 0.

%F b(n, m+2) = b(n, m)*b(n, m+1).

%F b(5, m) = 4^F(m)*5^F(m+1), where F(m) = A000045(m).

%F b(k, m) = (k-1)^F(m)*k^F(m+1), for all k where k+1 = a(k).

%o (PARI) listA(max_n) = {my(v=[1, 0], t=1); for(k=2, max_n, if(v[k]==0, t=1; if(k+t<#v, while(v[k+t]>0, t++)); v[k]=k+t); v=concat(v, vector(max(0, v[k]+1-#v))); if(v[v[k]]>0, print("The conjecture that a single forward check is sufficient failed at:", k)); v[v[k]]=k*v[k]); v[1..max_n]}

%Y Cf. A257794, A358793.

%Y Cf. A000045, A000304.

%Y Cf. A099267 ( a(n) = a(a(n))-n ).

%K nonn,easy,new

%O 1,2

%A _Thomas Scheuerle_, Feb 24 2025