A025547 Least common multiple of {1,3,5,...,2n-1}.

1, 3, 15, 105, 315, 3465, 45045, 45045, 765765, 14549535, 14549535, 334639305, 1673196525, 5019589575, 145568097675, 4512611027925, 4512611027925, 4512611027925, 166966608033225, 166966608033225, 6845630929362225, 294362129962575675, 294362129962575675

Offset: 1

Author

Clark Kimberling

Keywords

• easy
• nice
• nonn

Comments

This sequence coincides with the sequence f(n) = denominator of 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) iff n <= 38. But a(39) = 6414924694381721303722858446525, f(39) = 583174972216520118520259858775. - T. D. Noe, Aug 04 2004 [See A350670(n-1).]

Coincides for n=1..42 with the denominators of a series for Pi*sqrt(2)/4 and then starts to differ. See A127676.

a(floor((n+1)/2)) = gcd(a(n), A051426(n)). - Reinhard Zumkeller, Apr 25 2011

A051417(n) = a(n+1)/a(n).

Links

• T. D. Noe, Table of n, a(n) for n = 1..200
• Yue-Wu Li and Feng Qi, A New Closed-Form Formula of the Gauss Hypergeometric Function at Specific Arguments, Axioms (2024) Vol. 13, Art. No. 317. See p. 11 of 24.
• Eric Weisstein's World of Mathematics, Jeep Problem, Pi, Pi Continued Fraction, Least Common Multiple
• Wikipedia, Least common multiple
• Index entries for sequences related to lcm's

Crossrefs

Cf. A007509, A025550, A075135. The numerators are in A074599.

Cf. A003418 (LCM of {1..n}).

Cf. A005408, A350670.

Programs

Haskell

a025547 n = a025547_list !! (n-1)
a025547_list = scanl1 lcm a005408_list
-- Reinhard Zumkeller, Oct 25 2013, Apr 25 2011

Maple

A025547:=proc(n) local i,t1; t1:=1; for i from 1 to n do t1:=lcm(t1,2*i-1); od: t1; end;
f := n->denom(add(1/(2*k-1),k=0..n)); # a different sequence!

Mathematica

a = 1; Join[{1}, Table[a = LCM[a, n], {n, 3, 125, 2}]] (* Zak Seidov, Jan 18 2011 *)
nn=30;With[{c=Range[1,2*nn,2]},Table[LCM@@Take[c,n],{n,nn}]] (* Harvey P. Dale, Jan 27 2013 *)

PARI

a(n)=lcm(vector(n,k,2*k-1)) \\ Charles R Greathouse IV, Nov 20 2012

Python

# generates initial segment of sequence
from math import gcd
from itertools import accumulate
def lcm(a, b): return a * b // gcd(a, b)
def aupton(nn): return list(accumulate((2*i+1 for i in range(nn)), lcm))
print(aupton(23)) # Michael S. Branicky, Mar 28 2022