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

%I A366600 #28 Apr 23 2024 11:20:05
%S A366600 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,
%T A366600 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,
%U A366600 16,32,1,2,2,6,2,6,6,12,2,6,6,24,6,12,12,24,2
%N A366600 a(n) = (1 + A033264(n))*a(A053645(n)) for n > 0 with a(0) = 1.
%F A366600 a(2n + 1) = a(n).
%F A366600 a(4n) = a(2n) with a(0) = 1.
%F A366600 a(4n + 2) = 2*b(n), b(2n + 1) = 2*b(n), b(2n) = 3*c(n - 1, 1) with b(0) = 1.
%F A366600 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.
%F A366600 Another way to compute a(4n + 2):
%F A366600 a(2*(4^n - 1)/3) = (n + 1)!.
%F A366600 a(2^(2m)*(2k + 1) + 2*(4^m - 1)/3) = (m + 1)*a(2^(2m)*k + 2*(4^m - 1)/3).
%F A366600 a(2^(2m + 1)*(2k + 1) + 2*(4^(m + 1) - 1)/3) = a(2^(2m + 1)*k + 2*(4^(m + 1) - 1)/3).
%F A366600 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.
%e A366600 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.
%e A366600 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.
%e A366600 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.
%e A366600 For n=482, the bits of n and the resulting product for a(n) are
%e A366600   n    = 482 = binary 1 1 1 1 0 0 0 1 0
%e A366600   a(n) = 162 =        3*3*3*3      *2
%e A366600 n=3863 = binary 111100010111 is the same a(n) = 162 since its final trailing "111" has no effect.
%t A366600 A033264[n_] := SequenceCount[IntegerDigits[n, 2], {1, 0}];
%t A366600 A053645[n_] := n - 2^Floor@Log2@n;
%t A366600 a[n_] := a[n] = If[n == 0, 1, (1 + A033264[n]) a[A053645[n]]];
%t A366600 Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Nov 14 2023 *)
%o A366600 (PARI) 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
%Y A366600 Cf. A033264, A053645, A346422.
%K A366600 nonn,base
%O A366600 0,3
%A A366600 _Mikhail Kurkov_, Oct 14 2023