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%I #24 Oct 24 2020 11:40:59
%S 1381,16581,290217,1409635,1118176194,107792931954,673751534392,
%T 2587409974788
%N 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.
%C 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.
%C 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).
%C 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.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TauFunction.html">Ramanujan's Tau Function</a>
%e a(1) = 1381 because A046694(1381) = 0 is the first zero in A046694(n).
%e a(2) = 16581 because A046694(16581) = A046694(16582) = 0 are the first two consecutive zeros in A046694(n).
%p A134670 := proc(n)
%p option remember;
%p if n = 1 then
%p 1381 ;
%p else
%p for a from procname(n-1)+1 do
%p wrks := true;
%p for k from a to a+n-1 do
%p if A046694(k) <> 0 then
%p wrks := false ;
%p break;
%p end if;
%p end do:
%p if wrks then
%p return a;
%p end if;
%p end do:
%p end if;
%p end proc: # _R. J. Mathar_, Feb 01 2013
%Y Cf. A046694, A121733, A121734, A121742, A121743.
%K hard,nonn,more
%O 1,1
%A _Alexander Adamchuk_, Nov 05 2007
%E a(5) confirmed by _Jud McCranie_ Oct 17 2020
%E a(6) by _Jud McCranie_ Oct 17 2020
%E a(7) by _Jud McCranie_ Oct 19 2020
%E a(8) by _Jud McCranie_ Oct 22 2020
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