A322582 a(n) = n - A003958(n), where A003958 is fully multiplicative with a(p) = (p-1).

0, 1, 1, 3, 1, 4, 1, 7, 5, 6, 1, 10, 1, 8, 7, 15, 1, 14, 1, 16, 9, 12, 1, 22, 9, 14, 19, 22, 1, 22, 1, 31, 13, 18, 11, 32, 1, 20, 15, 36, 1, 30, 1, 34, 29, 24, 1, 46, 13, 34, 19, 40, 1, 46, 15, 50, 21, 30, 1, 52, 1, 32, 39, 63, 17, 46, 1, 52, 25, 46, 1, 68, 1, 38, 43, 58, 17, 54, 1, 76, 65, 42, 1, 72, 21, 44, 31, 78, 1

Offset: 1

Author

Antti Karttunen, Dec 17 2018

Keywords

• nonn

Comments

a(p*(n/p)) - (n/p) = (p-1)*a(n/p) holds for all prime divisors p of n, which can be seen by expanding the left hand side as p*(n/p) - A003958(p*(n/p)) - (n/p) = (p-1)*(n/p) - (p-1)*A003958(n/p) = (p-1)*((n/p) - A003958(n/p)) = (p-1)*a(n/p). This shows that this sequence gives a lower limit for arithmetic derivative (A003415) in the same way as A348507 gives an upper limit for it. - Antti Karttunen, Nov 07 2021

With n = Product_{i=1..k} p_i the prime factorization of n, if one constructs for each i a test with a probability of success equal to 1/p_i, and if the tests are independent, then a(n)/n is the probability that at least one of the k tests succeeds. - Luc Rousseau, Jan 14 2023

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A003415, A003958, A322581, A348507, A348928 [= gcd(n,a(n))], A348975 (difference from the arithmetic derivative).

Cf. A349139, A348980, A348981, A348982, A348983 (Dirichlet convolutions with other sequences).

Cf. A168065 (gives the arithmetic mean of this and A348507), A168066.

Programs

Mathematica

a[1] = 0; a[n_] := n - Times @@ ((First[#] - 1)^Last[#] & /@ FactorInteger[n]); Array[a, 60] (* Amiram Eldar, Dec 17 2018 *)

PARI

A003958(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]--); factorback(f); };
A322582(n) = (n-A003958(n));

PARI

A020639(n) = if(1==n, n, (factor(n)[1, 1]));
A322582(n) = { my(s=0, m=1, spf); while(n>1, spf = A020639(n); n /= spf; s += m*n; m *= (spf-1)); (s); }; \\ (Compare to the similar programs given in A003415 and A348507) - Antti Karttunen, Nov 07 2021

Formula

a(n) = n - A003958(n).

From Antti Karttunen, Nov 07 2021: (Start)

a(n) = A003415(n) - A348975(n).

For all n >= 1, a(n) <= A003415(n) <= A348507(n).

For n > 1, a(n) = a(A032742(n))*(A020639(n)-1) + A032742(n). [See the comment above and compare with Reinhard Zumkeller's May 09 2011 formula for A003415]

(End)