A269230 Indices for which A037053(n), the smallest prime with n digits '0', does not have n consecutive digits '0'.

32, 43, 46, 49, 50, 60, 69, 72, 73, 74, 78, 82, 84, 86, 88, 90, 91, 93, 94, 95, 98, 101, 107, 108, 110, 115, 116, 121, 123, 124, 125, 126, 130, 132, 136, 137, 139, 144, 147, 149, 152, 153, 154, 156, 158, 159, 160, 161, 163, 164, 166, 169, 170, 171, 172, 173, 176, 177, 178, 179

Offset: 1

Author

M. F. Hasler, Feb 20 2016

Keywords

• nonn
• base

Comments

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

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..771

Programs

PARI

for(n=1,999,n+2<#(t=digits(A037053(n))) && !t[#t-2] && print1(n","))

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