

A135736


Nearest integer to n*Sum_{k=1..n} 1/k = rounded expected coupon collection numbers.


2



0, 1, 3, 6, 8, 11, 15, 18, 22, 25, 29, 33, 37, 41, 46, 50, 54, 58, 63, 67, 72, 77, 81, 86, 91, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 161, 166, 171, 176, 182, 187, 192, 198, 203, 209, 214, 219, 225, 230, 236, 242, 247, 253, 258, 264, 269, 275
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OFFSET

0,3


COMMENTS

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.


LINKS

Adam Hugill, Table of n, a(n) for n = 0..10000 (terms 0..999 from G. C. Greubel)


FORMULA

a(n) = round(n*A001008(n)/A002805(n)) = either A052488(n) or A060293(n).
a(n) ~ A060293(n) ~ A052488(n) ~ A050502(n) ~ A050503(n) ~ A050504(n) (asymptotically)
Conjecture: a(n) = round(n*(log(n) + gamma) + 1/2) for n > 0, where gamma = A001620.  Ilya Gutkovskiy, Oct 31 2016
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


EXAMPLE

a(0) = 0 since nothing needs to be bought if nothing is to be collected.
a(1) = 1 since only 1 box needs to be bought if only 1 object is to be collected.
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.
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.


MAPLE

seq(round(n*harmonic(n)), n=1..100); # Robert Israel, Oct 31 2016


MATHEMATICA

Table[Round[n*Sum[1/k, {k, 1, n}]], {n, 0, 25}] (* G. C. Greubel, Oct 29 2016 *)


PROG

(PARI) A135736(n)=round(n*sum(i=1, n, 1/i))
(Python)
n=100 #set the number of terms you want to calculate here
ans=0
finalans = []
finalans.append(0) #continuity with A135736
for i in range(1, n+1):
ans+=(1/i)
finalans.append(int(round(ans*i)))
print(finalans)
# Adam Hugill, Feb 14 2022


CROSSREFS

Cf. A060293, A052488, A050502A050504.
Sequence in context: A085197 A164910 A276583 * A036434 A298795 A140400
Adjacent sequences: A135733 A135734 A135735 * A135737 A135738 A135739


KEYWORD

easy,nonn


AUTHOR

M. F. Hasler, Nov 29 2007


STATUS

approved



