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The largest infinitary divisor of n that is a term of A366243.
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%I #13 Jan 09 2025 15:44:26

%S 1,1,1,4,1,1,1,4,9,1,1,4,1,1,1,1,1,9,1,4,1,1,1,4,25,1,9,4,1,1,1,1,1,1,

%T 1,36,1,1,1,4,1,1,1,4,9,1,1,1,49,25,1,4,1,9,1,4,1,1,1,4,1,1,9,4,1,1,1,

%U 4,1,1,1,36,1,1,25,4,1,1,1,1,1,1,1,4,1

%N The largest infinitary divisor of n that is a term of A366243.

%C First differs from A335324 at n = 256.

%H Amiram Eldar, <a href="/A366245/b366245.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = p^A063695(e).

%F a(n) = n / A366244(n).

%F a(n) >= 1, with equality if and only if n is a term of A366242.

%F a(n) <= n, with equality if and only if n is a term of A366243.

%F From _Peter Munn_, Jan 09 2025: (Start)

%F a(n) = max({k in A366243 : A059895(k, n) = k}).

%F a(n) = Product_{k >= 0} A352780(n, 2k+1).

%F Also defined by:

%F - for n in A046100, a(n) = A008833(n);

%F - a(n^4) = (a(n))^4;

%F - a(A059896(n,k)) = A059896(a(n), a(k)).

%F Other identities:

%F a(n) = sqrt(A366244(n^2)).

%F a(A059897(n,k)) = A059897(a(n), a(k)).

%F a(A225546(n)) = A225546(A248101(n)).

%F (End)

%t f[p_, e_] := p^BitAnd[e, Sum[2^k, {k, 1, Floor@ Log2[e], 2}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) s(e) = -sum(k = 1, e, (-2)^k*floor(e/2^k));

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

%Y Cf. A063695, A335324, A366242, A366243, A366244.

%Y See the formula section for the relationships with A008833, A046100, A059895, A059896, A059897, A225546, A248101, A352780.

%K nonn,easy,mult

%O 1,4

%A _Amiram Eldar_, Oct 05 2023