cp's OEIS Frontend

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 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.

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.

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.

I conjecture that k < n+1 for all n (where k is function of n, of course).

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.

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)).

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

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

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, ...