%I #38 Feb 17 2021 20:25:52

%S 0,0,0,1,0,2,0,2,1,2,0,4,0,2,2,3,0,4,0,4,2,2,0,6,1,2,2,4,0,6,0,4,2,2,

%T 2,6,0,2,2,6,0,6,0,4,4,2,0,8,1,4,2,4,0,6,2,6,2,2,0,9,0,2,4,5,2,6,0,4,

%U 2,6,0,8,0,2,4,4,2,6,0,8,3,2,0,9,2,2,2,6,0,9,2,4,2,2,2,10,0,4,4,6,0,6,0,6

%N a(n) = (A001222(n) - 1)*A001221(n).

%C a(n) + 2 appears to differ from A000005 at n=1 and when n is a term of A320632. Verified up to n=3000.

%C If A320632 contains the numbers such that A001222(n) - A051903(n) > 1, then this sequence contains precisely the numbers p^k and p^k*q for distinct primes p and q. The comment follows, since d(p^k) = k+1 = (k-1)*1 + 2 and d(p^k*q) = 2k+2 = ((k+1)-1)*2 + 2. - _Charlie Neder_, May 14 2019

%C Positions of first appearances are A328965. - _Gus Wiseman_, Nov 05 2019

%C Regarding Neder's comment above, see also my comments in A322437. - _Antti Karttunen_, Feb 17 2021

%H Antti Karttunen, <a href="/A307409/b307409.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%F a(n) = (A001222(n) - 1)*A001221(n).

%F a(n) = binomial(A001222(n) - 1, 1)*binomial(A001221(n), 1).

%F a(n) = A307408(n) - 2.

%t a[n_] := (PrimeOmega[n] - 1)*PrimeNu[n];

%t aa = Table[a[n], {n, 1, 104}];

%o (PARI) a(n) = (bigomega(n) - 1)*omega(n); \\ _Michel Marcus_, May 15 2019

%Y Two less than A307408.

%Y A113901(n) is bigomega(n) * omega(n).

%Y A328958(n) is sigma_0(n) - bigomega(n) * omega(n).

%Y Cf. A000005, A001221, A001222, A060687, A070175, A071625, A113901, A124010, A303555, A320632, A323023, A328956, A328957, A328964, A328965, A322437.

%K nonn

%O 1,6

%A _Mats Granvik_, Apr 07 2019