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

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

Offset

1,1

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

References

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

Links

Table of n, a(n) for n=1..9.

Example

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.

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 *)

Prog

(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
(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

Crossrefs

Cf. A006530, A046731, A126282.

Sequence in context: A000256 A274396 A299113 * A138269 A228771 A370023

Adjacent sequences: A124199 A124200 A124201 * A124203 A124204 A124205

Keyword

base,more,nonn

Author

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

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

Status

approved