A060644 -id:A060644 - cp's OEIS frontend

A366451 The sum s > n which has the greatest probability of occurring when summing rolls of an n-sided die.

4, 6, 8, 9, 11, 13, 15, 16, 18, 20, 21, 23, 25, 27, 28, 30, 32, 34, 35, 37, 39, 40, 42, 44, 46, 47, 49, 51, 52, 54, 56, 58, 59, 61, 63, 64, 66, 68, 70, 71, 73, 75, 76, 78, 80, 82, 83, 85, 87, 88, 90, 92, 94, 95, 97, 99, 101, 102, 104, 106, 107, 109, 111, 113

Offset: 2

Nicholas Stefan Georgescu, Oct 10 2023

nonn

A player chooses a target t > n and rolls a fair n-sided die repeatedly, keeping a running total of the numbers rolled. The player wins if t occurs in the sequence of running totals. a(n) is the choice of t that maximizes the probability of winning
A fair die with sides numbered 1 through n is assumed.
For given n, sum s occurs by adding one die role to a sum s-n .. s-1; so the probability p(s) of s occurring is p(s) = Sum_{i=1..n} p(s-i)*(1/n), i.e., the mean of the preceding n probabilities, and starting from p(0) = 1 for s=0 always occurring and p(s) = 0 for s < 0 never occurring.
It is immediately apparent upon plotting the probabilities from sum s = n+1 onwards that the highest probability is achieved at s = a(n).
It can be shown that, given the equation of n < s <= 2n, the maximum near a(n) is the only zero of the first derivative, and the second derivative is always negative, therefore any following probability must necessarily be less the maximum of the previous n probabilities, and therefore less than p(a(n)). Therefore any subsequent terms must then be less than p(a(n)). Because this is appending a rolling average of the previous n terms the probabilities dampen in their oscillations about a value of 2/(n+1).

			For n=20, the probability of occurrence p(s) of sum s > n begins at p(21) = 0.08266... and rises to p(35) = 0.09767... before decreasing to p(36) = 0.09760... and never reaches p(35) again, so a(20) = 35 is the sum s with greatest probability of occurrence.
           s     p(s)
          --  ----------
          21  0.08266...
          ...
          34  0.09751...
  a(20) = 35  0.09767... <--- maximum probability
          36  0.09760...
          ...
          42  0.09350...

Paolo Xausa, Table of n, a(n) for n = 2..10000
James Monroe, Pi is Evil, Numberphile youtube video. Note the errata by the authors in the comments: a(20) is actually 35. In addition, the more accurate approximation of (e-1)(n+1/2)

Cf. A060644, A365443.

A366451[n_] := Floor[(n+1)^(n+1)/n^n] - n; Array[A366451, 100, 2] (* Paolo Xausa, Oct 12 2024 *)

from fractions import Fraction
a = lambda n: int(Fraction((n+1)*(n+1)**n-n**(n+1),n**n))

For n < s < 2n, the probability of sum s occurring is p(s) = ((n+1)^(s-1) - s*n^n*(n+1)^(s-n-2))/n^s, which means that, from the difference between s and s-1, the sequence value is given by:
a(n) = floor(((n+1)*(n+1)^n - n^(n+1))/n^n).
a(n) = A060644(n) - n.
An empirical observation is that a(n) is well approximated by (e-1)*(n+1/2).
From Javier Múgica, May 01 2025: (Start)
Taylor expansion of the maximum of p(s) gives a(n) ~ (e-1)*(n+1/2) - (e-2)/(24n).
The maximum for s > 2n is attained at s ~ (e-sqrt(-e^2+2e+2))*n, approx. 2.5*n. In general, the position of the maximum for s > kn is given by an algebraic equation in s and e of degree k in each of them. (End)

A069322 Square array read by antidiagonals of floor[(n+k)^(n+k)/(n^n*k^k)].

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 6, 6, 1, 1, 9, 16, 9, 1, 1, 12, 28, 28, 12, 1, 1, 14, 45, 64, 45, 14, 1, 1, 17, 65, 119, 119, 65, 17, 1, 1, 20, 89, 198, 256, 198, 89, 20, 1, 1, 23, 117, 307, 484, 484, 307, 117, 23, 1, 1, 25, 149, 449, 837, 1024, 837, 449, 149, 25, 1, 1, 28, 184, 629

Offset: 0

Henry Bottomley, Mar 14 2002

T(n,k)*sqrt(3)/(n*k*Pi) provides a rough approximation for A067059.

a(n,k) is an analog of the binomial coefficients over transformations instead of permutations. - Chad Brewbaker, Nov 25 2013

			Rows start: 1,1,1,1,1,1,...; 1,4,6,9,12,14,...; 1,6,16,28,45,65,...; 1,9,28,64,119,198,...; etc. T(3,5)=floor[8^8/(3^3*5^5)]=floor[16777216 /84375]=floor[198.84...]=198.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Initial columns and rows are A000012 and A060644, main diagonal is A000302.

