A060293 Expected coupon collection numbers rounded up; i.e., if aiming to collect a set of n coupons, the expected number of random coupons required to receive the full set.

0, 1, 3, 6, 9, 12, 15, 19, 22, 26, 30, 34, 38, 42, 46, 50, 55, 59, 63, 68, 72, 77, 82, 86, 91, 96, 101, 106, 110, 115, 120, 125, 130, 135, 141, 146, 151, 156, 161, 166, 172, 177, 182, 188, 193, 198, 204, 209, 215, 220, 225, 231, 236, 242, 248, 253, 259, 264, 270

Offset: 0

Author

Henry Bottomley, Mar 24 2001

Keywords

• easy
• nonn

Examples

			a(2)=3 since the probability of getting both coupons after two is 1/2, after 3 is 1/4, after 4 is 1/8, etc. and 2/2 + 3/2^2 + 4/2^3 + ... = 3.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• R. Wyss, Identitäten bei den Stirling-Zahlen 2. Art aus kombinatorischen Überlegungen beim Würfelspiel, Elem. Math. 51 (1996) 102-106, Eq (5). [From _R. J. Mathar_, Aug 02 2009]

Crossrefs

Cf. A052488.

Programs

Maple

H := proc(n)
    add(1/k,k=1..n) ;
end proc:
A060293 := proc(n)
    ceil(n*H(n)) ;
end proc: # R. J. Mathar, Aug 02 2009, Dec 02 2016
A060293:= n -> ceil(Psi(n+1)+gamma); # Robert Israel, May 19 2014

Mathematica

f[n_] := Ceiling[n*HarmonicNumber[n]]; Array[f, 60, 0] (* Robert G. Wilson v, Nov 23 2015 *)

PARI

vector(100, n, n--; ceil(n*sum(k=1, n, 1/k))) \\ Altug Alkan, Nov 23 2015

Python

from math import ceil
n=100 #number of terms
ans=0
finalans = [0]
for i in range(1, n+1):
    ans+=(1/i)
    finalans.append(ceil(ans*i))
print(finalans)
# Adam Hugill, Feb 14 2022

Formula

a(n) = ceiling(n*Sum_{k=1..n}(1/k)) = ceiling(n*A001008(n)/A002805(n)) = A052488(n) + 1 for n>2.