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a(n) is the number of distinct prime factors of A225546(n), or equally, number of distinct prime factors of A293442(n).
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%I #47 Dec 23 2023 09:33:53

%S 0,1,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,2,1,1,1,2,1,1,2,2,1,1,1,2,1,1,

%T 1,1,1,1,1,2,1,1,1,2,2,1,1,2,1,2,1,2,1,2,1,2,1,1,1,2,1,1,2,2,1,1,1,2,

%U 1,1,1,2,1,1,2,2,1,1,1,2,1,1,1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,2,2,1,1,1,1,2,1

%N a(n) is the number of distinct prime factors of A225546(n), or equally, number of distinct prime factors of A293442(n).

%C a(n) is the number of terms in the unique factorization of n into powers of squarefree numbers with distinct exponents that are powers of 2. See A329332 for a description of the relationship between this factorization, canonical (prime power) factorization and A225546.

%C The result depends only on the prime signature of n.

%C a(n) is the number of distinct bit-positions where there is a 1-bit in the binary representation of an exponent in the prime factorization of n. - _Antti Karttunen_, Feb 05 2020

%C The first 3 is a(128) = a(2^1 * 2^2 * 2^4) = 3 and in general each m occurs first at position 2^(2^m-1) = A058891(m+1). - _Peter Munn_, Mar 07 2022

%H Antti Karttunen, <a href="/A331591/b331591.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

%F a(n) = A001221(A293442(n)) = A001221(A225546(n)).

%F From _Peter Munn_, Jan 28 2020: (Start)

%F a(n) = A000120(A267116(n)).

%F a(n) = a(A007913(n)) + a(A008833(n)).

%F For m >= 2, a(A005117(m)) = 1.

%F a(n^2) = a(n).

%F (End)

%F a(n) <= A331740(n) <= A048675(n) <= A293447(n). - _Antti Karttunen_, Feb 05 2020

%F From _Peter Munn_, Mar 07 2022: (Start)

%F a(n) <= A299090(n).

%F a(A337533(n)) = A299090(A337533(n)).

%F a(A337534(n)) < A299090(A337534(n)).

%F max(a(n), a(k)) <= a(A059796(n, k)) = a(A331590(n, k)) <= a(n) + a(k).

%F (End)

%e From _Peter Munn_, Jan 28 2020: (Start)

%e The factorization of 6 into powers of squarefree numbers with distinct exponents that are powers of 2 is 6 = 6^(2^0) = 6^1, which has 1 term. So a(6) = 1.

%e Similarly, 40 = 10^(2^0) * 2^(2^1) = 10^1 * 2^2 = 10 * 4, which has 2 terms. So a(40) = 2.

%e Similarly, 320 = 5^(2^0) * 2^(2^1) * 2^(2^2) = 5^1 * 2^2 * 2^4 = 5 * 4 * 16, which has 3 terms. So a(320) = 3.

%e 10^100 (a googol) factorizes in this way as 10^4 * 10^32 * 10^64. So a(10^100) = 3.

%e (End)

%t Array[PrimeNu@ If[# == 1, 1, Times @@ Flatten@ Map[Function[{p, e}, Map[Prime[Log2@ # + 1]^(2^(PrimePi@ p - 1)) &, DeleteCases[NumberExpand[e, 2], 0]]] @@ # &, FactorInteger[#]]] &, 105] (* _Michael De Vlieger_, Jan 24 2020 *)

%t f[e_] := Position[Reverse[IntegerDigits[e, 2]], 1] // Flatten; a[n_] := CountDistinct[Flatten[f /@ FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100] (* _Amiram Eldar_, Dec 23 2023 *)

%o (PARI) A331591(n) = if(1==n,0,my(f=factor(n),u=#binary(vecmax(f[, 2])),xs=vector(u),m=1,e); for(i=1,u,for(k=1,#f~, if(bitand(f[k,2],m),xs[i]++)); m<<=1); #select(x -> (x>0),xs));

%o (PARI) A331591(n) = if(1==n, 0, hammingweight(fold(bitor, factor(n)[, 2]))); \\ _Antti Karttunen_, Feb 05 2020

%o (PARI) A331591(n) = if(n==1, 0, (core(n)>1) + A331591(core(n,1)[2])) \\ _Peter Munn_, Mar 08 2022

%Y Cf. A000120, A005117, A048675, A225546, A267116, A293442, A293447, A299090, A329332, A337533, A337534.

%Y Sequences with related definitions: A001221, A331309, A331592, A331593, A331740.

%Y Positions of records: A058891.

%Y Positions of 1's: A340682.

%Y Sequences used to express relationships between the terms: A007913, A008833, A059796, A331590.

%K nonn

%O 1,8

%A _Antti Karttunen_ and _Peter Munn_, Jan 21 2020