A051542 -id:A051542 - cp's OEIS frontend

A051538 Least common multiple of {b(1),...,b(n)}, where b(k) = k(k+1)(2k+1)/6 = A000330(k).

1, 5, 70, 210, 2310, 30030, 60060, 1021020, 19399380, 19399380, 446185740, 2230928700, 6692786100, 194090796900, 12033629407800, 12033629407800, 12033629407800, 445244288088600, 445244288088600, 18255015811632600

Offset: 1

Asher Auel

Also a(n) = lcm(1,...,(2n+2))/12. - Paul Barry, Jun 09 2006. Proof that this is the same sequence, from Martin Fuller, May 06 2007: k, k+1, 2k+1 are coprime so their lcm is the same as their product. Hence a(n) = lcm{k, k+1, 2k+1 | k=1..n}/6. {k, k+1, 2k+1 | k=1..n} = {1..2n+2 excluding even numbers >n+1}. Adding the higher even numbers to the set doubles the LCM. Hence lcm{k, k+1, 2k+1 | k=1..n}/6 = lcm{1..2n+2}/12.

			a(4) = lcm(1, 5, 14, 30) = 210.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Second column of A120101.
Cf. A000330.
Cf. A051542 (LCM), A025555.

a051538 n = a051538_list !! (n-1)
a051538_list = scanl1 lcm $ tail a000330_list
-- Reinhard Zumkeller, Mar 12 2014

[Lcm([1..2*n+2])/12: n in [1..30]]; // G. C. Greubel, May 03 2023

Table[LCM@@Range[2n+2]/12,{n,30}] (* Harvey P. Dale, Apr 25 2011 *)

def A051538(n):
    return lcm(range(1,2*n+3))/12
[A051538(n) for n in range(1,31)] # G. C. Greubel, May 03 2023

Corrected by James Sellers

Edited by N. J. A. Sloane, May 06 2007

A051543 Quotients of consecutive values of lcm of first n triangular numbers (A000217).

Original entry on oeis.org

3, 2, 5, 1, 7, 2, 3, 1, 11, 1, 13, 1, 1, 2, 17, 1, 19, 1, 1, 1, 23, 1, 5, 1, 3, 1, 29, 1, 31, 2, 1, 1, 1, 1, 37, 1, 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 7, 1, 1, 1, 53, 1, 1, 1, 1, 1, 59, 1, 61, 1, 1, 2, 1, 1, 67, 1, 1, 1, 71, 1, 73, 1, 1, 1, 1, 1, 79, 1, 3, 1, 83, 1, 1, 1, 1, 1, 89, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Asher Auel

			a(5) = A025555(6)/A025555(5) = 210/30 = 7

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A025555.

Cf. A051542.

a051543 n = a051542_list !! (n-1)
a051543_list = zipWith div (tail a025555_list) a025555_list
-- Reinhard Zumkeller, Mar 12 2014

a(n) = A025555(n+1)/A025555(n)

Corrected and extended by James Sellers

cp's OEIS Frontend

A051538 Least common multiple of {b(1),...,b(n)}, where b(k) = k(k+1)(2k+1)/6 = A000330(k).

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