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 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. 11
 1, 6, 120, 19399380, 195534950863140268380, 538938984694949877040715541221415046162838700, 216487559804430601784907786655491617909711008142914104790481010259258659171900 (list; graph; refs; listen; history; text; internal format)
 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: 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 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)). PROG (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))); 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. Sequence in context: A137149 A317888 A053710 * A126244 A336808 A296801 Adjacent sequences:  A336386 A336387 A336388 * A336390 A336391 A336392 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

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Last modified June 27 13:59 EDT 2022. Contains 354896 sequences. (Running on oeis4.)