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a(n) = lcm(tau(n+1), tau(n)), where tau = A000005.
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%I #16 Jan 27 2025 15:14:04

%S 2,2,6,6,4,4,4,12,12,4,6,6,4,4,20,10,6,6,6,12,4,4,8,24,12,4,12,6,8,8,

%T 6,12,4,4,36,18,4,4,8,8,8,8,6,6,12,4,10,30,6,12,12,6,8,8,8,8,4,4,12,

%U 12,4,12,42,28,8,8,6,12,8,8,12,12,4,12,6,12,8,8,10,10,20,4,12,12,4,4,8,8,12

%N a(n) = lcm(tau(n+1), tau(n)), where tau = A000005.

%H Harry J. Smith, <a href="/A060779/b060779.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = lcm(A000005(n+1), A000005(n)).

%t LCM@@#&/@Partition[DivisorSigma[0,Range[90]],2,1] (* _Harvey P. Dale_, Jan 27 2025 *)

%o (PARI) a(n) = { lcm(numdiv(n), numdiv(n+1)) } \\ _Harry J. Smith_, Jul 11 2009

%Y Cf. A057921, A000005.

%K nonn

%O 1,1

%A _Labos Elemer_, Apr 26 2001