A348444 If A348243(n) = i, then a(n) = 1 + number of copies of i that have already appeared in A348243.

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 1

Author

N. J. A. Sloane, Oct 21 2021

Keywords

• nonn

Comments

In other words, when we see i in A348243, which i is it? The first, second, third, ...? This gives a measure of how many attempts we have to make in A307730 and A349243 before all the n instances of n have been obtained.

The first 3's appear at n = 259, 490, 585, 627, ..., the first 4's at 3161, 4230, 5989, 8207, ...

Note that A348409 gives the index of the last occurrence of each k in A307730. So until we reach A348409(k), we will not know how many attempts are needed to obtain all copies of k.

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..750

Crossrefs

Cf. A307720, A307730, A348245, A348409.