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%I #24 Mar 14 2018 17:43:07
%S 32,43,46,49,50,60,69,72,73,74,78,82,84,86,88,90,91,93,94,95,98,101,
%T 107,108,110,115,116,121,123,124,125,126,130,132,136,137,139,144,147,
%U 149,152,153,154,156,158,159,160,161,163,164,166,169,170,171,172,173,176,177,178,179
%N Indices for which A037053(n), the smallest prime with n digits '0', does not have n consecutive digits '0'.
%C Sequence A085824 lists the indices n for which A037053(n) has only two nonzereo digits, i.e., A037053(n) = a*10^(n+1) + b, with 1 <= a,b <= 9.
%C It is conjectured that, apart from A037053(0) = 2, all other terms have three nonzero digits and are therefore of the form A037053(n) = a*10^(n+2) + b*10^k + c, where 1 <= a,b,c <= 9 and 1 <= k <= n+1.
%C Whenever 1 < k < n+1, the n digits '0' are not consecutive but separated in two "chunks" of length n-k+1 and k-1, respectively. These indices n are listed here.
%C I conjecture that k < n+1 for all n (where k is function of n, of course).
%C For most indices n listed here, the smallest prime with n consecutive digits '0' is of the above form with k = n+1, i.e., of the form ab0...0c = (10a+b)*10^(n+1) + c.
%C The first index n for which this is not the case remains to be found. It can be expected that for this index n, the least prime with n consecutive digits '0' is either of the form a0...0b0c = a*10^(n+3) + b*100 + c (in which case it equals A037053(n+1)) or of the form a0...0bc with a > 9 (in which case it equals A037053(n+1) if a = 0 (mod 10)).
%C Sequence A269260 lists the values a > 9 such that the least prime with (at least) n consecutive '0's equals nextprime(a*10^(n+1)), for the numbers n listed here. - _M. F. Hasler_, Feb 22 2016
%C The first two values of n that do not satisfy the above forms are 192 and 213. The least prime with 192 consecutive 0's is 11100...0007. The least prime with 213 consecutive 0's is 100...000499. - _Chai Wah Wu_, Mar 11 2018
%H Chai Wah Wu, <a href="/A269230/b269230.txt">Table of n, a(n) for n = 1..771</a>
%o (PARI) for(n=1,999,n+2<#(t=digits(A037053(n))) && !t[#t-2] && print1(n","))
%o (PARI) a269230=[32]; A269230(n)={my(t); while(n>#a269230, for(k=vecmax(a269230)+1,9e9, (t=A037053(k))>10^(k+2) && t%10^(k+2)>99 && (a269230=concat(a269230,k)) && break));a269230[n]} \\ _M. F. Hasler_, Feb 22 2016
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Feb 20 2016
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