Regarding Q.1 For D=2,3,5,6,7,8,
The Smallest Titanic Prime 10^999+7 is such that its square is free of
For D=4,9 , The Titanic Prime
10^999+3561 is such that its square is free of digit D.
A Titanic prime such that its
square is free of D for D=1. This Titanic prime is: 2*10^999+8567
... the solution for D=0 can not be
attempted without evolving some strategy to tackle most probable
solutions only. Lets try and hope to find a solution.
I have used Primeform/GW to find
solutions and Primo to prove a few primes. Numbers with special decimal
forms are requested and I have found the solutions by experimenting with
such numbers and not much analysis.
1. Find one titanic prime
(digits=>1000) such that its square is free of the digit D, for each of
the following values for D: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The digits here are covered by the
following cases but my first test was: p = 10^999+7 is well-known as the
smallest titanic prime. p^2 is free of the digits 2, 3, 5, 6, 7, 8.
p = 9*10^1058+1 is easily proven
prime by Primeform with p-1 factored. p^2 is free of the digits 2, 3, 4,
5, 6, 7, 9.
p = 3*10^1137+7 is proven prime with
p^2 is free of the digits 1, 3, 5, 6,
p = (10^1000-1)/3+49*10^132 is proven
prime with Primo.
p has exactly 1000 digits and p^2 is
only free of the digit 0. p also answers question 2.
The 3 previous primes are my own
finds covering all digits.
I also wanted a really big prime and
looked at the top 5000 list at primes.utm.edu. I did not have to look far.
Number 98 (list dated 18 Sep 2002): p = 105994*10^105994+1 is a
106000-digit Generalized Cullen prime found by Guenter Loeh and
Yves Gallot (presumably Loeh was running proth.exe). A quick test
shows that p^2 is missing the digit 5. I have not tested the bigger
numbers but would be extremely surprised to see a missing digit in a
p = 10^69882+3*10^34941+1 with 69883
digits is the record palprime. It was found by Daniel Heuer with
PrimeForm. p^2 is free of the seven digits 2, 3, 4, 5, 7, 8, 9.
2. Find one titanic prime of
exactly 1000 digits such that its square is only free of the digit "zero"
(10^1000-1)/3+49*10^132 from question
3. Find one titanic palprime
such that its square is free of the digit "zero"
p = (10^1867-1)/3+4*10^933 is a
p has 1867 digits, the middle is 7
and all others are 3.
p is a palprime - and also a
p^2 is free of the digit 0 (but also
of 4 and 6).
It would take around a day to prove p
prime with Primo.
I omitted the proof and proved the
p = (10^1287-1)/3+313*10^642 is
proven prime with Primo.
p is palindromic and p^2 is only free
of the digit 0.
Let p(n) = (10^(2*n+3)-1)/3+313*10^n
The previous solution is p(642).
p(n) is always palindromic and p(n)^2
is always free of only the digit 0.