

A361101


a(n) is the smallest positive number not among the terms in a(1..n1) with index a(n1)*k for any integer k; a(1)=1.


2



1, 2, 1, 3, 2, 1, 4, 1, 5, 1, 6, 2, 4, 4, 4, 4, 5, 3, 6, 4, 5, 3, 6, 4, 5, 3, 6, 5, 3, 7, 1, 8, 2, 6, 5, 3, 8, 2, 9, 1, 10, 2, 9, 1, 11, 4, 6, 5, 3, 8, 2, 9, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 3, 8, 2, 10, 2, 11, 4, 6, 6, 6, 8, 2, 20, 3, 8, 3, 8, 3
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OFFSET

1,2


COMMENTS

This sequence is partly defined by the following cases for a(353) onwards. But what is the pattern to the 1s and 2s?
For k >= 0:
a(353+4k) = 1 or 2;
a(354+2k) = (354+2k)/2  63;
a(355+4k) = 1.
Of the 14912 n equal to 1 (mod 4) between 352 and 60000, 4983 (33.416%) equal 1 and 9929 (66.584%) equal 2. From observation it appears that 1/3 of a(353+4k); k>=0 equal 1 and 2/3 equal 2 (see Figure).
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LINKS



EXAMPLE

To find a(13), we look at the last term in the sequence thus far: (1, 2, 1, 3, 2, 1, 4, 1, 5, 1, 6, 2). Since it is a 2, the next term will be the smallest not among the evenindexed terms of the sequence thus far, which are (2, 3, 1, 1, 1, 2). 4 is the smallest missing number, so a(13)=4.


MATHEMATICA

K = {1}; While[Length@K <= 85, A = {}; For[q = Last@K, q <= Length@K, q += Last@K, AppendTo[A, K[[q]]]]; k = 1; While[MemberQ[A, k], k++]; AppendTo[K, k]]; Print[K] (* Samuel Harkness, Mar 06 2023 *)


PROG

(PARI) { p = 1; for (n = 1, #a = vector(86), x = 2^0; forstep (k = p, n1, p, x = bitor(x, 2^a[k]); ); print1 (p = a[n] = valuation(1+x, 2)", "); ); } \\ Rémy Sigrist, Mar 02 2023


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



