A336389 The least positive integer k for which A336835(k) >= n, where A336835(k) is the number of iterations of x -> A003961(x) needed before the result is deficient (sigma(x) < 2x), when starting from x=k.

1, 6, 120, 19399380, 195534950863140268380, 538938984694949877040715541221415046162838700, 216487559804430601784907786655491617909711008142914104790481010259258659171900

Offset: 0

Author

Antti Karttunen, Aug 07 2020

Keywords

• nonn
• hard
• more

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:

a(4) <= 195534950863140268380 = A064989(A064989(A064989(20169691981106018776756331))) = A337202(3).

a(5) <= 538938984694949877040715541221415046162838700 = A064989^4((A047802(4)*17*19)/137).

(End)

From David A. Corneth, Aug 21 2020: (Start)

Subsequence of A025487.

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.

Links

• Index entries for sequences computed from indices in prime factorization
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A003961, A005100, A023196, A025487, A047802, A108951, A246277, A329900, A336834, A336835, A337202, A337205, A337477, A337478.

From term a(2) = 120 onward a subsequence of A337386.

Programs

PARI

A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961
A336835(n) = { my(i=0); while(sigma(n) >= (n+n), i++; n = A003961(n)); (i); };
A336389(n) = for(i=1,oo,if(A336835(i)>=n,return(i)));

Formula

For n >= 0, A336835(a(n)) >= n.

For all n >= 1, a(n) <= A337202(n-1) [= 2*A246277(A047802(n-1))].

a(n) = A025487(A337477(n)).

a(n) = A108951(A337478(n)).

Extensions

a(4) - a(6) from combined work of David A. Corneth and Peter Munn Aug 13-26 2020