A341392 a(n) = A284005(n) / (1 + A000120(n))!.

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 6, 2, 9, 3, 4, 1, 16, 8, 12, 4, 18, 6, 8, 2, 27, 9, 12, 3, 16, 4, 5, 1, 32, 16, 24, 8, 36, 12, 16, 4, 54, 18, 24, 6, 32, 8, 10, 2, 81, 27, 36, 9, 48, 12, 15, 3, 64, 16, 20, 4, 25, 5, 6, 1, 64, 32, 48, 16, 72, 24, 32, 8, 108, 36, 48, 12, 64, 16, 20, 4, 162, 54, 72, 18, 96, 24, 30, 6, 128

Offset: 0

Author

Mikhail Kurkov, Feb 10 2021 [verification needed]

Keywords

• nonn
• look

Comments

From Antti Karttunen, Feb 10 2021: (Start)

This sequence can be represented as a binary tree. Each child to the left is obtained by multiplying its parent with (1+{binary weight of its breadth-first-wise index in the tree}), while each child to the right is just a clone of its parent:

1

|

...................1...................

2 1

4......../ \........2 3......../ \........1

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

8 4 6 2 9 3 4 1

16 8 12 4 18 6 8 2 27 9 12 3 16 4 5 1

etc.

(End)

This sequence and A243499 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]

FindStat provides a sequence of mappings between this sequence and A000110 starting from collection [Set partitions] (see Links section for illustration). - Mikhail Kurkov, May 20 2023 [verification needed]

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16384
• Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 5.
• Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 8.
• FindStat, Illustration of the sequence of mappings

Crossrefs

Cf. A000120, A000142, A007814, A036987, A053645, A243499, A284005, A329369 (similar recurrence).

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1,
      a(iquo(n, 2, 'd'))*`if`(d=1, 1, add(i, i=Bits[Split](n+1))))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Jun 23 2021

Mathematica

Array[DivisorSigma[0, Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#1] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]]]/#2 & @@ {Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ #, (1 + Count[#, 1])!} &@ IntegerDigits[#, 2] &, 89, 0] (* Michael De Vlieger, Feb 24 2021 *)

PARI

A284005(n) = { my(k=if(n, logint(n, 2)), s=1); prod(i=0, k, s+=bittest(n, k-i)); }; \\ From A284005
A341392(n) = (A284005(n)/((1 + hammingweight(n))!)); \\ Antti Karttunen, Feb 10 2021

PARI

A341392(n) = if(!n,1,if(n%2, A341392((n-1)/2), (1+hammingweight(n))*A341392(n/2))); \\ Antti Karttunen, Feb 10 2021

Formula

a(n) = A284005(n) / (1 + A000120(n))! = A284005(n) / A000142(1 + A000120(n)).

a(2n+1) = a(n) for n >= 0.

a(2n) = (1 + A000120(n))*a(n) = A243499(2*A059894(n)) = a(n) + a(2n - 2^A007814(n)) for n > 0 with a(0) = 1.

[2*a(n) - 1 = A329369(n)] = A036987(A053645(n)).

From Mikhail Kurkov, Apr 24 2023: (Start)

a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1.

a(n) = a(f(n)) + Sum_{k=0..floor(log_2(n))-1} (1 - T(n, k))*a(f(n) + 2^k*(1 - T(n, k))) for n > 1 with a(0) = 1, a(1) = 1, where f(n) = A053645(n) and where T(n, k) = floor(n/2^k) mod 2. (End) [verification needed]