A124202 a(n) = median of the largest prime dividing a random n-digit number.

3, 12, 53, 229, 947, 3863, 15731, 63823, 258737

Offset: 1

Author

Mark Thornquist (mthornqu(AT)fhcrc.org), Dec 07 2006

Keywords

• base
• more
• nonn

Comments

A randomly selected n-digit number (uniformly distributed on 10^(n-1) to 10^n-1) has at least a 50% probability of having a prime factor at least as large as a(n).

For n >= 2 the number m = 9*10^(n-1) of n-digit numbers is even. The median is taken to be the average of the (m/2)-th and (m/2+1)-th of the sorted list of largest prime factors. - Robert Israel, Dec 11 2015

Examples

			The largest prime divisors of the nonunit 1-digit numbers are 2, 3, 2, 5, 3, 7, 2 and 3 respectively, with median 3.
Of the 90 2-digit numbers, there are 45 whose largest prime divisor is 11 or less and 45 whose largest prime divisor is 13 or greater, so any of 11, 12, or 13 could be used for the second term, although the arithmetic average of the endpoints is commonly used.
		

References

• D. E. Knuth, The Art of Computer Programming, Seminumerical Algorithms, Addison-Wesley, Reading, MA, 1969, Vol. 2.

Crossrefs

Cf. A006530, A046731, A126282.

Programs

GAUSS

n = 1;
a = 2 | 3 | 2 | 5 | 3 | 7 | 2 | 3;
meana = meanc(a);
mediana = median(a);
format /rdn 1,0;
print n;; "-digit numbers:";
print " Median = ";; mediana;
format /rdn 10,5;
print " Mean = ";; meana;
print;
b = 1 | a;
dim = 1;
_01: wait;
n = n+1;
dim = 10*dim;
a = b | zeros(9*dim,1);
i = dim;
do until i == 10*dim;
if i == 2*floor(i/2);
a[i] = a[i/2];
else;
p = firstp(i);
if p == i;
a[i] = i;
else;
a[i] = a[i/p];
endif;
endif;
i = i+1;
endo;
b = a[dim:10*dim-1];
meana = meanc(b);
mediana = median(b);
format /rdn 1,0;
print n;; "-digit numbers:";
print " Median = ";; mediana;
format /rdn 10,5;
print " Mean = ";; meana;
print;
b = a;
goto _01;
proc firstp(n);
local i;
i = 3;
do until i > sqrt(n);
if n == i*floor(n/i);
retp(i);
endif;
i = i+2;
endo;
retp(n);
endp;

MATLAB

P = primes(10^8);
L = zeros(1,10^8);
for p = P
     L([p:p:10^8]) = p;
end
A(1) = median(L(2:9));
for d = 2:8
    A(d) = median(L(10^(d-1):10^d-1));
end
A   % Robert Israel, Dec 11 2015

Maple

seq(Statistics:-Median([seq(max(numtheory:-factorset(n)),n=10^(d-1)..10^d-1)]),d=1..7); # Robert Israel, Dec 11 2015

Mathematica

f[n_] := Block[{k = If[n == 1, 1, 0], lst = {}, pt = 10^(n - 1)}, While[k < 9*pt, AppendTo[lst, FactorInteger[pt + k][[ -1, 1]]]; k++ ]; Median@ lst]; (* Robert G. Wilson v, Dec 14 2006 *)

Python

from sympy import factorint
from statistics import median
def a(n):
  lb, ub = max(2, 10**(n-1)), 10**n
  return int(round(median([max(factorint(i)) for i in range(lb, ub)])))
print([a(n) for n in range(1, 6)]) # Michael S. Branicky, Mar 12 2021

Extensions

Edited by Robert G. Wilson v, Dec 14 2006

a(8) from Robert Israel, Dec 11 2015

a(9) from Giovanni Resta, Apr 19 2016