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.
Upper bounds for a(4) and a(5) are:
a(5) <= 538938984694949877040715541221415046162838700 =
A064989^4((
A047802(4)*17*19)/137).
(End)
Let prime(n)# be the n-th primorial number,
A002110(n) =
A034386(prime(n)). Then:
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.
