A366245 The largest infinitary divisor of n that is a term of A366243.
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, 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, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
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]
-
PARI
s(e) = -sum(k = 1, 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^A063695(e).
a(n) = n / A366244(n).
a(n) >= 1, with equality if and only if n is a term of A366242.
a(n) <= n, with equality if and only if n is a term of A366243.
From Peter Munn, Jan 09 2025: (Start)
a(n) = Product_{k >= 0} A352780(n, 2k+1).
Also defined by:
- a(n^4) = (a(n))^4;
Other identities:
a(n) = sqrt(A366244(n^2)).
(End)
Comments