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%I #18 Aug 28 2019 16:09:33
%S 1,2,6,12,60,120,420,840,2520,5040,27720,55440,360360,720720,1081080,
%T 1441440,12252240,24504480,232792560,465585120,698377680,931170240,
%U 5354228880,10708457760,26771144400,53542288800,80313433200,160626866400,2329089562800,4658179125600,72201776446800
%N Lexicographically earliest increasing sequence such that n-th term is divisible by all positive integers up to n.
%H Rémy Sigrist, <a href="/A309233/b309233.txt">Table of n, a(n) for n = 1..2000</a>
%e The 5th term 60, is divisible by 1,2,3,4, and 5.
%e The 6th term 120, greater than 60, is divisible by 1,2,3,4,5, and 6.
%e The 10th term 5040, greater than 2520, is divisible by 1,2,3,4,5,6,7,8,9, and 10.
%o (Python)
%o x = [1] # To count the number in the sequence
%o y = [1] # To record the sequence itself
%o N = 20 # The number of terms to calculate
%o while len(x) < N:
%o x.append(x[-1] + 1)
%o # The next number in the sequence will be divisible by x[-1] so start
%o # looking at z which is the largest number divisible by x[-1] so far
%o z = y[-1]//x[-1] * x[-1] + x[-1]
%o # Now start counting up by x[-1] and check if the number is divisible by
%o # every integer less than x[-1]
%o while any([z%i != 0 for i in x]):
%o z += x[-1]
%o y.append(z)
%o print("{}\t {}\t".format(len(x),z))
%o (PARI) A003418(n) = if(n<1, n==0, 1/content(vector(n, k, 1/k)));
%o a(n) = {my(k=1);if(n>1,k=A003418(n)*(1+a(n-1)\A003418(n)));k} \\ _Jinyuan Wang_, Jul 22 2019
%Y Would be same as A003418 without the "increasing" condition.
%K nonn
%O 1,2
%A _Rian Rustvold_, Jul 16 2019
%E More terms from _Jinyuan Wang_, Jul 22 2019
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