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 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. 5
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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] FORMULA a(n) = ceiling(n*Sum_{k=1..n}(1/k)) = ceiling(n*A001008(n)/A002805(n)) = A052488(n) + 1 for n>2. EXAMPLE 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. 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 *) PROG (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 CROSSREFS Cf. A052488. Sequence in context: A070021 A083354 A156242 * A336803 A220657 A194273 Adjacent sequences: A060290 A060291 A060292 * A060294 A060295 A060296 KEYWORD easy,nonn AUTHOR Henry Bottomley, Mar 24 2001 STATUS approved

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Last modified December 7 01:40 EST 2022. Contains 358649 sequences. (Running on oeis4.)