A337468 -id:A337468 - cp's OEIS frontend

A337386 Numbers k for which A003973(k) >= 2*A003961(k).

120, 180, 240, 300, 360, 420, 480, 504, 540, 600, 630, 660, 720, 780, 840, 900, 924, 960, 990, 1008, 1020, 1050, 1080, 1092, 1140, 1170, 1200, 1260, 1320, 1380, 1440, 1470, 1500, 1512, 1560, 1620, 1650, 1680, 1740, 1800, 1848, 1860, 1890, 1920, 1980, 2016, 2040, 2100, 2160, 2184, 2220, 2280, 2310, 2340, 2400, 2460

Offset: 1

Antti Karttunen, Aug 27 2020

nonn

Provided that there are no odd perfect numbers, then these are equal to numbers k for which A003961(k) is in A005231, i.e., numbers that become odd abundant numbers when prime-shifted once.

Not all terms are even. The first odd term is a(8313165) = 334639305 = A064989(A115414(1)). (See A337385). For any odd term x present, A064989(x) is also present, for example, A064989(334639305) = 19399380 = a(482324).

Cf. A000203, A003961, A003973, A005231, A115414, A336389, A337468.
Subsequence of A005101, of A337381, and of A246282.
Subsequences: A337385 (odd terms), A337479 (primitive elements).

Select[Range[2500], If[# == 1, 1, DivisorSigma[1, # ]] >= 2# &@ Apply[Times, FactorInteger[#] /. {p_, e_} /; e > 0 :> Prime[PrimePi@ p + 1]^e] &] (* Michael De Vlieger, Aug 27 2020 *)

A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
isA337386(n) = (sigma(A003961(n))>=2*A003961(n));

A341508 a(n) = 0 if n is nonabundant, otherwise a(n) is the number of abundant divisors of the last abundant number in the iteration x -> A003961(x) (starting from x=n) before a nonabundant number is reached.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 5, 0, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0, 3, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 5, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1

Offset: 1

Antti Karttunen, Feb 20 2021

nonn

Question: Is a(A336389(n)) = 1 for all n >= 2? Note that all the terms of A047802 are obviously primitively abundant (in A091191).

			Starting from n = 120 = 2^3 * 3 * 5, the number of its abundant divisors is A080224(120) = 7. Then we apply a prime shift (A003961) to obtain the next number, 3^3 * 5 * 7 = 945, which has A080224(945) = 1 abundant divisors (as 945 is a term of A091191). The next prime shift gives 5^3 * 7* 11 = 9625, which has zero abundant divisors (as it is nonabundant, in A263837), so A080224(9625) = 0, and a(120) = 1, the last nonzero value encountered.

Cf. A003961, A047802, A091191, A336389, A336835, A337468.
Cf. A263837 (positions of zeros), A005101 (and of nonzeros).
Differs from A080224 for the first time at n=120, with a(120) = 1, while A080224(120) = 7.

A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A080224(n) = sumdiv(n, d, sigma(d)>2*d);
A341508(n) = { my(t, u=0); while((t=A080224(n))>0, u=t; n = A003961(n)); (u); };

cp's OEIS Frontend

A337386 Numbers k for which A003973(k) >= 2*A003961(k).

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