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%I #15 Feb 19 2019 18:16:51
%S 5,41,257,2309,14621,48821,125429,177269,1595417,5278001,10596137,
%T 15146069,21523361,63993929,83629517,111321257,363526421,375805589,
%U 427518041,446072909,552010829,752665649,1980098177
%N Numbers n such that with b=n*(2n-1) two remainders x and y are defined via x = 2^(b-1) -1 mod b and y = (2*n-1)^(b-1) - 1 mod b which satisfy x==y==0 (mod n) and y-x=n.
%C The first 23 terms (at least) are primes.
%C Conjecture: All terms have the form 12*k+5.
%C The first composite numbers in the sequence are 2^80+1 and 2^512+1.
%C If we modify one of the conditions to y - x = 2*n, the sequence changes to 3, 7, 19, 31, 79, 139, 199, 211, 271, 283, 307, 331, 367, 379, 439, 499, 511, ...
%C or if we modify it to y - x = 64*n, the sequence becomes 89, 101, 197, 269, 317, 341, 461, 521, 569, 701, 821, 857, 881, 929, 1109, 1181, 1217, ...
%C There seem to be no solutions n if the condition is modified to any y - x <= 0.
%e For n=41, b = 41*(2*41-1)=3321. So 2^3320 == 3199 (mod 3321) leads to x = 3199 - 1 = 3198 which satisfies x == 0 (mod 41), and 81^3320 == 3240 (mod 3321) leads to y = 3240 - 1 = 3239 which satisfies y == 0 (mod 41) and y - x = 41. Therefore n=41 is in the sequence.
%p isA190638 := proc(n) local b,x,y; b := n*(2*n-1) ; x := modp( 2 &^ (b-1),b) -1; y := modp( (2*n-1) &^ (b-1),b) -1; if y-x =n and modp(x,n) = 0 and modp(y,n) = 0 then true; else false; end if; end proc:
%p for n from 2 do if isA190638(n) then print(n); end if; end do: # _R. J. Mathar_, Jun 04 2011
%K nonn
%O 1,1
%A _Alzhekeyev Ascar M_, May 15 2011
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