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a(n) = denominator(Sum_{k=1..n} 1/(prime(k)-1)).
8

%I #13 Jun 11 2021 09:43:16

%S 1,2,4,12,60,10,80,720,7920,55440,55440,18480,18480,18480,425040,

%T 5525520,160240080,53413360,160240080,160240080,480720240,480720240,

%U 19709529840,19709529840,39419059680,197095298400,3350620072800

%N a(n) = denominator(Sum_{k=1..n} 1/(prime(k)-1)).

%C A120271(n) = numerator(Sum_{k=1..n} 1/(prime(k)-1)); A128648(n) = denominator(Sum_{k=1..n} (-1)^(k+1)/(prime(k)-1)); numbers m such that a(m) = A128648(m) are listed in A128649.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeSums.html">Prime Sums</a>

%F a(n) = denominator(Sum_{k=1..n} 1/(prime(k)-1)).

%t Table[Denominator[Sum[1/(Prime[k]-1),{k,1,n}]],{n,1,36}]

%Y Cf. A120271 (numerator(Sum_{k=1..n} 1/(prime(k)-1))).

%Y Cf. A128649, A128647, A128648 (denominator(Sum_{k=1..n} (-1)^(k+1)/(prime(k)-1))).

%Y Cf. A119686, A006093, A000040.

%K frac,nonn

%O 1,2

%A _Alexander Adamchuk_, Mar 18 2007