A284005 a(0) = 1, and for n > 1, a(n) = (1 + A000120(n))*a(floor(n/2)); also a(n) = A000005(A283477(n)).
1, 2, 4, 6, 8, 12, 18, 24, 16, 24, 36, 48, 54, 72, 96, 120, 32, 48, 72, 96, 108, 144, 192, 240, 162, 216, 288, 360, 384, 480, 600, 720, 64, 96, 144, 192, 216, 288, 384, 480, 324, 432, 576, 720, 768, 960, 1200, 1440, 486, 648, 864, 1080, 1152, 1440, 1800, 2160, 1536, 1920, 2400, 2880, 3000
Offset: 0
Links
Programs
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Mathematica
Table[DivisorSigma[0, #] &@ Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]], {n, 0, 71}] (* Michael De Vlieger, Mar 18 2017 *) -
PARI
A284005(n) = numdiv(A283477(n)); \\ edited by Michel Marcus, May 01 2019, M. F. Hasler, Nov 10 2019
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PARI
a(n) = my(k=if(n,logint(n,2)),s=1); prod(i=0,k, s+=bittest(n,k-i)); \\ Kevin Ryde, Jan 20 2021
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Scheme
(define (A284005 n) (A000005 (A283477 n)))
Formula
Conjecture: a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) + 2^k*(1 - T(n,k))) for n > 1 with a(0) = 1, a(1) = 2, f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2. - Mikhail Kurkov, Nov 10 2019
From Mikhail Kurkov, Aug 23 2021: (Start)
a(2n+1) = a(n) + a(2n) for n >= 0.
a(2n) = a(n) + a(2n - 2^A007814(n)) for n > 0 with a(0) = 1. (End)
Conjecture: a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*A329369(k). In other words, this sequence is modulo 2 binomial transform of A329369. - Mikhail Kurkov, Mar 10 2023
Conjecture: a(2^m*(2n+1)) = Sum_{k=0..m+1} binomial(m+1, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1. - Mikhail Kurkov, Apr 24 2023
Extensions
Made Mikhail Kurkov's Nov 10 2019 formula the new primary name of this sequence - Antti Karttunen, Dec 30 2020