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A135736 Nearest integer to n*Sum_{k=1,..n}(1/k) = rounded expected coupon collection numbers. 1
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 (list; graph; refs; listen; history; text; internal format)
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

G. C. Greubel, Table of n, a(n) for n = 0..999

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=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))

CROSSREFS

Cf. A060293, A052488, A050502-A050504.

Sequence in context: A085197 A164910 A276583 * A036434 A140400 A190430

Adjacent sequences:  A135733 A135734 A135735 * A135737 A135738 A135739

KEYWORD

easy,nonn

AUTHOR

M. F. Hasler, Nov 29 2007

STATUS

approved

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Last modified November 20 12:23 EST 2017. Contains 294971 sequences.