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The largest term of A054743 that divide n.
5

%I #6 Dec 21 2023 21:15:04

%S 1,1,1,1,1,1,1,8,1,1,1,1,1,1,1,16,1,1,1,1,1,1,1,8,1,1,1,1,1,1,1,32,1,

%T 1,1,1,1,1,1,8,1,1,1,1,1,1,1,16,1,1,1,1,1,1,1,8,1,1,1,1,1,1,1,64,1,1,

%U 1,1,1,1,1,8,1,1,1,1,1,1,1,16,81,1,1,1,1

%N The largest term of A054743 that divide n.

%C First differs from A360540 at n = 27.

%C The largest divisor d of n such that e > p for all prime powers p^e in the prime factorization of d (i.e., e >= 1 and p^(e+1) does not divide d).

%H Amiram Eldar, <a href="/A368329/b368329.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = 1 if e <= p, and a(p^e) = p^e if e > p.

%F A034444(a(n)) = A368330(n).

%F a(n) >= 1, with equality if and only if n is in A207481.

%F a(n) <= n, with equality if and only if n is in A054743.

%F Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^((p+2)*s-1) - 1/p^((p+2)*(s-1)+1) - 1/p^((p+1)*s) + 1/p^((p+1)*(s-1))).

%t f[p_, e_] := If[e <= p, 1, p^e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

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

%Y Cf. A054743, A034444, A207481, A327939, A360540, A368329, A368330, A368331, A368333.

%K nonn,easy,mult

%O 1,8

%A _Amiram Eldar_, Dec 21 2023