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Robert Israel, <a href="/A090519/b090519_1.txt">Table of n, a(n) for n = 1..1800</a>
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Robert Israel, <a href="/A090519/b090519_1.txt">Table of n, a(n) for n = 1..1800</a>
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Robert Israel, <a href="/A090519/b090519_1.txt">Table of n, a(n) for n = 1..1800</a>
f:= proc(n) local t, p;
t:= 10^n;
p:= 1;
while p < t/2 do
p:= nextprime(p);
if isprime(floor(t/p)) then return p fi
od;
0
end proc:
map(f, [$1..50]); # Robert Israel, Jul 30 2023
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Smallest prime p such that floor[((10^n)/p] ) is prime, or 0 if no such number exists.
Conjecture: No term is zero. Subsidiary Sequence: Number of primes in floor[((10^n)/p], ), p is a prime. a(1) = 3, the primes are 10/2, floor[(10/3] ) and 10/5.
a(5) = 89, as floor[((10^5)/89]) = 1123 is the largest such prime.
<<NumberTheory`; Do[k = 2; While[ !PrimeQ[Floor[10^n / k]], k = NextPrime[k]]; Print[k], {n, 1, 50}] (* _Ryan Propper_, Jun 19 2005 *)
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_Amarnath Murthy (amarnath_murthy(AT)yahoo.com), _, Dec 07 2003
Corrected and extended by _Ryan Propper (rpropper(AT)stanford.edu), _, Jun 19 2005
Conjecture: No term is zero. Subsidiary Sequence: Number of primes in floor[(10^n)/p], p is a prime. a(1) = 3, the primes are 10/2, floor[10/3], and 10/5.
base,nonn,new