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For indices k not listed in A269230, the least prime with k digits '0', A037053(k), has these digits consecutively, in a single run. If k is listed in A269230, this is not the case (e.g., A037053(32) = 10...0603), and the most economical way to make a prime with k consecutive digits 0 is to put two (a priori nonzero) digits in front of the string of k '0's, i.e., p = a*10^(k+1) + b with a > 9.

This sequence lists these numbers a, and the corresponding prime (least prime with k consecutive digits 0) is simply nextprime(a*10^(k+1)).

If a is a multiple of 10, then b can have two nonzero digits, 11 <= b <= 99. Otherwise (b < 10), this prime is also the least prime with k+1 (consecutive) digits '0', A037053(k+1), and k+1 is listed in A085824 (unless a > 90). It is then obviously not the smallest prime with *exactly* k consecutive digits 0, but with *at least* k consecutive digits 0. This happens for (n,k,a,b) = (2,43,10,9), (24,108,10,7), (28,121,10,3), (34,132,10,7), (38,144,10,9), ...

Certain entries require n+3 digits such as a(13). Those which do not require three digits besides n zeros are in A085824. Conjecture: a prime of this form, A037053, requires at most three judiciously placed nonzero digits, two on the exterior and one inside. - Robert G. Wilson v

The zeros in a(n) do not need to be consecutive. If the zeros must be consecutive we get a new sequence which agrees with this one up though a(31), but then here we have a(32) = 10000000000000000000000000000000603, whereas the smallest prime with exactly 32 consecutive zeros is 19000000000000000000000000000000009 = 19*10^33+9. - N. J. A. Sloane, Feb 20 2016

Sequence A269230 lists the indices (32, 43, 46, 49, 50, 60, 69, ...) for which a(n) does not have n consecutive digits '0', and gives more information about the smallest prime which has n consecutive digits '0': Sequence A269260 lists the values a > 9 such that the least prime with n *consecutive* '0's equals nextprime(a*10^(n+1)), for these indices n. - M. F. Hasler, Feb 20 2016 and Feb 22 2016

Since the definition requires "exactly" and not "at least" n 0's, the sequence is not increasing, e.g., a(22) = 10^24 + 49 > a(23) = 10^24 + 7. However, it seems that no term has more digits than its successor. - M. F. Hasler, Feb 20 2016

From Bob Selcoe, Feb 20 2016: (Start)

Conjecture 1: (following Robert G. Wilson v's conjecture above and insight provided by Hans Havermann): terms with two substrings of zeros have first digit of 1.

Conjecture 2: as n increases, there are more terms with two substrings of zeros than one consecutive string.

The logic is as follows: Let n = number of zeros and z = number of substrings of consecutive zeros contained in candidate primes. Candidates with two substrings of zeros (z=2) must be considered after only 72 smaller z=1 candidates have been considered and excluded, i.e., numbers of the form a[n zeros]b and 1[n zeros]ab, where a in {1..9}, b in {1,3,7,9}. After these 72 candidates are excluded, 36*n z=2 candidates are considered before having to consider only 36 additional z=1 candidates (i.e., 2[n zeros]ab), followed by 36*n additional z=2 candidates, etc. So as n increases, it becomes increasingly unlikely that any z=1 term appears. Additionally, the number of candidates increases as n increases for z>=3. For a given n, 1044 + 324*(n+1) candidates must be excluded before considering the smallest z=3: 1[n-2 zeros]10101. Since the probability p of n-digit primes occurring is p ~ 1/(n log 10) by the Prime Number Theorem, and the number of z in {1,2} candidates which must be excluded increases exponentially as n and z increase, it is unlikely that any z >= 3 term appears in A037053 and that first digit will be > 1 for any z=2 term.

Sequence A269233 lists the number of candidate primes < a(n); i.e., the number of excluded candidates.

(End)

Sequence agrees with A037053 up to a(31) (see comment in A037053). A269230 lists indices where these 2 sequences differ.

For the first 1001 terms of this sequence, the number of nonzero digits of each term is 4 or less. This differs from A037053 for which the number of nonzero digits is 3 or less for the first 12000 terms. Does there exist n such that a(n) has 5 or more nonzero digits?

a(n) has 3 nonzero digits for n = 13, 22, 29, 31, 32, 33, 40, 42, 43, ...

a(n) has 4 nonzero digits for n = 192, 213, 238, 250, 252, 257, 268, 293, 297, 303, ...

a(n) <> A037053(n) and a(n) = A037053(m) for some m > n for n = 436, 780, 845, 866, 894, 911, 945, 957, 967, ... In all these cases so far, a(n) has n+1 zero digits. Are there n satisfying these conditions such that a(n) has more than n+1 zero digits?

Sequence is not monotonically increasing; indices for which a(n) > a(n+1) are 22, 43, 47, 58, 67, 105, 108, 121, 132, 144, 192, 220, 238, 250, 252, 257, 261, 270, ...