

A359416


Write n as 2^m  k, where 2^m is the least power of 2 >= n (0 <= k <= 2^(m1)1). For n a power of 2 (k = 0), a(n) = n. For numbers with k > 0, a(n) is the least p*a(k) which has not occurred previously, the count of k being taken from right to left (backwards) from k = 1 at 2^m  1.


1



1, 2, 3, 4, 9, 6, 5, 8, 25, 18, 27, 12, 15, 10, 7, 16, 49, 50, 105, 36, 81, 54, 75, 24, 35, 30, 45, 20, 21, 14, 11, 32, 121, 98, 231, 100, 495, 210, 175, 72, 225, 162, 243, 108, 315, 150, 147, 48, 77, 70, 165, 60, 135, 90, 125, 40, 55, 42, 63, 28, 33, 22, 13, 64
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OFFSET

1,2


COMMENTS

A variant of the recursive definition of the Doudna sequence A005940, and A356886. Whereas a sequence is normally computed in natural order A000027) of its indices (a(1), a(2), a(3), etc.), in this case terms with indices n other than powers of 2 are computed backwards, right to left from the least power of 2 exceeding n (ordering as in A122155). For example, with terms between a(4) and a(8) the order of computation is a(7), then a(6), then a(5), each time choosing a least novel number matching the definition, see Example. Compare with similar sequence A356886, where a similar definition is used but the count is conventional: left to right.
Conjectured to be a permutation of the positive integers in which primes appear in natural order. The even bisection, when divided by 2, reproduces the sequence.


LINKS

Michael De Vlieger, Labeled fanstyle binary tree showing a(n), n = 1..2^12 in levels k such that n = 2^k+1..2^(k+1), with a color function indicating a(2^k) and a(n), where a(n) = a(2^k) in green, a(n) < a(2^k) in blues, and a(n) > a(2^k) in yellows, oranges, and reds.


FORMULA

a(2*n)/2 = a(n); n >= 1.
a(2^n  1) = prime(n); n >= 2.
a(2^n + 1) = prime(n)^2; n >= 2.
At the occurrence of 2^n (n >= 3) the following pattern of five successive terms is observed: 3*prime(n1), 2*prime(n1), prime(n), 2^n, prime(n)^2, ....
For n >= 2, a(2^n + 1)/a(2^n  1) = prime(n); compare with A357057).


EXAMPLE

a(3) = 3 because 3 = 2^2  1, so k = 1 and 3 is the least odd prime multiple of a(1).
a(7) = 5 because 7 = 2^3  1, k = 1, a(1) = 1 and 5 is the least multiple of 1 not seen already. (At this point a(5), a(6) have not been found.)
a(6) = 6 since k = 2, a(2) = 2, and 3*2 is the least number not seen already.
a(5) = 9 since k = 3, a(3) = 3, so we choose 3*3.


MATHEMATICA

nn = 2^6; c[_] = False; Do[Set[{m, k}, {2, n  2^Floor[Log2[n]]}]; If[k == 0, Set[{a[n], c[n]}, {n, True}], While[Set[t, Prime[m] a[k]]; c[t], m++]; Set[{a[2^#/(1 + Boole[IntegerQ[Log2[n]]])  n + 2^(#  1) &@ IntegerLength[n, 2]], c[t]}, {t, True}]], {n, 2^Ceiling[Log2[nn]] }]; Array[a, nn] (* Michael De Vlieger, Jan 02 2023 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



