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A309233
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Lexicographically earliest increasing sequence such that n-th term is divisible by all positive integers up to n.
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1
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1, 2, 6, 12, 60, 120, 420, 840, 2520, 5040, 27720, 55440, 360360, 720720, 1081080, 1441440, 12252240, 24504480, 232792560, 465585120, 698377680, 931170240, 5354228880, 10708457760, 26771144400, 53542288800, 80313433200, 160626866400, 2329089562800, 4658179125600, 72201776446800
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OFFSET
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1,2
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LINKS
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EXAMPLE
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The 5th term 60, is divisible by 1,2,3,4, and 5.
The 6th term 120, greater than 60, is divisible by 1,2,3,4,5, and 6.
The 10th term 5040, greater than 2520, is divisible by 1,2,3,4,5,6,7,8,9, and 10.
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PROG
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(Python)
x = [1] # To count the number in the sequence
y = [1] # To record the sequence itself
N = 20 # The number of terms to calculate
while len(x) < N:
x.append(x[-1] + 1)
# The next number in the sequence will be divisible by x[-1] so start
# looking at z which is the largest number divisible by x[-1] so far
z = y[-1]//x[-1] * x[-1] + x[-1]
# Now start counting up by x[-1] and check if the number is divisible by
# every integer less than x[-1]
while any([z%i != 0 for i in x]):
z += x[-1]
y.append(z)
print("{}\t {}\t".format(len(x), z))
(PARI) A003418(n) = if(n<1, n==0, 1/content(vector(n, k, 1/k)));
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CROSSREFS
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Would be same as A003418 without the "increasing" condition.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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