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a(0) = 1; for n > 0, a(n) = lcm(a(n-1), a(n-2)+1, a(n-3)+2, ..., a(0)+n-1).
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%I #44 Mar 19 2024 13:47:54

%S 1,1,2,6,12,420,10920,13791960,31781240978760,

%T 2475081972959523523910954640,

%U 3442004402022863055515137969749021701676249608093520,45446117889526225044218072480163823127876682090382722344419488836545177458721523373548206735891390800

%N a(0) = 1; for n > 0, a(n) = lcm(a(n-1), a(n-2)+1, a(n-3)+2, ..., a(0)+n-1).

%F a(n) = lcm_{k=0..n-1} a(n-1 - k) + k with a(0)=1.

%e Prepend a(n-1)

%e after incrementing the value

%e in each of these columns

%e | | | |

%e v v v v

%e n=0: 1 = a(0),

%e n=1: lcm( a(0) ) = lcm( 1) = 1 = a(1),

%e n=2: lcm( a(1), a(0)+1) = lcm( 1,2) = 2 = a(2),

%e n=3: lcm( a(2), a(1)+1, a(0)+2) = lcm( 2,2,3) = 6 = a(3),

%e n=4: lcm(a(3), a(2)+1, a(1)+2, a(0)+3) = lcm(6,3,3,4) = 12 = a(4).

%o (Python)

%o import math

%o def a(n):

%o p=[1]

%o for i in range(n):

%o p.append(math.lcm(*p))

%o for x in range(0, len(p)-1):

%o p[x]+=1

%o return p[-1]

%K nonn

%O 0,3

%A _Justin Dugan_, Nov 13 2023