A381464 Lexicographically earliest positive integer sequence satisfying a(n) = a(a(n))/n.

1, 3, 6, 5, 20, 18, 8, 56, 10, 90, 12, 132, 14, 182, 16, 240, 19, 108, 323, 100, 22, 462, 24, 552, 26, 650, 28, 756, 30, 870, 32, 992, 34, 1122, 36, 1260, 38, 1406, 40, 1560, 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

Offset: 1

Author

Thomas Scheuerle, Feb 24 2025

Keywords

• nonn
• easy

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.

Crossrefs

Cf. A257794, A358793.

Cf. A000045, A000304.

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

Programs

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]}

Formula

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

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

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

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

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