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%I #47 Feb 16 2022 18:38:46
%S 0,1,3,6,8,11,15,18,22,25,29,33,37,41,46,50,54,58,63,67,72,77,81,86,
%T 91,95,100,105,110,115,120,125,130,135,140,145,150,155,161,166,171,
%U 176,182,187,192,198,203,209,214,219,225,230,236,242,247,253,258,264,269,275
%N Nearest integer to n*Sum_{k=1..n} 1/k = rounded expected coupon collection numbers.
%C Somewhat more realistic than A052488 but more optimistic than A060293, the expected number of boxes that must be bought to get the full collection of N objects, if each box contains any one of them at random. See also comments in A060293, A052488.
%H Adam Hugill, <a href="/A135736/b135736.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..999 from G. C. Greubel)
%F a(n) = round(n*A001008(n)/A002805(n)) = either A052488(n) or A060293(n).
%F a(n) ~ A060293(n) ~ A052488(n) ~ A050502(n) ~ A050503(n) ~ A050504(n) (asymptotically)
%F Conjecture: a(n) = round(n*(log(n) + gamma) + 1/2) for n > 0, where gamma = A001620. - _Ilya Gutkovskiy_, Oct 31 2016
%F The conjecture is false: a(2416101) = 36905656 while round(2416101*(log(2416101) + gamma) + 1/2) = 36905657, with the unrounded numbers being 36905656.499999982... and 36905656.500000016.... Heuristically this should happen infinitely often. - _Charles R Greathouse IV_, Oct 31 2016
%e a(0) = 0 since nothing needs to be bought if nothing is to be collected.
%e a(1) = 1 since only 1 box needs to be bought if only 1 object is to be collected.
%e a(2) = 3 since the chance of getting the other object at the second purchase is only 1/2, so it takes 2 boxes on the average to get it.
%e a(3) = 6 since the chance of getting a new object at the second purchase is 2/3 so it takes 3/2 boxes in the mean, then the chance becomes 1/3 to get the 3rd, i.e., 3 other boxes on the average to get the full collection and the rounded value of 1 + 3/2 + 3 = 11/2 = 5.5 is 6.
%p seq(round(n*harmonic(n)), n=1..100); # _Robert Israel_, Oct 31 2016
%t Table[Round[n*Sum[1/k, {k, 1, n}]], {n,0,25}] (* _G. C. Greubel_, Oct 29 2016 *)
%o (PARI) A135736(n)=round(n*sum(i=1,n,1/i))
%o (Python)
%o n=100 #set the number of terms you want to calculate here
%o ans=0
%o finalans = []
%o finalans.append(0) #continuity with A135736
%o for i in range(1, n+1):
%o ans+=(1/i)
%o finalans.append(int(round(ans*i)))
%o print(finalans)
%o # _Adam Hugill_, Feb 14 2022
%Y Cf. A060293, A052488, A050502-A050504.
%K easy,nonn
%O 0,3
%A _M. F. Hasler_, Nov 29 2007
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