A334204 a(n) = A329697(A163511(n)).

0, 0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 2, 0, 4, 3, 3, 2, 3, 2, 4, 1, 3, 2, 3, 1, 3, 2, 2, 0, 5, 4, 4, 3, 4, 3, 6, 2, 4, 3, 5, 2, 5, 4, 4, 1, 4, 3, 4, 2, 4, 3, 4, 1, 4, 3, 3, 2, 3, 2, 2, 0, 6, 5, 5, 4, 5, 4, 8, 3, 5, 4, 7, 3, 7, 6, 6, 2, 5, 4, 6, 3, 6, 5, 6, 2, 6, 5, 5, 4, 5, 4, 4, 1, 5, 4, 5, 3, 5, 4, 6, 2, 5

Offset

0,6

Comments

As the underlying sequence A163511 can be represented as a binary tree, so can be this:

0

|

...................0...................

0 1

0......../ \........2 1......../ \........1

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

0 3 2 2 1 2 1 2

0 4 3 3 2 3 2 4 1 3 2 3 1 3 2 2

etc.

The nodes at the left edge are all zeros, and their right-hand children give positive integers, A000027.

Each left-hand leaning branch stays constant, because A329697(2n) = A329697(n).

The right-hand leaning branches are not necessarily monotonic. For example, a((2^6)-1) = 2 > 1 = a((2^7)-1), because A000040(7) = 17 is a Fermat prime (but A000040(6) = 13 is not), and therefore the latter is only one step away from a power of 2.

Links

Antti Karttunen, Table of n, a(n) for n = 0..65537

Index entries for sequences related to binary expansion of n

Formula

a(n) = A329697(A163511(n)).

a(n) = A334109(A334860(n)).

a(n) = a(2n) = a(A000265(n)).

For all n >= 0, a(2^n) = 0, a(2^n + 1) = n.

Prog

(PARI)
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
A054429(n) = ((3<<#binary(n\2))-n-1); \\ From A054429
A163511(n) = if(!n, 1, A005940(1+A054429(n)));
A329697(n) = if(!bitand(n, n-1), 0, 1+A329697(n-(n/vecmax(factor(n)[, 1]))));
A334204(n) = A329697(A163511(n));

Crossrefs

Cf. A000079, A163511, A329697, A334109, A334860, A334867, A334873.

Sequence in context: A080791 A336361 A364260 * A336362 A336363 A308451

Adjacent sequences: A334201 A334202 A334203 * A334205 A334206 A334207

Keyword

nonn

Author

Antti Karttunen, Jun 09 2020

Status

approved