t[n_, 0] := 1; t[n_, n_] := 1; t[n_, k_] := Floor[(n^n)/((k^k)*((n - k)^(n - k)))];  Table[t[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 22 2018 *)

for(n=0,15, for(k=0,n, print1(if(k==0, 1, if(k==n,1,floor((n^n)/(( k^k)*((n - k)^(n - k)))))), ", "))) \\ G. C. Greubel, Apr 22 2018

def transitorial(n)
    return n**n
end
def transnomial(n,k)
       return transitorial(n)/(transitorial(k) *transitorial(n-k))
end
0.upto(15) do |i|
   0.upto(i) do |j|
       print transnomial(i,j).to_s + " "
   end
   puts ""
end # Chad Brewbaker, Nov 25 2013

a(n,k) = (n^n) /((k^k)*((n-k)^(n-k))). - Chad Brewbaker, Nov 25 2013

A184588 floor[(n+1/2)*e/(e-1)].

Original entry on oeis.org

2, 3, 5, 7, 8, 10, 11, 13, 15, 16, 18, 19, 21, 22, 24, 26, 27, 29, 30, 32, 34, 35, 37, 38, 40, 41, 43, 45, 46, 48, 49, 51, 52, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 71, 73, 75, 76, 78, 79, 81, 83, 84, 86, 87, 89, 90, 92, 94, 95, 97, 98, 100, 102, 103, 105, 106, 108, 109, 111, 113, 114, 116, 117, 119, 121, 122, 124, 125, 127, 128, 130, 132, 133, 135, 136, 138, 140, 141, 143, 144, 146, 147, 149, 151, 152, 154, 155, 157, 158, 160, 162, 163, 165, 166, 168, 170, 171, 173, 174, 176, 177, 179, 181, 182, 184, 185, 187, 189, 190

Offset: 1

Clark Kimberling, Jan 17 2011

nonn

The complement of A184588 appears to be A060644.

Cf. A060644.

r=E; c=1/2; s=r/(r-1);
Table[Floor[n*r-c*r],{n,1,120}] (* A060644? *)
Table[Floor[n*s+c*s],{n,1,120}]  (* A184588 *)

a(n)=floor[(n+1/2)*e/(e-1)].

A062458 Nearest integer to (n+1)^(n+1)/n^n.

Original entry on oeis.org

1, 4, 7, 9, 12, 15, 18, 20, 23, 26, 29, 31, 34, 37, 39, 42, 45, 48, 50, 53, 56, 58, 61, 64, 67, 69, 72, 75, 77, 80, 83, 86, 88, 91, 94, 96, 99, 102, 105, 107, 110, 113, 116, 118, 121, 124, 126, 129, 132, 135, 137, 140, 143, 145, 148, 151, 154, 156, 159, 162, 164

Offset: 0

Olivier Gérard, Jun 23 2001

Cf. A060644.

Join[{1},Table[Round[(1+n)^(1+n)/n^n],{n,60}]] (* Harvey P. Dale, Aug 24 2025 *)

a(n)=if(n<1,1,round(exp(1)*(n+1/2-1/24/n))) \\ Benoit Cloitre, Apr 24 2008

For n>0: a(n)=round(exp(1)*(n+1/2-1/24/n)). Coming from: (n+1)^(n+1)/n^n=exp(1)*(n+1/2-1/24/n)+O(1/n^2) - Benoit Cloitre, Apr 24 2008

Previous Mathematica program replaced by Harvey P. Dale, Aug 24 2025

A078111 a(n) = floor((n+2)^(n+2)/n^n).

Original entry on oeis.org

4, 27, 64, 115, 182, 263, 359, 470, 596, 736, 891, 1061, 1246, 1445, 1660, 1889, 2132, 2391, 2664, 2953, 3256, 3573, 3906, 4253, 4615, 4992, 5384, 5790, 6211, 6647, 7098, 7563, 8044, 8539, 9049, 9573, 10113, 10667, 11236, 11820, 12418, 13031, 13659

Offset: 0

Benoit Cloitre, Dec 03 2002

nonn

Cf. A060644.

a(n)=(1+2./n)^n*(n+2)^2\1 \\ Charles R Greathouse IV, May 07 2013

a(n) - e^2(n+1)^2 is bounded. Conjecture: there's a constant c = 2.462... such that a(n) = floor(e^2*(n+1)^2-c).

I believe (n+2)^(n+2)/n^n = e^2(n^2 + 2n + 2/3 + o(1)), so the conjecture should be correct (with perhaps finitely many exceptions) with c = e^2/3 = 2.463.... - Charles R Greathouse IV, May 07 2013

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A366451 The sum s > n which has the greatest probability of occurring when summing rolls of an n-sided die.

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A069322 Square array read by antidiagonals of floor[(n+k)^(n+k)/(n^n*k^k)].

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A184588 floor[(n+1/2)*e/(e-1)].

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A062458 Nearest integer to (n+1)^(n+1)/n^n.

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A078111 a(n) = floor((n+2)^(n+2)/n^n).

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