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A309233 Lexicographically earliest increasing sequence such that n-th term is divisible by all positive integers up to n. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..2000

EXAMPLE

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.

PROG

(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)));

a(n) = {my(k=1); if(n>1, k=A003418(n)*(1+a(n-1)\A003418(n))); k} \\ Jinyuan Wang, Jul 22 2019

CROSSREFS

Would be same as A003418 without the "increasing" condition.

Sequence in context: A004490 A224078 A309811 * A135060 A185021 A191836

Adjacent sequences:  A309230 A309231 A309232 * A309234 A309235 A309236

KEYWORD

nonn

AUTHOR

Rian Rustvold, Jul 16 2019

EXTENSIONS

More terms from Jinyuan Wang, Jul 22 2019

STATUS

approved

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Last modified November 18 22:50 EST 2019. Contains 329305 sequences. (Running on oeis4.)