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%I #37 Feb 16 2022 18:37:12
%S 0,1,3,6,9,12,15,19,22,26,30,34,38,42,46,50,55,59,63,68,72,77,82,86,
%T 91,96,101,106,110,115,120,125,130,135,141,146,151,156,161,166,172,
%U 177,182,188,193,198,204,209,215,220,225,231,236,242,248,253,259,264,270
%N 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.
%H Robert Israel, <a href="/A060293/b060293.txt">Table of n, a(n) for n = 0..10000</a>
%H R. Wyss, <a href="http://dx.doi.org/10.5169/seals-46962">Identitäten bei den Stirling-Zahlen 2. Art aus kombinatorischen Überlegungen beim Würfelspiel</a>, Elem. Math. 51 (1996) 102-106, Eq (5). [From _R. J. Mathar_, Aug 02 2009]
%F a(n) = ceiling(n*Sum_{k=1..n}(1/k)) = ceiling(n*A001008(n)/A002805(n)) = A052488(n) + 1 for n>2.
%e 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.
%p H := proc(n)
%p add(1/k,k=1..n) ;
%p end proc:
%p A060293 := proc(n)
%p ceil(n*H(n)) ;
%p end proc: # _R. J. Mathar_, Aug 02 2009, Dec 02 2016
%p A060293:= n -> ceil(Psi(n+1)+gamma); # _Robert Israel_, May 19 2014
%t f[n_] := Ceiling[n*HarmonicNumber[n]]; Array[f, 60, 0] (* _Robert G. Wilson v_, Nov 23 2015 *)
%o (PARI) vector(100, n, n--; ceil(n*sum(k=1, n, 1/k))) \\ _Altug Alkan_, Nov 23 2015
%o (Python)
%o from math import ceil
%o n=100 #number of terms
%o ans=0
%o finalans = [0]
%o for i in range(1, n+1):
%o ans+=(1/i)
%o finalans.append(ceil(ans*i))
%o print(finalans)
%o # _Adam Hugill_, Feb 14 2022
%Y Cf. A052488.
%K easy,nonn
%O 0,3
%A _Henry Bottomley_, Mar 24 2001
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