A182010 -id:A182010 - cp's OEIS frontend

A014088 Minimal number of people to give a 50% probability of having at least n coincident birthdays in one year.

1, 23, 88, 187, 313, 460, 623, 798, 985, 1181, 1385, 1596, 1813, 2035, 2263, 2494, 2730, 2970, 3213, 3459, 3707, 3959, 4213, 4470, 4728, 4989, 5252, 5516, 5783, 6051, 6320, 6592, 6864, 7138, 7413, 7690, 7968, 8247, 8527, 8808, 9090, 9373, 9657, 9942, 10228

Offset: 1

Steven Finch

nonn

Stig Blücher Brink, Table of n, a(n) for n = 1..10000 (terms 1..61 from Hiroaki Yamanouchi, terms 62..250 from Rob Cook)
Patrice Le Conte, Coincident Birthdays.
P. Diaconis and F. Mosteller, Methods of studying coincidences, J. Amer. Statist. Assoc. 84 (1989), pp. 853-861.
Bruce Levin, Exact Solutions of the Generalized Birthday Problem.
B. Martin, Coincidence:Remarkable or Random, Skeptical Inquirer Volume 22.5, September / October 1998.
I. Peterson, Mathtrek, Birthday Surprises [Archived version from Jun 28 2013]
Maksym Petkus, Efficient (Non-)Membership Tree from Multicollision-Resistance with Applications to Zero-Knowledge Proofs, Cryptology ePrint Archive (2024). See pp. 6, 34.
Eric Weisstein's World of Mathematics, Birthday Problem.

Cf. A033810 (2 people on n days), A225852 (3 people on n days), A225871 (4 people on n days).

Cf. A088141, A182008, A182009, A182010.

q[1][n_, d_] := q[1][n, d] = d!/((d-n)!*d^n) // N; q[k_][n_, d_] := q[k][n, d] = Sum[ n!*d!/(d^(i* k)*i!*(k!)^i*(n-i*k)!*(d-i)!)*Sum[ q[j][n-i*k, d-i]*(d-i)^(n-i* k)/d^(n-i*k), {j, 1, k-1}], {i, 1, Floor[n/k]}] // N; p[k_][n_, d_] := 1 - Sum[q[i][n, d], {i, 1, k-1}]; a[1] = 1; a[k_] := a[k] = For[n = a[k-1], True, n++, If[p[k][n, 365] >= 1/2, Return[n]]]; Table[ Print["a(", k, ") = ", a[k]]; a[k], {k, 1, 15}] (* Jean-François Alcover, Jun 12 2013, after Eric W. Weisstein *)

Broken links corrected by Steven Finch, Jan 27 2009

a(16)-a(45) from Hiroaki Yamanouchi, Mar 19 2015

A033810 Number of people needed so that probability of at least two sharing a birthday out of n possible days is at least 50%.

Original entry on oeis.org

2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12

Offset: 1

Eric W. Weisstein

a(365) = 23 is the solution to the Birthday Problem.

T. D. Noe and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (366 terms from T. D. Noe)
S. E. Ahmed and R. J. McIntosh, An Asymptotic Approximation for the Birthday Problem, Crux Mathematicorum 26(3) 151-5 2000 CMS.
D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, June 2012, Volume 28, Issue 2, pp. 223-238. - From _N. J. A. Sloane_, Oct 08 2012
Mathforum, The Birthday Problem.
Eric Weisstein's World of Mathematics, Birthday Problem.

Essentially the same as A088141. See also A182008, A182009, A182010.
Cf. A014088 (n people on 365 days), A225852 (3 on n days), A225871 (4 people on n days).
Cf. A380129 (Strong Birthday Problem).

lst = {}; s = 1; Do[Do[If[Product[(n - i + 1)/n, {i, j}] <= 1/2, If[j > s, s = j]; AppendTo[lst, j]; Break[]], {j, s, s + 1}], {n, 86}]; lst (* Arkadiusz Wesolowski, Apr 29 2012 *)
A033810[n_] := Catch@Do[If[1/2 >= n!/(n - m)!/n^m, Throw[m]], {m, 2, Infinity}]; Array[A033810, 86] (* JungHwan Min, Mar 27 2017 *)

from math import comb, factorial
def A033810(n):
    def p(m): return comb(n,m)*factorial(m)<<1
    kmin, kmax = 0, 1
    while p(kmax) > n**kmax: kmax<<=1
    while kmax-kmin > 1:
        kmid = kmax+kmin>>1
        if p(kmid) <= n**kmid:
            kmax = kmid
        else:
            kmin = kmid
    return kmax # Chai Wah Wu, Jan 21 2025

a(n) = ceiling(sqrt(2*n*log(2)) + (3 - 2*log(2))/6 + (9 - 4*log(2)^2) / (72*sqrt(2*n*log(2))) - 2*log(2)^2/(135*n)) for all n up to 10^18. It is conjectured that this formula holds for all n.

A225852 The number of people required for there to be at least a 50% chance that at least 3 share a birthday in a year with n days.

