OFFSET
1,4
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
FORMULA
a(n) = n - A003958(n).
From Antti Karttunen, Nov 07 2021: (Start)
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)
MATHEMATICA
a[1] = 0; a[n_] := n - Times @@ ((First[#] - 1)^Last[#] & /@ FactorInteger[n]); Array[a, 60] (* Amiram Eldar, Dec 17 2018 *)
PROG
(PARI)
A003958(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]--); factorback(f); };
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Dec 17 2018
STATUS
approved