cp's OEIS Frontend

A163582 Smallest pseudoprimes in ascending bases b of the form pq, where p = (b^k - 1) / (b - 1) and q = (b^k + 1) / (b + 1), both prime, with k a prime less than 100.

%I A163582 #4 Jun 02 2025 01:48:39
%S A163582 341,597871,1010101010101010101010101010101010101,432988561,
%T A163582 584645231109031,54989488181,
%U A163582 48793204382746801501446610630739608190006929723969,11694525061301
%N A163582 Smallest pseudoprimes in ascending bases b of the form pq, where p = (b^k - 1) / (b - 1) and q = (b^k + 1) / (b + 1), both prime, with k a prime less than 100.
%C A163582 These numbers can be factored in O(log n) time by iterating over their multiplicative Sylow p-subgroups.
%C A163582 Let n be an odd integer. Its maximal multiplicative order is n - 1; the largest power of 2 dividing n - 1 is o, the maximal order of its multiplicative Sylow 2-subgroup.
%C A163582 Let t = (n - 1) / o.
%C A163582 Pick random integers x and y such that the Jacobi symbols (x/n) and (y/n) are -1; that is, x and y are quadratic nonresidues modulo n. Exponentiate both by t to produce generators of the Sylow 2-subgroup. Retain the value of y as r.
%C A163582 Repeatedly multiply x by y and y by r, modulo n, giving a parabolic sequence of x values. A factor may be obtained from gcd(x - 1, n), gcd(y - 1, n), gcd(x - y, n), or gcd(x + y, n). If a factor is not found quickly, choose new values of x and y.
%C A163582 It is often the case that one iteration is sufficient, instead of log n iterations.
%e A163582 341 = 11 * 31 is a base 2 pseudoprime; 2^340 = 1 (mod 341).
%e A163582 597871 = 547 * 1093 is a base 3 pseudoprime; 3^597870 = 1 (mod 597871).
%e A163582 1010101010101010101010101010101010101 = 909090909090909091 * 1111111111111111111 is a base 10 pseudoprime.
%e A163582 The smallest base 387 pseudoprime semiprime has 1204 bits:
%e A163582 190343478807499085058031516398268680442601127373980882552883668761244360084075072419711216782718751807174818426029099795926922432206385551671790497449073768776989824173201266255008090697631436472577273835739136689804694203609505130893771033656337490070783749133621893887506391690839509492668015407074108567267922714146861065256735761674160812989129563106060165551
%e A163582 which has factors
%e A163582 13760898475567760339045070218774452423864352937859851193617152180919304736064745532601237230112112091064203139709556823171465678472351610172571294148439637693965101978819764531439203
%e A163582 and
%e A163582 13832198467669147698314733795037532488236707098159643168713614109317850356458863385101761775345853604489406264785772143498778972143192810225278917434182848251964921160057172637819717
%e A163582 The large factor can be found in 8 iterations, taking six seconds on a 2GHz processor. In general, factorization requires less than a minute for numbers having fewer than 2000 bits.
%Y A163582 Cf. A000040, A001358, A007535
%K A163582 nonn
%O A163582 2,1
%A A163582 _Reikku Kulon_, Jul 31 2009