Original entry on oeis.org

3, 4, 5, 6, 7, 7, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 28, 28, 28, 29, 29, 29, 29, 30, 30

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 17 2013

nonn

a(365) = 88.

For n <= 1000, a(n) = 1.436 + 1.812*n^0.654 - 0.817/n^3 provides an estimate accurate to 0.6 units.

			The probability that out of 87 people 3 share a birthday in a year with 365 days is 0.4994549. The corresponding probability for 88 people is 0.5110651. Therefore a(365)=88.

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000
Christian N. K. Anderson, Table of n and exact probabilities of a(n)-1 and a(n) for n = 1..1000
Patrice Le Conte, Coincident Birthdays

Cf. A014088 (n people on 365 days), A033810 (2 people on n days), A225871 (4 people on n days).

Cf. A088141, A182008, A182009, A182010.

library(gmp);#prob of a maximum of exactly k coincident birthdays is
BigQ<-function(nday,p,k) { #nday=days in a year; p=people
    if(p1,sum(sapply(2:k-1,function(j) BigQ(nday-i,p-k*i,j))),1)
    }
    tot
}
BDaySharedByAtLeast<-function(nday,people,k) {
    if(nday<1 | people

A225871 Number of people required for there to be a 50% probability that at least 4 share a birthday in a year with n days.

Original entry on oeis.org

4, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 28, 29, 30, 31, 32, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 51, 52, 53, 53, 54, 54, 55, 56

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 18 2013

nonn

a(365)=187.

For n<1000, the formula a(n) = 2.79 + 2.456*n^0.732 - 1.825/n provides an estimate of a(n) accurate to 0.82.

			For a year with 365 days, a(365), the probability that out of 186 people 4 of them share a birthday is 0.495825. The corresponding probability for 187 people is 0.502685, and therefore a(365)=187.

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000
Christian N. K. Anderson, Table of n, exact probabilities of a(n)-1 and a(n) for n = 1..1000.
Patrice Le Conte, Coincident Birthdays

Cf. A014088 (n people on 365 days), A033810 (2 people on n days), A225852 (3 on n days).

Cf. A088141, A182008, A182009, A182010.

library(gmp);#prob of a maximum of exactly k coincident birthdays is
BigQ<-function(nday,p,k) { #nday=days in a year; p=people
    if(p1,sum(sapply(2:k-1,function(j) BigQ(nday-i,p-k*i,j))),1)
    }
    tot
}
BDaySharedByAtLeast<-function(nday,people,k) {
    if(nday<1 | people

A109901 a(n) = binomial(n^2, n*(n+1)/2).

Original entry on oeis.org

1, 1, 4, 84, 8008, 3268760, 5567902560, 39049918716424, 1118770292985239888, 130276394656770614583240, 61448471214136179596720592960, 117118180539414377821494470432491764, 900390992257782351906806257139068209113040, 27883369051325994219981405855549095749234579210080

Offset: 0

Amarnath Murthy, Jul 14 2005

8*a(2*n+1)^4 = A182010(n) = number of potential group developed cocyclic Hadamard matrices over (the group) Z_{(2*n+1)^2} X Z^2_2 [Baliga, et al., p. 130]. - L. Edson Jeffery, Apr 10 2012

			a(6) = 36!/(21!*15!) = 5567902560.

Andrew Howroyd, Table of n, a(n) for n = 0..50
A. Baliga and K. J. Horadam, Cocyclic Hadamard matrices over Z_t X Z^2_2, Australas. J. Combin. 11(1995), 123-134.

Cf. variants: A014062 (C(n^2,n*(n-1))), A135757 (C(n*(n+1),n*(n+1)/2)).

Cf. A182010.

seq(binomial(n^2,n*(n+1)/2),n=0..12); # Emeric Deutsch, Jul 16 2005

Table[Binomial[n^2,(n(n+1))/2],{n,20}] (* Harvey P. Dale, Jun 04 2011 *)

PARI
```
a(n)=binomial(n^2,n*(n+1)/2)
```

a(n) = C(n^2, n*(n+1)/2) = (n^2!)/((n(n+1)/2)!*(n(n-1)/2)!).

a(n) = C(n^2, n*(n-1)/2).

More terms from Emeric Deutsch, Jul 16 2005
Offset changed to 0 (with a(0)=1), and name changed slightly by Paul D. Hanna, Jun 24 2011
Terms a(12) and beyond from Andrew Howroyd, Nov 09 2019

cp's OEIS Frontend

A014088 Minimal number of people to give a 50% probability of having at least n coincident birthdays in one year.

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A033810 Number of people needed so that probability of at least two sharing a birthday out of n possible days is at least 50%.

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A225852 The number of people required for there to be at least a 50% chance that at least 3 share a birthday in a year with n days.

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A225871 Number of people required for there to be a 50% probability that at least 4 share a birthday in a year with n days.

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A109901 a(n) = binomial(n^2, n*(n+1)/2).

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