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The largest divisor of n whose number of divisors is a power of 2.
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%I #13 Apr 30 2024 15:57:06

%S 1,2,3,2,5,6,7,8,3,10,11,6,13,14,15,8,17,6,19,10,21,22,23,24,5,26,27,

%T 14,29,30,31,8,33,34,35,6,37,38,39,40,41,42,43,22,15,46,47,24,7,10,51,

%U 26,53,54,55,56,57,58,59,30,61,62,21,8,65,66,67,34,69,70

%N The largest divisor of n whose number of divisors is a power of 2.

%C First differs from A350390 at n = 32.

%C The largest term of A036537 dividing n.

%C The largest divisor of n whose exponents in its prime factorization are all of the form 2^k-1 (A000225).

%H Amiram Eldar, <a href="/A372379/b372379.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = p^(2^floor(log_2(e+1)) - 1).

%F a(n) = n if and only if n is in A036537.

%F a(A162643(n)) = A282940(n).

%F Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 0.7907361848... = Product_{p prime} (1 + Sum_{k>=1} (p^f(k) - p^(f(k-1)+1))/p^(2*k)), f(k) = 2^floor(log_2(k))-1 for k >= 1, and f(0) = 0.

%t f[p_, e_] := p^(2^Floor[Log2[e + 1]] - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i, 1]^(2^exponent(f[i, 2]+1)-1));}

%o (Python)

%o from math import prod

%o from sympy import factorint

%o def A372379(n): return prod(p**((1<<(e+1).bit_length()-1)-1) for p, e in factorint(n).items()) # _Chai Wah Wu_, Apr 30 2024

%Y Cf. A000225, A000523, A036537, A138302, A162643, A282940, A350390, A353897, A372380, A372381.

%K nonn,easy,mult

%O 1,2

%A _Amiram Eldar_, Apr 29 2024