Primorials are like Factorials except that only the prime numbers are multiplied:
2, 6, 30, 210, 2310, 30030, ...
Starting with the first primorial which is at least two digits, consider a sequence composed of the second to last digit of each primorial:
3, 1, 1, 3 ...
The first digit of this sequence is '3'.
Please provide the 999th and 9999th digits of this sequence.
(In reply to
computer solution by Charlie)
http://oeis.org/A137728
<table width="100%" cellspacing="0" cellpadding="0" border="0" style="font-family: "Times New Roman"; background-color: rgb(238, 238, 255);"><tbody><tr><td valign="top" align="left">Second digit from the end of product of first n primes.</td></tr></tbody></table>is almost the same as the sequence you found for this problem but is starts with 0,0,3,1,1,3... with two initial zeros.
Thus the 999th term of this problems sequence is a(1001)
and the 9999th term is a(10001)
The linked table shows a(1001)=9 but maddeningly it stops at 10000.
a(10000)=3
So we Google:
What is the 10,001st prime number? The answer is: 104743.
And so 3*3=9
a(10001)=9
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Posted by Jer
on 2022-06-14 18:55:36 |
re: computer solution (confirmation)
