A366244 The largest infinitary divisor of n that is a term of A366242.

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 16, 17, 2, 19, 5, 21, 22, 23, 6, 1, 26, 3, 7, 29, 30, 31, 32, 33, 34, 35, 1, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 48, 1, 2, 51, 13, 53, 6, 55, 14, 57, 58, 59, 15, 61, 62, 7, 16, 65, 66, 67, 17, 69, 70, 71

Offset: 1

Author

Amiram Eldar, Oct 05 2023

Keywords

• nonn
• easy
• mult

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A063694, A366242, A366243, A366245.

See the formula section for the relationships with A007913, A046100, A059895, A059896, A059897, A225546, A247503, A352780.

Programs

Mathematica

f[p_, e_] := p^BitAnd[e, Sum[2^k, {k, 0, Floor@ Log2[e], 2}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

s(e) = sum(k = 0, e, (-2)^k*floor(e/2^k));
a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^s(f[i,2]));}

Formula

Multiplicative with a(p^e) = p^A063694(e).

a(n) = n / A366245(n).

a(n) >= 1, with equality if and only if n is a term of A366243.

a(n) <= n, with equality if and only if n is a term of A366242.

Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (1-1/p)*(Sum_{k>=1} p^(A063694(k)-2*k)) = 0.35319488024808595542... .

From Peter Munn, Jan 09 2025: (Start)

a(n) = max({k in A366242 : A059895(k, n) = k}).

a(n) = Product_{k >= 0} A352780(n, 2k).

Also defined by:

- for n in A046100, a(n) = A007913(n);

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

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

Other identities:

a(n) = sqrt(A366245(n^2)).

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

a(A225546(n)) = A225546(A247503(n)).

(End)