A342669 Even numbers which are either primitively nondeficient (A006039), or become such after applying prime shift A003961 some number of times to them.

6, 20, 28, 70, 88, 104, 120, 180, 272, 300, 304, 368, 420, 464, 496, 504, 550, 572, 630, 650, 660, 748, 780, 836, 924, 990, 1020, 1050, 1092, 1140, 1170, 1184, 1312, 1376, 1380, 1430, 1470, 1504, 1650, 1696, 1740, 1860, 1870, 1888, 1952, 2002, 2090, 2210, 2220, 2310, 2460, 2470, 2530, 2580, 2584, 2730, 2820, 2856, 2990

Offset: 1

Author

Antti Karttunen, Mar 20 2021

Keywords

• nonn

Comments

Even numbers k for which A341624(k) = 1.

Even numbers whose closure under map x -> A003961(x) contains a primitive non-deficient number (one of the terms of A006039). Shifting each term k exactly A336835(k)-1 times with A003961 towards larger primes gives those numbers, but not in monotonic order, producing instead a permutation of A006039.

Sequence 2*A246277(A006039(.)), sorted into ascending order.

If there are any two terms, x and y, such that the other is a multiple of the other, then A336835(x) != A336835(y), and furthermore, for any term k present here, for all its proper divisors (d|k, dA336835(d) < A336835(k), in other words, they reach the deficiency earlier (by prime shifting) than k itself.

Examples

			For n = 120 = 2^3 * 3 * 5, A341620(120) = 8, so it is not primitive nondeficient. However, prime-shifting it once gives A003961(120) = 945 = 3^3 * 5 * 7, which is one of the terms of A006039 as A341620(945) = 1. Therefore 120 is included in the sequence.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to sigma(n)
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A000396, A006039 (even terms of these form a subsequence).

Cf. A003961, A246277, A336835, A341605/A341606, A341620, A341624.

Programs

PARI

A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A341620(n) = sumdiv(n,d,(sigma(d)>=(2*d)));
A341624(n) = { my(t, u=0); while((t=A341620(n))>0, u=t; n = A003961(n)); (u); };
isA342669(n) = (!(n%2)&&(1==A341624(n)));