A366600 - OEIS

A366600

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

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

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OFFSET

0,3

LINKS

Table of n, a(n) for n=0..79.

FORMULA

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.

EXAMPLE

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.

MATHEMATICA

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 *)

PROG

(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

CROSSREFS