A060293 -id:A060293 - cp's OEIS frontend

A052488 a(n) = floor(n*H(n)) where H(n) is the n-th harmonic number, Sum_{k=1..n} 1/k (A001008/A002805).

1, 3, 5, 8, 11, 14, 18, 21, 25, 29, 33, 37, 41, 45, 49, 54, 58, 62, 67, 71, 76, 81, 85, 90, 95, 100, 105, 109, 114, 119, 124, 129, 134, 140, 145, 150, 155, 160, 165, 171, 176, 181, 187, 192, 197, 203, 208, 214, 219, 224, 230, 235, 241, 247, 252, 258, 263, 269

Offset: 1

Tomas Mario Kalmar (TomKalmar(AT)aol.com), Mar 15 2000

Floor(n*H(n)) gives a (very) rough approximation to the n-th prime.

a(n) is the integer part of the solution to the Coupon Collector's Problem. For example, if there are n=4 different prizes to collect from cereal boxes and they are equally likely to be found, then the integer part of the average number of boxes to buy before the collection is complete is a(4)=8. - Ron Lalonde (ronronronlalonde(AT)hotmail.com), Feb 04 2004

John D. Barrow, One Hundred Essential Things You Didn't Know You Didn't Know, Ch. 3, 'On the Cards', W. W. Norton & Co., NY & London, 2008, pp. 30-32.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A001008, A002805, A006218, A060293.

Cf. A001620, A073004.

[Floor(n*HarmonicNumber(n)): n in [1..60]]; // G. C. Greubel, May 14 2019

for n from 1 to 100 do printf(`%d,`,floor(n*sum(1/k, k=1..n))) od:
# Alternatively:
A052488:= n -> floor(n*(Psi(n+1)+gamma));
seq(A052488(n),n=1..100); # Robert Israel, May 19 2014

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

a(n) = floor(n*sum(k=1, n, 1/k)) \\ Altug Alkan, Nov 23 2015

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

from fractions import Fraction
from itertools import count, islice
def agen():
    Hn = 0
    for n in count(1):
        Hn += Fraction(1, n)
        yield (n*Hn.numerator)//Hn.denominator
print(list(islice(agen(), 60))) # Michael S. Branicky, Aug 10 2022

from sympy import harmonic
def A052488(n): return int(n*harmonic(n)) # Chai Wah Wu, Oct 24 2023

[floor(n*harmonic_number(n)) for n in (1..60)] # G. C. Greubel, May 14 2019

More terms from James Sellers, Mar 17 2000

A135736 Nearest integer to n*Sum_{k=1..n} 1/k = rounded expected coupon collection numbers.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Nov 29 2007

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.

			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.

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

Cf. A060293, A052488, A050502-A050504.

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

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

PARI
```
A135736(n)=round(n*sum(i=1,n,1/i))
```

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

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

A071417 Triangle of expected coupon collection numbers rounded up; i.e., if aiming to collect part of a set of n coupons, the expected number of random coupons required to receive first the set with exactly k missing.

Original entry on oeis.org

0, 1, 0, 3, 1, 0, 6, 3, 1, 0, 9, 5, 3, 1, 0, 12, 7, 4, 3, 1, 0, 15, 9, 6, 4, 3, 1, 0, 19, 12, 8, 6, 4, 3, 1, 0, 22, 14, 10, 8, 6, 4, 3, 1, 0, 26, 17, 12, 9, 7, 5, 4, 3, 1, 0, 30, 20, 15, 11, 9, 7, 5, 4, 3, 1, 0, 34, 23, 17, 14, 11, 9, 7, 5, 4, 3, 1, 0, 38, 26, 20, 16, 13, 10, 8, 7, 5, 4, 3, 1, 0, 42

Offset: 0

Henry Bottomley, May 29 2002

			Rows start
0;
1,0;
3,1,0;
6,3,1,0;
9,5,3,1,0;
etc.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150).

Cf. A060293 (left hand column), A067176.

Table[Ceiling[n Sum[1/j, {j, k + 1, n}]], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 30 2017 *)

a(n, k) = ceiling(n*Sum_{j=k+1..n} 1/j) = ceiling(A067176(n, k)*k!/(n-1)!) = ceiling(A008279(n, n-k)*Sum_{j>=n-k} j*A008277(j-1, n-k-1)/n^j).

A085813 Number of cards that need to be drawn (with replacement) from a deck of n cards to have a 95% or greater chance of seeing each card at least once.

Original entry on oeis.org

1, 6, 11, 16, 21, 27, 33, 38, 44, 51, 57, 63, 70, 76, 83, 90, 96, 103, 110, 117, 124, 131, 138, 145, 152, 159, 167, 174, 181, 189, 196, 203, 211, 218, 226, 233, 241, 248, 256, 264, 271, 279, 287, 294, 302, 310, 318, 326, 333, 341, 349, 357, 365, 373, 381, 389

Offset: 1

Alfred Heiligenbrunner, Jul 25 2003

nonn

The probability that there are at most k different cards in t drawings is (k/m)^t * binomial(m,k). This also includes the cases with k-1 different cards, which we want to subtract. Inclusion and exclusion leads to the formula Sum_{k=1..m} (-1)^(m-k) * (k/m)^t * binomial(m,k).

			a(2)=6 because you have to throw a coin 6 times to get both sides at least once with probability greater than or equal to 0.95. (The probability of getting only one side in a series of 6 throws is (1/2)^6 * 2 = 1/32 = 0.03125 < 0.05.)
a(6)=27 because you have to roll a die 27 times to see all 6 possible outcomes with a probability over 0.95. (If you roll a die 27 times, the probability of getting all 6 sides at least once is 0.95658638... . If you roll the die only 26 times, the probability is 0.94798274... .)

Cf. A073593 (number of drawings for a 50% probability of seeing each card, = median) and A060293 (expected value of the number of drawings until each card is drawn once).

Cf. A090582.

f[1] = 1; f[n_] := f[n] = Block[{k = f[n - 1]}, While[ 2StirlingS2[k, n]*n!/n^k < 19/10, k++ ]; k]; Table[ f[n], {n, 1, 56}]

More terms from Robert G. Wilson v, Sep 07 2003

cp's OEIS Frontend

A052488 a(n) = floor(n*H(n)) where H(n) is the n-th harmonic number, Sum_{k=1..n} 1/k (A001008/A002805).

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A135736 Nearest integer to n*Sum_{k=1..n} 1/k = rounded expected coupon collection numbers.

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A071417 Triangle of expected coupon collection numbers rounded up; i.e., if aiming to collect part of a set of n coupons, the expected number of random coupons required to receive first the set with exactly k missing.

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Crossrefs

Programs

Mathematica

Formula

A085813 Number of cards that need to be drawn (with replacement) from a deck of n cards to have a 95% or greater chance of seeing each card at least once.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions