cp's OEIS Frontend

For n in A004023, a(n) = A002275(n). For all other n > 1, a(n) has at least n+1 digits and is (for small n) often of the form a*10^n + b*(10^n-1)/9 or a*(10^n-1)/9*10 + b, with 1 <= a <= 9 and b in {1, 3, 7, 9}. However, for n = 24, 46, 48, 58, 60, 64, 66, ..., more digits are required. Only then a(n) can have a digit 0, and if it has, '0' is often the repeated digit. The first indices where a(n) has more than n+2 digits are n = 208, 277, 346, ... - M. F. Hasler, Feb 25 2016; corrected by Robert Israel, Feb 26 2016

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)

We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*9 with one of the (first) digits 9 replaced by a digit among {1, 2, 4, 5, 7, 8}. - M. F. Hasler, Feb 22 2016

We conjecture that for all n >= 2, a(n) equals floor(10^(n+1)/9)*7 with one of the (first) digits 7 replaced by a digit among {0, ..., 6}. - M. F. Hasler, Feb 22 2016

The conjecture is false: a(668) = 7*(10^669-1)/9 + 10^276. - Robert Israel, Jul 13 2016

For n > 1, A037055 is conjectured to be identical to A084673. - Robert G. Wilson v, Jul 04 2003

a(n) = A002275(n) for n in A004023. For all other n < 900, a(n) has n+1 digits. - Robert Israel, Feb 21 2016

For n > 1, the last digit cannot be 2, so a(n) has at least n+1 digits. The probability is big that none of [10^n/9]*20 + {1,3,7,9} is prime, in which case a(n) must have at least n+2 digits. This is the most frequent case. We can even conjecture that for all n > 1, a(n) equals [10^(n+1)/9]*20 + b with 1 <= b <= 9 and one of the (first) digits 2 replaced by 0 or 1. - M. F. Hasler, Feb 22 2016

For n > 1, the last digit of n cannot be 5, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*50 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n > 1, a(n) equals [10^(n+1)/9]*50 + b with 1 <= b <= 9 and one of the (first) digits 5 replaced by a 0, 1, 2, 3 or 4. - M. F. Hasler, Feb 22 2016

Note that a(11) is palindromic.

The last digit of n cannot be 6, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*60 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*60 + b with 1 <= b <= 9 and one of the (first) digits 6 replaced by a digit among {0, 1, 2, 3, 4, 5}. - M. F. Hasler, Feb 22 2016

The last digit of n cannot be 4, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*40 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*40 + b with 1 <= b <= 9 and one of the (first) digits 4 replaced by a 0, 1, 2 or 3. - M. F. Hasler, Feb 22 2016

The last digit of n cannot be 8, therefore a(n) must have at least n+1 digits. It is probable (using [] for floor) that none among [10^n/9]*80 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*80 + b with 1 <= b <= 9 and one of the (first) digits 8 replaced by a digit among {0, ..., 7}. - M. F. Hasler, Feb 22 2016