cp's OEIS Frontend

A072989 lists the indices for which a(n) differs from A050399(n), e.g., in n = 20, 40, 52, ... in addition to the zeros in this sequence (n = 30, 42, 66, 70, 78, 90, ...). See also A009195 vs. A072994. [Corrected and extended by M. F. Hasler, Feb 23 2014]

The sequence seems difficult to extend, as the next term a(30) is larger than 5100. However, a(32)=64, a(64)=128 and a(128)=256 can be easily calculated. It thus appears that a(2^k)=2^(k+1), for k=1,2,3,.... Is this known to be true? - John W. Layman, Aug 05 2003 -- Answer: It's true. One could have defined the sequence so that a(1)=2: then it would be true for 2^0 also. - Don Reble, Feb 23 2014

a(30), if it exists, is greater than 400000. - Ryan Propper, Sep 10 2005

a(30) doesn't exist: If N is even, and divisible by D different odd primes, but not divisible by 2^D, then a(N) doesn't exist. - Don Reble, Feb 23 2014 [This and the preceding comment refer to the former definition lacking the clause "0 if no such k exists". - Ed.]

Conjecture: a(n)=0 iff n/2 is in A061346. - Robert G. Wilson v, Feb 23 2014

[n=420 seems to be a counterexample to the above conjecture. - M. F. Hasler, Feb 24 2014]

From Robert G. Wilson v, Mar 05 2014: (Start)

Observation:

If n = 1 then a(n) = 1 by definition;

If, but not iff, n (an even number) is a member of A238367 then a(n) = 0;

If n (an even number not in A238367) is {684, 954, ...}, then a(n) = 0;

If n (an odd number) is {273, 399, 651, 741, 777, 903, ...}, then a(n) = 0;

If p is a prime [A000040] and e is its exponent, then a(p^e) = p^(e+1);

If p is a prime then a(2p^e) = 2p^(e+1);

If p is a prime then a(n) # p since the f(p)=1.

(End)

Often A072995(n) equals A050399(n). They differ at n: 20, 30, 40, 42, 52, 60, 66, 68, 70, 78, 80, 84, 90, 100, 102, 104, 110, 114, 116, 120, 126, 130, 132, ... - Robert G. Wilson v, Dec 06 2014

When A072995(n)>0 and does not equal A050399(n): 20, 40, 52, 60, 68, 80, 84, 100, 104, 116, 120, 132, 136, 140, 148, 156, 160, 164, 168, 171, 180, 200, ... - Robert G. Wilson v, Dec 06 2014

When a(n) > 1, then 2n <= a(n) <= n^2. - Robert G. Wilson v, Dec 10 2014