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Number of divisors of A025487(n) that are at most sqrt(A025487(n)).
2

%I #15 Mar 25 2026 00:00:03

%S 1,1,2,2,2,3,3,4,4,3,5,5,6,4,6,6,8,4,8,9,7,8,8,10,5,9,12,8,12,10,12,5,

%T 11,15,9,16,12,14,14,6,16,12,18,13,18,10,20,14,18,16,6,20,14,16,24,15,

%U 21,11,24,16,23,18,7,24,15,24,30,18,24,24,12,27,25,28,18,27,32,20,18,7

%N Number of divisors of A025487(n) that are at most sqrt(A025487(n)).

%H Amiram Eldar, <a href="/A108504/b108504.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A038548(A025487(n)).

%o (Python)

%o from itertools import count

%o from functools import lru_cache

%o from sympy import prime, integer_log, primorial, divisor_count

%o from oeis_sequences.OEISsequences import bisection

%o def A108504(n):

%o if n == 1: return 1

%o @lru_cache(maxsize=None)

%o def g(x, m, j): return sum(g(x//(prime(m)**i), m-1, i) for i in range(j,integer_log(x, prime(m))[0]+1)) if m-1 else max(0,x.bit_length()-j)

%o def f(x):

%o c = n-1+x

%o for k in count(1):

%o if primorial(k)>x:

%o break

%o c -= g(x,k,1)

%o return c

%o return divisor_count(bisection(f,n,n))+1>>1 # _Chai Wah Wu_, Mar 24 2026

%Y Cf. A025487, A038548.

%K nonn

%O 1,3

%A _Christian G. Bower_, Jun 06 2005

%E Offset corrected by _Charles R Greathouse IV_, Sep 28 2012