A281889
a(n) is the least integer k such that more than half of all integers are divisible by a product of n integers chosen from 2..k.
Original entry on oeis.org
3, 7, 433, 9257821
Offset: 1
For n=1, we have a(1) = 3 since for all m > 1, more than half of the integers in -m..m are divisible by an integer chosen from 2..3, i.e., either 2 or 3. We must have a(1) > 2, because the only integer in 2..2 is 2, but in each interval -2m-1..2m+1, only 2m+1 integers are even, so 2 is not a divisor of more than half of all integers in the precise sense given above.
A126283
Largest number k for which the n-th prime is the median of the largest prime dividing the first k integers.
Original entry on oeis.org
4, 18, 40, 76, 116, 182, 246, 330, 426, 532, 652, 770, 904, 1058, 1210, 1386, 1560, 1752, 1956, 2162, 2394, 2640, 2894, 3150, 3422, 3680, 3984, 4302, 4628, 4974, 5294, 5650, 5914, 6006, 6372, 6746, 7146, 7536, 7938, 8386, 8794, 9222, 9702, 10156
Offset: 1
a(1)=4 because the median of {2,3,2} = {2, *2*,3} is 2 (the * surrounds the median) and for any number greater than 4 the median is greater than 2.
a(1)=18 because the median of {2,3,2,5,3,7,2,3,5,11,3,13,7,5,2,17,3} = {2,2,2,2,3,3,3,3, *3*,5,5,5,7,7,11,13,17}.
-
t = Table[0, {100}]; lst = {}; Do[lpf = FactorInteger[n][[ -1, 1]]; AppendTo[lst, lpf]; mdn = Median@lst; If[PrimeQ@ mdn, t[[PrimePi@mdn]] = n], {n, 2, 10^4}]; t
A124202
a(n) = median of the largest prime dividing a random n-digit number.
Original entry on oeis.org
3, 12, 53, 229, 947, 3863, 15731, 63823, 258737
Offset: 1
Mark Thornquist (mthornqu(AT)fhcrc.org), Dec 07 2006
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.
- D. E. Knuth, The Art of Computer Programming, Seminumerical Algorithms, Addison-Wesley, Reading, MA, 1969, Vol. 2.
-
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;
-
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
-
seq(Statistics:-Median([seq(max(numtheory:-factorset(n)),n=10^(d-1)..10^d-1)]),d=1..7); # Robert Israel, Dec 11 2015
-
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 *)
-
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
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