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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.

A236852 Remultiply n first "downward", from N to GF(2)[X], and then remultiply that result back "upward", from GF(2)[X] to N: a(n) = A234742(A234741(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 6, 7, 8, 9, 18, 11, 12, 13, 14, 27, 16, 81, 18, 19, 36, 21, 22, 39, 24, 81, 26, 27, 28, 33, 54, 31, 32, 33, 162, 63, 36, 37, 38, 39, 72, 41, 42, 75, 44, 81, 78, 47, 48, 49, 162, 243, 52, 57, 54, 99, 56, 57, 66, 59, 108, 61, 62, 63, 64, 117, 66, 67, 324
Offset: 0

Views

Author

Antti Karttunen, Feb 02 2014

Keywords

Comments

This sequence appears to be completely multiplicative with a(p) = A234742(p) although neither A234741 or A234742 are even multiplicative. Terms tested up to n = 10^7. - Andrew Howroyd, Aug 01 2018
Yes, this is true. Consider for example n = p*q*r*r, where p, q, r are primes in N. Then A234741(n) = p1 x q1 x q2 x q3 x r1 x r2 x r1 x r2, where p1, q1, r1, ..., are the irreducible factors of p, q, r when factored over GF(2), and x stands for multiplication in ring GF(2)[X] (A048720). [Note that these irreducible factors are not necessarily primes in N, except p1 (= p), which must be a term of A091206. Also, A234741(p) = p for any prime p.] Next, a(n) = A234742(p1 x q1 x q2 x q3 x r1 x r2 x r1 x r2) = p1 * q1 * q2 * q3 * r1 * r2 * r1 * r2, which can be obtained also as a(p)*a(q)*a(r)*a(r), thus proving the multiplicativity. - Antti Karttunen, Aug 02 2018

Examples

			From _Antti Karttunen_, Aug 02 2018: (Start)
For n = 3, we have A234741(3) = 3 = 11 in binary, which encodes a (0,1)-polynomial x+1, which is irreducible over GF(2) thus A234742(3) = 3 and  a(3) = 3.
For n = 5, we have A234741(5) = 5 = 101 in binary, which encodes a (0,1)-polynomial x^2 + 1, which factorizes as (x+1)(x+1) when factored over GF(2), that is 5 = A048720(3,3), thus it follows that A234742(5) = 3*3 = 9, and a(5) = 9.
For n = 9 = 3*3, we have A234741(9) = A048720(3,3) = 5, and A234742(5) = 9 as shown above. Also by multiplicativity, we have a(3*3) = a(3)*a(3) = 3*3 = 9.
(End)
		

Crossrefs

Programs

Formula

a(n) = A234742(A234741(n)).

Extensions

Keyword mult added after Andrew Howroyd's observation. - Antti Karttunen, Aug 02 2018