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a(n) is the number of times A324862(d) attains the maximal value it obtains among the divisors d of n.
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%I #6 Mar 27 2019 18:56:31

%S 1,2,2,1,2,1,2,1,1,4,2,3,2,1,1,1,2,2,2,1,1,1,2,3,3,1,1,3,2,2,2,2,4,1,

%T 1,1,2,1,1,1,2,2,2,3,2,1,2,2,3,6,1,3,2,2,1,3,1,1,2,2,2,1,2,1,4,3,2,3,

%U 4,2,2,1,2,1,1,3,1,2,2,2,1,1,2,2,1,1,1,3,2,1,1,3,4,1,1,1,2,2,1,1,2,2,2,3,1

%N a(n) is the number of times A324862(d) attains the maximal value it obtains among the divisors d of n.

%H Antti Karttunen, <a href="/A324869/b324869.txt">Table of n, a(n) for n = 1..10000</a> (based on Hans Havermann's factorization of A156552)

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>

%F a(n) = Sum_{d|n} [A324862(d) = A324864(n)], where [ ] is the Iverson bracket.

%F a(p) = 2 for all primes p.

%e Divisors of 9 are [1, 3, 4]. A324862 applied to these gives values [0, 0, 3], of which the largest (3) occurs just once, thus a(9) = 1.

%e Divisors of 10 are [1, 2, 5, 10]. A324862 applied to these gives values [0, 0, 0, 0], of which the largest (0) occurs just four times, thus a(10) = 4.

%e Divisors of 88 are [1, 2, 4, 8, 11, 22, 44, 88]. A324862 applied to these gives values [0, 0, 1, 0, 0, 1, 1, 0], of which the largest (which is 1) occurs three times, thus a(88) = 3.

%o (PARI) A324869(n) = { my(m=0,w,c=0); fordiv(n,d,w=A324862(d); if(w>=m,if(w==m,c++,c=1;m=w))); (c); };

%Y Cf. A324862, A324864.

%K nonn

%O 1,2

%A _Antti Karttunen_, Mar 21 2019