1,3

Not the same as A064380.

With n having a unique factorization as A050376(i) * A050376(j) * ... * A050376(k), with i, j, ..., k all distinct, a(n) = (A050376(i)-1) * (A050376(j)-1) * ... * (A050376(k)-1). (Cf. the first formula). - Antti Karttunen, Jan 15 2019

Antti Karttunen, Table of n, a(n) for n = 1..21011

Graeme L. Cohen and Peter Hagis, Arithmetic functions associated with the infinitary divisors of an integer, Internat. J. Math. Math. Sci. 16 (1993) 373-383.

Steven R. Finch, Unitarism and infinitarism, 2004.

Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]

Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000

Consider the set, I, of integers of the form p^(2^j), where p is any prime and j >= 0. Let n > 1. From the fundamental theorem of arithmetic and the fact that the binary representation of any integer is unique, it follows that n can be uniquely factored as a product of distinct elements of I. If n = P_1*P_2*...*P_t, where each P_j is in I, then iphi(n) = Product_{j=1..t} (P_j - 1).

From Vladimir Shevelev, Feb 20 2011: (Start)

Thus we have the following analog of the formula phi(n) = n*Product_{p prime divisors of n} (1-1/p): if the factorization of n over distinct terms of A050376 is n = Product(q) (this factorization is unique), then a(n) = n*Product(1-1/q). Thus a(n) is infinitary multiplicative, i.e., if n_1 and n_2 have no common i-divisors, then a(n_1*n_2) = a(n_1)*a(n_2). Now we see that this property is stronger than the usual multiplicativity, therefore a(n) is a multiplicative arithmetic function.

Add that Sum_{d runs i-divisors of n} a(d)=n and a(n) = n*Sum_{d runs i-divisors of n} A064179(d)/d. The latter formulas are analogs of the corresponding formulas for phi(n): Sum_{d|n} phi(d) = n and phi(n) = n*Sum_{d|n} mu(d)/d. (End).

a(n) = n - A323413(n). - Antti Karttunen, Jan 15 2019

a(n) <= A064380(n), with equality if and only if n is in A050376. - Amiram Eldar, Feb 18 2023

a(6)=2 since 6=P_1*P_2, where P_1=2^(2^0) and P_2=3^(2^0); hence (P_1-1)*(P_2-1)=2.

12=3*4 (3,4 are in A050376). Therefore, a(12) = 12*(1-1/3)*(1-1/4) = 6. - Vladimir Shevelev, Feb 20 2011

A091732 := proc(n) local f, a, e, p, b; a :=1 ; for f in ifactors(n)[2] do e := op(2, f) ; p := op(1, f) ; b := convert(e, base, 2) ; for i from 1 to nops(b) do if op(i, b) > 0 then a := a*(p^(2^(i-1))-1) ; end if; end do: end do: a ; end proc:

seq(A091732(n), n=1..20) ; # R. J. Mathar, Apr 11 2011

f[p_, e_] := p^(2^(-1 + Position[Reverse @ IntegerDigits[e, 2], 1])); a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) - 1); Array[a, 100] (* Amiram Eldar, Feb 28 2020 *)

(PARI)

ispow2(n) = (n && !bitand(n, n-1));

A302777(n) = ispow2(isprimepower(n));

A091732(n) = { my(m=1); while(n > 1, fordiv(n, d, if((d<n)&&A302777(n/d), m *= ((n/d)-1); n = d; break))); (m); }; \\ Antti Karttunen, Jan 15 2019

Cf. A037445, A049417, A050376, A064380, A300841, A323413.

Sequence in context: A117744 A366481 A361697 * A299439 A109746 A384247

Adjacent sequences: A091729 A091730 A091731 * A091733 A091734 A091735

nonn,mult

Steven Finch, Mar 05 2004

approved