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URL: https://oeis.org/A367408

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A367408
a(n) = n for n a power of 2; otherwise let 2^r be the greatest power of 2 which does not exceed n, then a(n) = the least novel m*a(k) where k = n - 2^r, and m is not a prior term.
1
1, 2, 3, 4, 5, 12, 18, 8, 6, 14, 21, 28, 35, 84, 126, 16, 7, 20, 27, 36, 45, 108, 162, 72, 54, 140, 189, 252, 315, 756, 1134, 32, 9, 22, 30, 40, 50, 120, 180, 80, 60, 154, 210, 280, 350, 840, 1260, 160, 70, 200, 270, 360, 450, 1080, 1620, 720, 540, 1400, 1890, 2520, 3150, 7560, 11340, 64, 10
OFFSET
1,2
COMMENTS
Based on a recursion similar to that which produces the Doudna sequence, A005940, using the same definition of k, the "distance" from the greatest power of 2 less than n (compare with A365436).
Sequence is conjectured to be a permutation of the positive integers, A000027.
LINKS
Michael De Vlieger, Plot prime(i)^m | a(n) at (x,y) = (n,i), 64X vertical exaggeration, with a color function showing m = 1 in black, m = 2 in red, m = 3 in orange, ..., m = 16 in magenta.
FORMULA
a(2^k) = 2^k for all k >= 0.
a(2^k + 1) = m, the least unused term up to a(2^k), where multiples (other than 1) of m have been used to generate terms between a(2^(k-1)) and a(2^k) except for those which have occurred earlier; see Example.
EXAMPLE
a(3) = 3 since k = 1, a(1) = 1, and 3 is the smallest number which has not already occurred.
a(7) = 18, since k = 3, a(3) = 3, m = 6 is the least unused number and 18 has not already occurred.
For n = 18, k = 2, a(2) = 2, m = 9 is the least unused number, so we should expect a(18) = 2*9 = 18, but 18 has already occurred at a(7). Therefore we increment to m = 10, the next smallest unused number, and find a(18) = 20 (which has not occurred previously).
MATHEMATICA
Block[{a, c, k, m, t, nn},
nn = 128; c[_] := False; c[1] = True; m[_] := 1; a[1] = 1;
Do[If[IntegerQ[#], Set[k, i],
While[Or[c[m[#]], c[Set[k, # m[#]]]], m[#]++] &[
a[i - 2^Floor[#]]]] &@ Log2[i];
Set[{a[i], c[k]}, {k, True}], {i, nn}];
Array[a, nn] ] (* Michael De Vlieger, Jul 01 2025 *)
PROG
(PARI) lista(nn) = my(va=vector(nn)); for (n=1, nn, my(p=2^logint(n, 2)); if (p == n, va[n] = n, my(k=n-p, m=1); while (#select(x->(x==m), va) || #select(x->(x==m*va[k]), va), m++); va[n] = m*va[k]; ); ); va; \\ Michel Marcus, Dec 17 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved