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a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).
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%I #15 Jan 08 2020 18:40:52

%S 1,6,66,300,8220,8820,457380,1917720,17965080,18600120,2320468920,

%T 2384502120,412970037480,422245703880,430902992520,1756076802480,

%U 516336630329520,524676485052720,192260441419366320,194970060218934000,197550649551855600,200013939369644400

%N a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).

%H Andrew Howroyd, <a href="/A175455/b175455.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) = (A001008(n) / A002805(n)) * (A003418(n))^2.

%F a(n) = A000142(n) * A025529(n) / A025527(n) = A025529(n) * A003418(n).

%F a(n) = (1/1 + 1/2 + ... + 1/n) * (lcm(1,2,...,n))^2.

%e For n = 3, a(3) = (1/1 + 1/2 + 1/3) * (1*2*3)^2 = (11/6) * 36 = 66.

%o (PARI) a(n)={sum(k=1, n, 1/k)*lcm([1..n])^2} \\ _Andrew Howroyd_, Jan 08 2020

%Y Cf. A000142, A001008, A002805, A003418, A025527, A025529.

%K nonn

%O 1,2

%A _Jaroslav Krizek_, May 17 2010

%E Terms a(13) and beyond from _Andrew Howroyd_, Jan 08 2020