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A345683 a(n) = n! * Sum_{k=1..n} 1/floor(n/k). 7

%I #35 Oct 27 2023 09:42:15

%S 1,3,14,66,444,2880,25080,216720,2247840,24071040,304335360,

%T 3752179200,54965433600,810550540800,13176376012800,219079045785600,

%U 4078723532083200,75227891042304000,1550619342784512000,31871016307113984000,710529031487987712000,16180987966182014976000

%N a(n) = n! * Sum_{k=1..n} 1/floor(n/k).

%H Seiichi Manyama, <a href="/A345683/b345683.txt">Table of n, a(n) for n = 1..448</a>

%H Vaclav Kotesovec, <a href="/A345683/a345683.jpg">Plot of a(n) / (n*n!) for n = 1..1000000</a>

%H Mathoverflow, <a href="https://mathoverflow.net/questions/338432/an-asymptotic-formula-for-a-sum-involving-powers-of-floor-functions">An asymptotic formula for a sum involving powers of floor functions</a>, 2019.

%F a(n) ~ c * n * n!, where c = Sum_{j>=1} 1/(j^2*(j+1)) = Pi^2/6 - 1 = 0.644934... [proved by Harry Richman, see Mathoverflow link]

%F E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k). - _Seiichi Manyama_, Jul 23 2022

%t Table[n! * Sum[1/Floor[n/k], {k, 1, n}], {n, 1, 25}]

%t Table[n!*(Sum[(Floor[n/j] - Floor[n/(j + 1)])/j, {j, 1, n}]), {n, 1, 25}]

%o (PARI) a(n) = n!*sum(k=1, n, 1/(n\k)); \\ _Michel Marcus_, Jun 24 2021

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k))/(1-x))) \\ _Seiichi Manyama_, Jul 23 2022

%o (Python)

%o from math import factorial, isqrt

%o def A345683(n): return (m:=factorial(n))*(n-1)+m//n+sum((q:=n//k)*(m//k-m//(k-1))+m//q for k in range(2,isqrt(n)+1)) # _Chai Wah Wu_, Oct 27 2023

%Y Cf. A006218, A010786, A092143, A117871, A345682, A345684, A345686, A355987.

%K nonn

%O 1,2

%A _Vaclav Kotesovec_, Jun 23 2021

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Last modified March 29 08:13 EDT 2024. Contains 371265 sequences. (Running on oeis4.)