%I #56 Oct 21 2024 20:10:14
%S 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,
%T 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,
%U 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
%N a(n) = A284005(n) / (1 + A000120(n))!.
%C From _Antti Karttunen_, Feb 10 2021: (Start)
%C 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:
%C 1
%C |
%C ...................1...................
%C 2 1
%C 4......../ \........2 3......../ \........1
%C / \ / \ / \ / \
%C / \ / \ / \ / \
%C / \ / \ / \ / \
%C 8 4 6 2 9 3 4 1
%C 16 8 12 4 18 6 8 2 27 9 12 3 16 4 5 1
%C etc.
%C (End)
%C 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]
%C 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]
%H Alois P. Heinz, <a href="/A341392/b341392.txt">Table of n, a(n) for n = 0..16384</a>
%H Michael De Vlieger, <a href="/A341392/a341392.png">This sequence as a binary tree</a> showing rows 0 <= r <= 5.
%H Michael De Vlieger, <a href="/A341392/a341392_1.png">This sequence as a binary tree</a> showing rows 0 <= r <= 8.
%H FindStat, <a href="https://www.findstat.org/StatisticsDatabase/St000289oMp00105oMp00130oMp00236oMp00080/">Illustration of the sequence of mappings</a>
%F a(n) = A284005(n) / (1 + A000120(n))! = A284005(n) / A000142(1 + A000120(n)).
%F a(2n+1) = a(n) for n >= 0.
%F 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.
%F [2*a(n) - 1 = A329369(n)] = A036987(A053645(n)).
%F From _Mikhail Kurkov_, Apr 24 2023: (Start)
%F 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.
%F 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]
%p a:= proc(n) option remember; `if`(n=0, 1,
%p a(iquo(n, 2, 'd'))*`if`(d=1, 1, add(i, i=Bits[Split](n+1))))
%p end:
%p seq(a(n), n=0..120); # _Alois P. Heinz_, Jun 23 2021
%t 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 *)
%o (PARI)
%o A284005(n) = { my(k=if(n, logint(n, 2)), s=1); prod(i=0, k, s+=bittest(n, k-i)); }; \\ From A284005
%o A341392(n) = (A284005(n)/((1 + hammingweight(n))!)); \\ _Antti Karttunen_, Feb 10 2021
%o (PARI) A341392(n) = if(!n,1,if(n%2, A341392((n-1)/2), (1+hammingweight(n))*A341392(n/2))); \\ _Antti Karttunen_, Feb 10 2021
%Y Cf. A000120, A000142, A007814, A036987, A053645, A243499, A284005, A329369 (similar recurrence).
%K nonn,look
%O 0,3
%A _Mikhail Kurkov_, Feb 10 2021 [verification needed]