cp's OEIS Frontend

Equally: after 1, numbers n such that, if the prime factorization of 2n-1 = Product_{k >= 1} (p_k)^(c_k) then Product_{k >= 1} (p_{k-1})^(c_k) = n.

Factorization of the initial terms: 1, 2, 3, 5^2, 2*13, 3*11, 3*31, 2*11*47, 5^2*197^2, 2*11*47*8269, 3*11*797*12373, 5*11^2*433*1583, 2*23*59*101*6803, 2*11*53*1201*1741.

The only 3-cycle of permutation A048673 in range 1 .. 402653184 is (2821 3460 5639).

For 2-cycles, take setwise difference of A245449 and this sequence.

Numbers k for which A336853(k) = k-1. - Antti Karttunen, Nov 26 2021

Numbers k such that A252748(k) divides A286385(k).

Conjecture: Apart from a(5)=6, this is a subsequence of A319630, i.e., for all terms k<>6, gcd(k, A003961(k)) = 1. See also A372562, A372566.

No other terms below 3221225472.

Numbers k such that A252748(k) [= A003961(k) - 2*k] <> 1 (i.e., k is not in A348514), and A286385(k) [= A003961(k) - A000203(k)] = m*A252748(k) for some positive integer m. Note that this entails that k is nonabundant (A000203(k) <= 2*k) and primeshift-abundant (A252748(k) > 2), thus this is a subsequence of A341614. - revised Dec 13 2024

This is a subsequence of A378980, see further comments there. - Antti Karttunen, Dec 13 2024

Numbers k for which A326057(k) = gcd(A003961(k)-2k, A003961(k)-sigma(k)) is equal to abs(A252748(k)) = |A003961(k)-2k|.

The odd terms of A326134 form a subsequence of this sequence. Unlike in A326134, here we don't constrain the value of A252748(k) = A003961(k)-2k, thus allowing also values <= +1. Because of that, the odd terms of A048674 and A348514 are all included here, for example 57 and 68727 that occur in A348514, and 1, 3, 25, 33, 93, 970225, 325422273, 414690595 that occur in A048674.

Conjecture (1): This is a subsequence of A319630, in other words, for all terms k, gcd(k, A003961(k)) = 1.

Conjecture (2): Apart from 1, there are no common terms with A349169, which would imply that no odd perfect numbers exist.

None of the 36 initial terms is Zumkeller, in A083207, because all are deficient (in A005100). See also A337372. - Antti Karttunen, Dec 05 2024

Like in many sequences of this type, the criterion seems to select for numbers with a long tail of trailing 1-bits. Terms that are not in A004767 are: 1, 4, 10, 18, 57, 348, 1054, 2626, 675348, 1869741, 12371554, 14070141, 1158654378, 1673018314, etc.

Conjecture: A202274 gives all terms of A028982 that occur in this sequence.

Among the initial 69 terms, there are eleven +1's and eleven -1's. The former correspond in A378980 with those of its terms that are in A048674 (1, 2, 3, 25, 26, 33, 93, 1034, ...), while the latter here correspond in A378980 with those of its terms that are in A348514 (4, 10, 57, 1054, 2626, ...).

6 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 3 occurs only once in A379231. Proof: If k is not a multiple of 3 and k is in A104210, then there are primes p > 3 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 3, therefore the equation 2k +- 3 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 3, which immediately entails that k must be also even, for A003961(k) to be a multiple of 3. Let x = k/6; then the equation can be rewritten as 2*6*x +- 3 = A003961(6)*A003961(x) <=> 12x +- 3 = 15*A003961(x) <=> 3*(4x +- 1) = 3*5*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=6.

If it exists, abs(a(15)) > 2^32.

15 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 5 occurs only once in A379231. Proof: If k is not a multiple of 5 and k is in A104210, then there are primes p (either p=2 or p > 5 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 5, therefore the equation 2k +- 5 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 5, which immediately entails that k must be also a multiple of 3, for A003961(k) to be a multiple of 5. Let x = k/15; then the equation can be rewritten as 2*15*x +- 5 = A003961(15)*A003961(x) <=> 30x +- 5 = 35*A003961(x) <=> 5*(6x +- 1) = 5*7*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=15.

If it exists, abs(a(14)) > 2^32.

35 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 7 occurs only once in A379231. Proof: If k is not a multiple of 7 and k is in A104210, then there are primes p (either p=2, p=3 or p > 7 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 7, therefore the equation 2k +- 7 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 7, which immediately entails that k must be also a multiple of 5, for A003961(k) to be a multiple of 7. Let x = k/35; then the equation can be rewritten as 2*35*x +- 7 = A003961(35)*A003961(x) <=> 70x +- 7 = 77*A003961(x) <=> 7*(10x +- 1) = 7*11*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=35.

If it exists, abs(a(21)) > 2^32.