cp's OEIS Frontend

Indices of even terms in A307437; A341886 is the main entry.

Numbers k such that 2^k is a term: 8, 9, 10, 12, 13, 16, 18, 19, 20, 21, 22, 23, ... Are all sufficiently large powers of 2 terms? - Chai Wah Wu, Feb 24 2021

We don't know. From the comments in A341886, 2^k is a term if and only if t*2^(k+1)+1 is composite for t = 1, 2, 3. But we don't know if there are infinitely many Fermat primes. - Jianing Song, Feb 26 2021

All terms are divisible by 256. See my comment in A341886. - Jianing Song, Feb 27 2021

a(n) is the first k such that the smallest m such that C_(2k) is a subgroup of (Z/mZ)* is m = prime(n), where C_(2k) is the cyclic group of order 2k and (Z/mZ)* is the multiplicative group of integers modulo m.

a(n) is well-defined since A307437((p-1)/2) = p for odd primes p.

Except for a(1) = 1 and a(13) = 706, the fractional part of A307437(k)/k appears to be monotonically strictly decreasing. - Chai Wah Wu, Feb 24 2021

Indices of terms of A307437 that is divisible by 3.

For e > 0, 3^e is a term if and only if 2*3^e+1 is composite. Hence this sequence is infinite.

All terms > 1 are divisible by 27. Proof: Write a term k = 3^a*r > 1, 3 does not divide r. Suppose A307437(k) = 3^e*s, 3 does not divide s, e >= 1.

i) a = 0. If s = 1, then 2k = 2r divides psi(3^e) = 2*3^(e-1) => k = r = 1, a contradiction. Hence s > 1, then 2k = 2r divides psi(s).

ii) a = 1. If s has a prime factor congruent to 1 modulo 3, then r | psi(s) => 2k = 6r divides psi(s). Otherwise, we must have 3 | psi(3^e) => e >= 2, then 2k = 6r divides psi(7*s), a contradiction.

iii) a = 2. If s has a prime factor congruent to 1 modulo 9, then r | psi(s) => 2k = 18r divides psi(s). Otherwise, we must have 9 | psi(3^e) => e >= 3, then 2k = 18r divides psi(19*s), a contradiction.

Different from A342038, here a nontrivial multiple of p is required.

a(n) exists for all n: from the formula in A307437 we know a((p-1)/2*p^e) = p^(e+1) if p is an odd prime and (p-1)*p^e+1 is composite. For fixed p, there exist infinitely many e such that (p-1)*p^e+1 is composite.

Conjecture: a(n) >= prime(n)-1 for all n.

a(n) exists for all n: by Dirichlet's theorem on arithmetic progressions, there exists two primes p, q congruent to 1 modulo 2n, in which case C_(2n) X C_(2n) is a subgroup of (Z/(p*q)Z)*.

a(n) is the smallest k such that there exists some x, y such that ord(x,k) = ord(y,k) = 2n and the set of powers of x and the set of powers of y modulo k have trivial intersection {1}, where ord(x,k) is the multiplicative order of x modulo n.

Even k such that A307437(k/2) is even. Note that k must be divisible by 4.

Write k = r*2^e with odd r. Let s be the smallest odd number such that k | psi(s), t be the smallest number such that r | psi(t), v2(psi(t)) = a, then k is a term <=> a < e, t*2^(e+2) < s, where v2 = A007814 is the 2-adic valuation.

Proof: Let m be the smallest number such that k | psi(k).

"<=" If m is odd, then m >= s > t*2^(e+2), but k | psi(t*2^(e+2)), contradicting with minimality of m.

"=>" If a >= e, then m = t is odd. Write m = l*2^b, b >= 4, l odd, then k | lcm(psi(l), 2^(b-2)) => r | psi(l) => l >= t. If(v2(psi(l))) < b-2, then b-2 >= e => t*2^(e+2) <= l*2^b = m < s. If(v2(psi(l))) > b-2, then k | psi(l), contradicting with minimality of m. QED.

A special case: suppose k = p*2^e where p is an odd prime. Let q be the smallest number such that p | psi(q), suppose that q = p*2^a*l + 1 < p^2, l odd. Then k is a term <=> a < e; t*2^e + 1 is composite for t = 1, 2, 3; t'*p*2^e + 1 is composite for t < 4*q/p.

Let s be the smallest odd number such that k | psi(s), then k is a term <=> a < e, q*2^(e+2) < s. Suppose that a < e, there are two cases:

(i) s = (p_1)^(e_1)*(p_2)^(e_2), p | psi((p_1)^(e_1)), 2^e | psi((p_2)^(e_2)). Since a < e, p_2 != q, so (p_1)^(e_1) = q, s = q*(t*2^e + 1) with t*2^e + 1 prime.

(ii) s = (p_1)^(e_1), n | psi((p_1)^(e_1)), so s = t'*p*2^e + 1 with t'*p*2^e + 1 prime.

Hence q*2^(e+2) < s <=> t*2^e + 1 is composite for t = 1, 2, 3; t'*p*2^e + 1 is composite for t < 4*q/p. QED.

2^e is a term <=> t*2^e + 1 is composite for t = 1, 2, 3. It follows that this sequence is infinite.

3*2^e is a term <=> t*2^e + 1 is composite for t = 1, 2, 3, 6, 9, 12, 15, 18, 21, 24, 27.

From Jianing Song, Feb 27 2021: (Start)

All terms are divisible by 512. Proof: Write a term k = 2^a*r > 2 with odd r. Suppose the smallest m such that k divides psi(m) is m = 2^e*s with odd s, e >= 1.

i) a <= 1. If s = 1, then k = r divides psi(2^e) => k <= 2r = 2, a contradiction. Hence s > 1, then k = r or 2r divides psi(s).

ii) a = 2. If s has a prime factor congruent to 1 modulo 3, then r | psi(s) => k = 4r divides psi(s). Otherwise, we must have 4 | psi(2^e) => e >= 4, then k = 4r divides psi(5*s), a contradiction.

iii) a = 3. If s has a prime factor congruent to 1 modulo 8, then r | psi(s) => k = 8r divides psi(s). Otherwise, we must have 8 | psi(3^e) => e >= 5, then k = 8r divides psi(17*s), a contradiction.

The cases a = 4, 5, 6, 7, 8 are similar. (End)

A061026(n) = A061026(2n) for odd n > 1 since phi(m) is even for m >= 3. In this sequence the redundant values are omitted.

We have the obvious inequality A070846(n) >= A307437(n) >= a(n). For odd p = prime(k), A307437(p) = a(p), and if A341861(k) > 0 we have A070846(p) = a(p).

The smallest n such that A070846(n) > A307437(n) > a(n) is n = 40, where A070846(40) = 241, A307437(40) = 187 and a(40) = 123.