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A307642 a(n) = n!*Sum_{i=1..n} (Sum_{j=1..i} (i/j)). 0

%I #43 Sep 08 2022 08:46:21

%S 1,8,57,428,3510,31644,312984,3380544,39664080,502927200,6858181440,

%T 100135491840,1559197261440,25797280723200,452046655872000,

%U 8364495012249600,162994310248089600,3336683369519001600,71596721810396160000,1606993396943155200000

%N a(n) = n!*Sum_{i=1..n} (Sum_{j=1..i} (i/j)).

%F a(n) = n! * Sum_{i=1..n} (Sum_{j=1..i} (i/j)).

%F a(n) = n * A182541(n+2).

%F a(n) = (1/4) * n * (n+1)! * (2*harmonic(n+1) - 1).

%e a(3) = 57 because a(3) = 3!*Sum_{i=1..3} (Sum_{j=1..i} (i/j)).

%t Array[#!*Sum[Sum[i/j, {j, i}], {i, #}] &, 25] (* _Michael De Vlieger_, Apr 21 2019 *)

%t Table[n*(n+1)!*(2*HarmonicNumber[n+1] -1)/4, {n, 25}] (* _G. C. Greubel_, Jul 15 2019 *)

%o (PARI) a(n)=n!*sum(i=1, n, sum(j=1, i, i/j)); \\ _Michel Marcus_, Apr 20 2019

%o (Magma) [n*Factorial(n+1)*(2*HarmonicNumber(n+1)-1)/4: n in [1..25]]; // _G. C. Greubel_, Jul 15 2019

%o (Sage) [n*factorial(n+1)*(2*harmonic_number(n+1)-1)/4 for n in (1..25)] # _G. C. Greubel_, Jul 15 2019

%o (GAP) List([1..25], n-> n*Factorial(n+1)*(1+2*Sum([2..n+1], j-> 1/j))/4 ); # _G. C. Greubel_, Jul 15 2019

%Y Cf. A001008/A002805 (harmonic), A182541.

%K nonn

%O 1,2

%A _Pedro Caceres_, Apr 19 2019

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Last modified April 23 13:51 EDT 2024. Contains 371914 sequences. (Running on oeis4.)