A346422 -id:A346422 - cp's OEIS frontend

A347188 a(n) = A346422(4*A003754(n-1) + 3) for n > 1 with a(1) = 1.

1, 2, 6, 4, 24, 18, 12, 120, 8, 96, 72, 48, 720, 54, 36, 600, 24, 480, 360, 240, 5040, 16, 384, 288, 192, 4320, 216, 144, 3600, 96, 2880, 2160, 1440, 40320, 162, 108, 3000, 72, 2400, 1800, 1200, 35280, 48, 1920, 1440, 960, 30240, 1080, 720, 25200, 480, 20160

Offset: 1

Mikhail Kurkov, Aug 21 2021 [verification needed]

nonn

To get the distinct terms of A346422 in the order of their appearance up to A346422(2^n - 1), just take the first A000045(n+1) terms of this sequence and remove the duplicates.

Cf. A003754, A130312, A200649, A346422.

a(n) = (1 + A200649(n))*a(n - A130312(n-1)) for n > 1 with a(1) = 1.

a(n) = A346422(4*A003754(n-1) + 3) for n > 1 with a(1) = 1.

A366600 a(n) = (1 + A033264(n))*a(A053645(n)) for n > 0 with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 6, 2, 4, 4, 8, 1, 2, 2, 6, 2, 6, 6, 12, 2, 4, 4, 18, 4, 8, 8, 16, 1, 2, 2, 6, 2, 6, 6, 12, 2, 6, 6, 24, 6, 12, 12, 24, 2, 4, 4, 18, 4, 18, 18, 36, 4, 8, 8, 54, 8, 16, 16, 32, 1, 2, 2, 6, 2, 6, 6, 12, 2, 6, 6, 24, 6, 12, 12, 24, 2

Offset: 0

Mikhail Kurkov, Oct 14 2023

			a(6) = 4 because the binary expansion of 6 is 110 and we have [(10), 1(10)] -> [1, 1]. Increasing these values by 1 gives us 2*2 = 4.
a(18) = 6 because the binary expansion of 18 is 10010 and we have [(10), (10)0(10)] -> [1, 2]. Increasing these values by 1 gives us 2*3 = 6.
a(26) = 18 because the binary expansion of 26 is 11010 and we have [(10), (10)(10), 1(10)(10)] -> [1, 2, 2]. Increasing these values by 1 gives us 2*3*3 = 18.
For n=482, the bits of n and the resulting product for a(n) are
  n    = 482 = binary 1 1 1 1 0 0 0 1 0
  a(n) = 162 =        3*3*3*3      *2
n=3863 = binary 111100010111 is the same a(n) = 162 since its final trailing "111" has no effect.

Cf. A033264, A053645, A346422.

A033264[n_] := SequenceCount[IntegerDigits[n, 2], {1, 0}];
A053645[n_] := n - 2^Floor@Log2@n;
a[n_] := a[n] = If[n == 0, 1, (1 + A033264[n]) a[A053645[n]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 14 2023 *)

a(n) = my(A = 1, B = 1); if(n, for(i=1, logint(n, 2), if(bittest(n, i), A *= (B += !bittest(n, i-1))))); A

a(2n + 1) = a(n).
a(4n) = a(2n) with a(0) = 1.
a(4n + 2) = 2*b(n), b(2n + 1) = 2*b(n), b(2n) = 3*c(n - 1, 1) with b(0) = 1.
c(2n + 1,  k) = c(n, k), c(4n + 2, k) = (k + 2)*c(2n, k), c(4n, k) = (k + 3)*c(n - 1, k + 1) with c(0, k) = 1.
Another way to compute a(4n + 2):
a(2*(4^n - 1)/3) = (n + 1)!.
a(2^(2m)*(2k + 1) + 2*(4^m - 1)/3) = (m + 1)*a(2^(2m)*k + 2*(4^m - 1)/3).
a(2^(2m + 1)*(2k + 1) + 2*(4^(m + 1) - 1)/3) = a(2^(2m + 1)*k + 2*(4^(m + 1) - 1)/3).
Note that a(4n + 2) is completely defined by these 3 last formulas. However, it looks like that it is not so easy to identify m and k for a given n, which makes these formulas useless for computing this sequence.

cp's OEIS Frontend

A347188 a(n) = A346422(4*A003754(n-1) + 3) for n > 1 with a(1) = 1.

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A366600 a(n) = (1 + A033264(n))*a(A053645(n)) for n > 0 with a(0) = 1.

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