OFFSET
1,2
COMMENTS
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:
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.
LINKS
Thomas Scheuerle, Table of n, a(n) for n = 1..5000
Thomas Scheuerle, Scatter plot log(a(n)/n^2) for n = 1..800.
FORMULA
PROG
(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]}
CROSSREFS
KEYWORD
nonn,easy,new
AUTHOR
Thomas Scheuerle, Feb 24 2025
STATUS
approved