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A359416 Write n as 2^m - k, where 2^m is the least power of 2 >= n (0 <= k <= 2^(m-1)-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 (list; graph; refs; listen; history; text; internal format)
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, Log log scatterplot of a(n), n = 1..2^12.
Michael De Vlieger, Labeled fan-style 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(n-1), 2*prime(n-1), 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
Sequence in context: A091205 A106447 A356886 * A222248 A352728 A236852
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified April 23 02:53 EDT 2024. Contains 371906 sequences. (Running on oeis4.)