A052488 -id:A052488 - cp's OEIS frontend

A278586 Start with X = n^2. Repeatedly replace X with X - ceiling(X/n); a(n) is the number of steps to reach 0.

1, 3, 5, 8, 11, 14, 17, 21, 24, 28, 32, 36, 40, 44, 49, 53, 57, 62, 66, 71, 75, 80, 84, 90, 94, 99, 103, 109, 113, 118, 123, 128, 133, 139, 143, 149, 154, 159, 164, 170, 175, 180, 185, 191, 196, 201, 207, 212, 217, 223, 229, 234, 240, 246, 251, 256, 262, 268, 273, 279, 284, 290, 296, 302, 308

Offset: 1

N. J. A. Sloane, Dec 02 2016, based on discussions about the Pythagoras article in the Sequence Fans Mailing List, Dec 01 2016. Jack Brennen provided the definition given here

Robert G. Wilson v, Table of n, a(n) for n = 1..1000
Matthijs Coster, Een Eigen Rij - Uitslag Prijsvraag, Pythagoras, Number 6, June 2016, pp. 20-21. The sequence was discovered by Pim Spelier.

Cf. A052488.

a:=[]; for n in [1..58] do k:=n^2; count:=0; while k gt 0 do count+:=1; k-:=Ceiling(k/n); end while; a[n]:=count; end for; a; // Jon E. Schoenfield, Dec 01 2016

A278586 := proc(n)
    local x,a;
    x := n^2 ;
    a := 0 ;
    while x <> 0 do
        x:= x-ceil(x/n) ;
        a := a+1 ;
    end do:
    a;
end proc: # R. J. Mathar, Dec 02 2016

f[n_] := Length@ NestWhileList[# - Ceiling[#/n] &, n^2, # > 1 &]; Array[f, 65] (* Robert G. Wilson v, Dec 01 2016 *)

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.

Original entry on oeis.org

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

Henry Bottomley, Mar 24 2001

			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.

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]

Cf. A052488.

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

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

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

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

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

A102722 Given n, sum all division remainders {n/k}, with k=1,...,n. The value a(n) is given by the floor of that sum. Note that {x}:=x-[x].

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 2, 1, 2, 2, 4, 2, 4, 4, 4, 4, 6, 4, 7, 5, 6, 7, 9, 6, 8, 9, 10, 8, 11, 8, 11, 10, 11, 13, 14, 10, 13, 14, 15, 13, 16, 13, 17, 16, 15, 17, 20, 16, 18, 17, 19, 18, 22, 20, 21, 19, 20, 22, 26, 19, 23, 25, 24, 23, 25, 23, 26, 26, 28, 26, 30, 23, 27, 29, 29, 29, 31, 29, 33

Offset: 1

Carlos Alves, Feb 06 2005

Conjecture: a(n) ~ (1-EulerGamma)n.

			a(5) = [{5/1}+{5/2}+{5/3}+{5/4}+{5/5}]=[0+0.5+0.6666+0.25+0]=[1.4166]=1 (division by 1 or by the number itself is to be avoided).

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

Cf. A052488, A006218.

Cf. A000005.

N:= 100:
H:= ListTools:-PartialSums([seq(1/n,n=1..N)]):
S:= ListTools:-PartialSums(map(numtheory:-tau,[$1..N])):
seq(floor(n*H[n])-S[n],n=1..N); # Robert Israel, Mar 20 2016

Resto = Function[n, Sum[n/k - Floor[n/k], {k, 2, n - 1}]]; Floor[Map[Resto, Range[1, 1000]]]
Table[Floor[n*HarmonicNumber[n]] - Sum[DivisorSigma[0, k], {k, 1, n}], {n, 1, 200}] (* Enrique Pérez Herrero, Aug 25 2009 *)
Table[Floor[Sum[FractionalPart[n/k], {k, 1, n}]], {n, 1, 200}] (* Enrique Pérez Herrero, Aug 25 2009 *)

from math import isqrt
from sympy import harmonic
def A102722(n): return int(n*harmonic(n))+(s:=isqrt(n))**2-(sum(n//k for k in range(1,s+1))<<1) # Chai Wah Wu, Oct 24 2023

a(n) = floor(n*H(n)) - Sum_{j=1..n} d(j), where d(n)=A000005(n) is the number of divisors of n, and H(n) is the n-th Harmonic Number. [Enrique Pérez Herrero, Aug 25 2009; corrected by Robert Israel, Mar 20 2016]

a(n) = A052488(n) - A006218(n). [Enrique Pérez Herrero, Aug 25 2009]

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

A079447 Primes p such that there is an integer k satisfying p = floor(k*H(k)) where H(k) denotes the k-th harmonic number (i.e., H(k) = 1 + 1/2 + 1/3 + ... + 1/k).

Original entry on oeis.org

3, 5, 11, 29, 37, 41, 67, 71, 109, 181, 197, 241, 263, 269, 349, 367, 373, 379, 397, 409, 421, 433, 439, 457, 463, 487, 587, 593, 599, 631, 701, 727, 773, 911, 971, 991, 1039, 1093, 1223, 1237, 1279, 1307, 1321, 1433, 1447, 1489, 1553, 1567, 1667, 1747, 1783

Offset: 1

Benoit Cloitre, Jan 13 2003

nonn

			10*Sum_{j=1..10} 1/j = 29.2896825..., hence 29 = floor(10*H(10)) and 29 is in the sequence.

Primes in A052488.

Cf. A001008/A002805.

Data corrected by Sean A. Irvine, Aug 13 2025

A090541 a(n) = floor((Sum_{r=1..n} r)*(Sum_{r=1..n} 1/r)).

Original entry on oeis.org

1, 4, 11, 20, 34, 51, 72, 97, 127, 161, 199, 242, 289, 341, 398, 459, 526, 597, 674, 755, 842, 933, 1030, 1132, 1240, 1352, 1470, 1594, 1723, 1857, 1997, 2142, 2293, 2450, 2612, 2780, 2953, 3132, 3317, 3508, 3704, 3907, 4115, 4328, 4548, 4774, 5006, 5243

Offset: 1

Amarnath Murthy, Dec 09 2003

nonn

For large n, a(n) = floor(n*(n+1)/2)*(log(n) + 0.5772...,(Euler's constant)).

Cf. A052488.

a:=n->floor(sum(k,k=1..n)*sum(1/k,k=1..n)):seq(a(n),n=1..56); # Emeric Deutsch, Apr 18 2005

a(n) = floor((n*(n+1)/2)*sum(k=1, n, 1/k)); \\ Michel Marcus, Aug 18 2019

More terms from Emeric Deutsch, Apr 18 2005

cp's OEIS Frontend

A278586 Start with X = n^2. Repeatedly replace X with X - ceiling(X/n); a(n) is the number of steps to reach 0.

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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.

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A102722 Given n, sum all division remainders {n/k}, with k=1,...,n. The value a(n) is given by the floor of that sum. Note that {x}:=x-[x].

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

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A079447 Primes p such that there is an integer k satisfying p = floor(k*H(k)) where H(k) denotes the k-th harmonic number (i.e., H(k) = 1 + 1/2 + 1/3 + ... + 1/k).

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A090541 a(n) = floor((Sum_{r=1..n} r)*(Sum_{r=1..n} 1/r)).

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