A378662 Number of divisors d of n such that sigma(d) <= 2*d < A003961(d), where A003961 is fully multiplicative with a(p) = nextprime(p).

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

Offset: 1

Author

Antti Karttunen, Dec 06 2024

Keywords

• nonn

Comments

Number of terms of A341614 that divide n.

Claim: a(n) > 0 if and only if A003961(n) > 2*n [i.e., n is in A246282]. That a(n) must be zero when n is in A246281 is obvious, as is also that a(n) > 0 when n is a term of A341614 [as then A378664(n) = n], but that a(n) > 0 for all abundant numbers (A005101) is slightly less clear. So the claim boils down to this: All abundant numbers have at least one (by necessity a proper) divisor d|n such that it is in A341614, i.e., sigma(d) <= 2*d < A003961(d), i.e., that for abundant numbers n, A337345(n) is always strictly greater than A080224(n). Equivalently, of the all nonabundant divisors d of an abundant number, at least one is primeshift-abundant, i.e., A003961(d) > 2*d. This has been proved Dec 11 2024 by Jianing Song in A337372. The claim given in A378658 also follows from that proof.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000
• Index entries for sequences related to prime indices in the factorization of n
• Index entries for sequences related to sigma(n)

Crossrefs

Inverse Möbius transform of A341612.

Cf. A000203, A003961, A005101, A080224, A080225, A337345, A341614, A378663, A378664.

Cf. A246281 (positions of 0's), A246282 (of terms > 0).

Cf. also A337372, A378658.

Programs

Mathematica

Table[Length@ Select[Divisors[n], DivisorSigma[1, #] <= 2 # < (Times @@ Map[Power @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi[p] + 1], e}] - Boole[# == 1]) &], {n, 105}] (* Michael De Vlieger, Dec 06 2024 *)

PARI

A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A341612(n) = ((sigma(n)<=(2*n))&&((2*n)<A003961(n)));
A378662(n) = sumdiv(n,d,A341612(d));

Formula

a(n) = Sum_{d|n} A341612(d).

a(n) = A337345(n) - A080224(n).

a(n) = A080225(n) + A378663(n).