

A134670


First position k such that A046694(k) = A046694(k+1) =.. 0 are n consecutive zeros starting with A046694(k), where A046694 = Ramanujan tau numbers mod 691.


3




OFFSET

1,1


COMMENTS

Most probably a(5) = 1118176194, because it is a starting point of a string of 5 zeros, but the fact that this is the least such number needs to be confirmed.
Note that zeros of A046694(n) have the indices equal to the terms of arithmetic progressions of the type k*p, where primes p belong to A134671. Thus: a(1) = 1381 = 2*691  1, a(2) = 16581 = 3*5527 = 3*(8*691  1), a(3) = 290217 = 3*96739 = 3*(140*691  1), a(4) = 1409635 = 5*281927 = 5*(408*691  1), a(5) = 1118176194 = 6*186362699 = 6*(269700*691  1).
Also, note that all listed terms have the form a(n) = k*p  1, where prime p is a prime of the form p = 2m*691  1 that belong to A134671. a(1) = 2*691  1, a(2) = 2*8291  1, a(3) = 2*145109  1, a(4) = 4*352409  1, a(5) = 5*223635239  1.


LINKS



EXAMPLE

a(1) = 1381 because A046694(1381) = 0 is the first zero in A046694(n).
a(2) = 16581 because A046694(16581) = A046694(16582) = 0 are the first two consecutive zeros in A046694(n).


MAPLE

option remember;
if n = 1 then
1381 ;
else
for a from procname(n1)+1 do
wrks := true;
for k from a to a+n1 do
wrks := false ;
break;
end if;
end do:
if wrks then
return a;
end if;
end do:
end if;


CROSSREFS



KEYWORD

hard,nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



