cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A328926 Numbers k such that A328925(k) = 1; numbers k such that if we write k = Product_{i=1..t} p_i^e_i, then lcm_{1<=i,j<=t,i!=j} ord(p_i,p_j^e_j) = A002322(k), where ord(a,r) is the multiplicative order of a modulo r, and A002322 is the Carmichael lambda (usually written as psi).

Original entry on oeis.org

1, 2, 6, 10, 12, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 35, 36, 38, 40, 42, 44, 45, 48, 50, 51, 52, 54, 56, 57, 58, 60, 63, 66, 69, 70, 72, 74, 75, 76, 77, 78, 80, 84, 85, 87, 88, 90, 91, 92, 93, 96, 99, 100, 102, 104, 105, 106, 108, 110, 114, 115, 116, 118, 119, 120, 122
Offset: 1

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Author

Jianing Song, Oct 31 2019

Keywords

Comments

Numbers such that A118106(k) = A002322(k).
{a(n)} intersect A000961 = {1, 2}.
{a(n)} union A329062 = N* \ A000961 U {1, 2}.
If k = Product_{i=1..t} p_i^e_i (t > 1), where {p_i} are primes such that p_i is a lambda primitive root modulo p_j^e_j for all i != j (that is to say, ord(p_i,p_j^e_j) = A002322(p_j^e_j), where ord(a,r) is the multiplicative order of a modulo r), then k is here, as A118106(k) = A002322(k). For example, k = 2^5 * 5 * 13.

Examples

			For k = 115 = 5 * 23, A118106(115) = lcm(ord(23,5),ord(5,23)) = lcm(4,22) = 44 = A002322(115), so 115 is a term.
For k = 973 = 7 * 139, A118106(973) = lcm(ord(139,7),ord(7,139)) = lcm(6,69) = 138 = A002322(973), so 973 is a term.
		

Crossrefs

Programs

A329062 Numbers k that are not prime powers (i.e., not in A000961) such that A328925(k) > 1; numbers k such that if we write k = Product_{i=1..t} p_i^e_i , then t > 1, and lcm_{1<=i,j<=t, i!=j} ord(p_i,p_j^e_j) < A002322(k), where ord(a,r) is the multiplicative order of a modulo r, and A002322 is the Carmichael lambda (usually written as psi).

Original entry on oeis.org

14, 34, 39, 46, 55, 62, 65, 68, 82, 86, 94, 95, 98, 111, 112, 117, 123, 124, 133, 136, 142, 145, 146, 153, 155, 158, 164, 172, 175, 178, 183, 194, 201, 203, 205, 206, 209, 218, 219, 221, 224, 226, 248, 253, 254, 259, 272, 274, 275, 287, 291, 292, 295, 299, 301, 302, 305, 309
Offset: 1

Views

Author

Jianing Song, Nov 02 2019

Keywords

Comments

Write psi = A002322, b = A118106. These are numbers k that are not prime powers such that b(k) < psi(k).
If k = Product_{i=1..t} p_i^e_i (t > 1), where {p_i} are distinct odd primes such that p_i is a quadratic residue modulo p_j for all i != j, then k is here, as b(k) | psi(k)/2.
If k = 2 * Product_{i=1..t} p_i^e_i, where {p_i} are distinct odd primes such that p_i is a quadratic residue modulo p_j for all i != j, and p_i == 1, 7 (mod 8), then k is here. For example, k = 2p, where p == 1, 7 (mod 8).
If k = 2^e * Product_{i=1..t} p_i^e_i (e > 1), where {p_i} are distinct odd primes such that p_i is a quadratic residue modulo p_j for all i != j, and p_i == 1 (mod 8), then k is here. For example, k = 2^e*p, where p == 1 (mod 8).
If k = p_1^e_1 * p_2^e_2, where p_1 is not a primitive root modulo p_2^e_2, p_2 is not a primitive root modulo p_1^e_1, then k is not necessarily here: for k = 3^2 * 67 = 603, 3 is not a primitive root modulo 67, 67 is not a primitive root modulo 3^2, but b(603) = psi(603) = 66. Conversely, if p_1 is a primitive root modulo p_2^e_2, then k can still be here: for k = 3^2 * 17 = 153, 3 is a primitive root modulo 17, but b(153) = 16 while psi(153) = 48.
If k is an odd number, then 4k is here if and only if 8k is also here. Write k = Product_{i=1..t} p_i^e_i, then b(4k) = lcm(lcm_{i=1..t} ord(p_i,4),lcm_{i=1..t} ord(2,p_i^e_i),b(k)), b(8k) = lcm(lcm_{i=1..t} ord(p_i,8),lcm_{i=1..t} ord(2,p_i^e_i),b(k)). It is easy to see that lcm(ord(p_i,4),ord(2,p_i^e_i)) = lcm(ord(p_i,8),ord(2,p_i^e_i)), so b(4k) = b(8k). Note that psi(4k) = psi(8k).
If k is an odd number such that 4k is here, then 16k is also here (but the converse is not true). Write k = Product_{i=1..t} p_i^e_i, N = lcm(lcm_{i=1..t} ord(2,p_i^e_i),b(k)), then b(4k) = lcm(lcm_{i=1..t} ord(p_i,4),N), b(16k) = lcm(lcm_{i=1..t} ord(p_i,16),N) = b(4k) or b(4k)*2 or b(4k)*4. Note that psi(16k) = lcm(psi(4k),4). If we have b(4k) < psi(4k) and b(16k) = psi(16k), let M = psi(k), then:
Case (a): M == 2 (mod 4), then b(16k) = psi(16k) = 2*psi(4k) = 2M, and b(4k) = b(16k)/4 = M/2 which is an odd number, so ord(2,p_i^e_i) is odd for all i, so p_i == 1, 7 (mod 8), which gives ord(p_i,16) <= 2*ord(p_i,4). As a result, b(16k) <= 2*b(4k), a contradiction!
Case (b): M == 0 (mod 4), then b(16k) = psi(16k) = psi(4k) = M. If N is divisible by 4, then b(4k) = b(16k) = N, a contradiction. So 4 does not divide N, we have v(M,2) = v(lcm(lcm_{i=1..t} ord(p_i,16),N),2) <= 2, but 4 | M, so v(M,2) = 2, where v(,2) is the 2-adic valuation. As a result, there must exist p_i congruent to 5 mod 8 to make v(M,2) = 2, then 4 | ord(2,p_i^e_i), so 4 | N, a contradiction!
{a(n)} union A328926 = N* \ A000961 U {1, 2}.

Examples

			Let ord(a,r) be the multiplicative order of a modulo r.
For k = 175 = 5^2 * 7, b(175) = lcm(ord(7,5^2),ord(5,7)) = lcm(4,6) = 12, while psi(175) = lcm(20,6) = 60, so 175 is a term.
For k = 410 = 2 * 5 * 41, b(410) = lcm(ord(5,2),ord(41,2),ord(2,5),ord(41,5),ord(2,41),ord(5,41)) = 20, while psi(410) = lcm(1,4,40) = 40, so 410 is a term.
		

Crossrefs

Programs

Showing 1-2 of 2 results.