OFFSET
0,2
COMMENTS
For n > 0, the least k such that for at least n-1 iterations of map x -> A003961(x), starting from x=k, x stays nondeficient. In other words, from each a(n) starts a chain of at least n nondeficient numbers (A023196) obtained by successive prime shifts, e.g, for a(3) we have: 19399380 -> 334639305 -> 5391411025, where -> stands for applying A003961, the prime shift towards larger primes.
After 1 all other terms here are even, because if an odd number k is nondeficient, then A064989(k) is nondeficient also, where A064989 is the prime shift towards smaller primes. Moreover, because A047802 is defined for every n >= 0, also this sequence is.
From Peter Munn, Aug 13 2020 (Start)
Upper bounds for a(4) and a(5) are:
(End)
From David A. Corneth, Aug 21 2020: (Start)
Subsequence of A025487.
a(6) <= 191# * 7#;
a(7) <= 311# * 5#;
a(8) <= 457# * 5#.
(End)
That each term occurs in A025487 follows because (1), the abundancy index of prime(i)^e is larger than that of prime(i+1)^e, that is, sigma(prime(i)^e)/prime(i)^e > sigma(prime(i+1)^e)/prime(i+1)^e, and (2) because the abundancy index of p^(e+d) * q^e is larger than that of p^e * q^(e+d), where p and q are distinct primes, p < q, and e, d > 0. Thus, for any n, we can first find a "prime-factorization compressed version" of it, A071364(n), and then sort the exponents to the non-ascending order with A046523 (and actually, A046523(A071364(n)) = A046523(n), so we need to apply just A046523), to get a term x of A025487, that certainly have the abundancy index >= n [and this inequivalence stays same for their successive prime shifts as well, the abundancy index of A003961(x) being at least that of A003961(n), etc.], and as A046523(n) <= n for all n, it is guaranteed that the least k for which A336835(k) >= n are found from A025487, which is the range of A046523.
LINKS
FORMULA
PROG
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Antti Karttunen, Aug 07 2020
EXTENSIONS
a(4) - a(6) from combined work of David A. Corneth and Peter Munn Aug 13-26 2020
STATUS
approved