A370452 - OEIS

A370452

Prime powers of the form 2*p^k-1, where p is prime and k >= 1.

3, 5, 7, 9, 13, 17, 25, 31, 37, 49, 53, 61, 73, 81, 97, 121, 127, 157, 193, 241, 277, 313, 337, 361, 397, 421, 457, 541, 577, 613, 625, 661, 673, 733, 757, 841, 877, 997, 1093, 1153, 1201, 1213, 1237, 1249, 1321, 1381, 1453, 1621, 1657, 1681, 1753, 1873, 1933, 1993, 2017, 2137, 2341, 2401, 2473, 2557, 2593, 2797, 2857

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OFFSET

1,1

COMMENTS

Also, sizes of finite fields such that halving the size of their unit group is also the unit group of a field. - Keith J. Bauer, Jun 20 2024

Original motivation for this sequence: let k be a term of this sequence. Then consider the finite field of k elements, denoted by F_k. Adjoin the hyperbolic unit j^2 = 1 to F_k to form a ring whose elements are of the form a + bj for a, b in F_k. Let M be the multiplication monoid of F_k[j] and let ~ be the equivalence relation on the elements of M defined by a + bj ~ b + aj (with no further unnecessary equivalences). Then M/~ is isomorphic to the multiplication monoid of the ring F_k x F_(k+1)/2 and therefore there exists a ring with M/~ as its multiplication. For prime powers k not in this sequence, no such ring will exist. See the link for a proof of this fact.

LINKS

Table of n, a(n) for n=1..63.

Keith J. Bauer, The monoid (F_n[j], *)/(a + bj ~ b + aj) is isomorphic to (F_n X F_(n+1)/2, *), when such fields exist.

EXAMPLE

3 = 2*2^1 - 1 = 3^1;

5 = 2*3^1 - 1 = 5^1;

7 = 2*2^2 - 1 = 7^1;

9 = 2*5^1 - 1 = 3^2.

MAPLE

filter:= n -> nops(numtheory:-factorset(n))=1 and nops(numtheory:-factorset((n+1)/2))=1:

select(filter, [seq(i, i=3..10000, 2)]); # Robert Israel, Feb 20 2024

MATHEMATICA