A364857 a(n) = n^n/E, where E is the expected number of rolls of a fair n-sided die before obtaining 3 consecutive strictly increasing rolls.

1, 15, 225, 3781, 72078, 1550016, 37259191, 991980099, 29008029501, 924873082849, 31944725060988, 1188568865803032, 47403638535874501, 2017753008682107135, 91309129890388047873, 4377769140759352823773, 221687675024545322612226

Offset

3,2

Comments

a(n) is the determinant of (nI-M+M^2/n-OM/n), where O is the n X n matrix of ones, M is the lower triangular of O, and I is the n X n identity matrix.

a(n) is asymptotically equal to n^n / L, where L = sqrt(3e) / (cos(C)*sqrt(3) - sin(C)) and C = sqrt(3)/2. L=7.9243724345... is the expected number of uniform random samples of a real number between 0 and 1 until obtaining 3 increasing values.

gcd(n^n,a(n)) is n^2 if n == {5, 8, 11} (mod 12), 1 if n == {0, 1} (mod 3), and 2^(k+1)n^2 otherwise, where 2^k is the highest power of 2 that divides floor(n/12).

The graph of n^n / a(n) against n appears to follow a shifted reciprocal.

Links

Table of n, a(n) for n=3..19.

Formula

a(n) = (n-1+A)/2(n-1/2+B)^(n-1) + (n-1-A)/2(n-1/2-B)^(n-1), where A=i*(n+1)/sqrt(3) and B=i*(sqrt(3)/2).

a(n) = f(n), where f(2) = n-1, f(3) = n*(n-2) and f(m+1) = (2n-1)*f(m)-(n^2-n+1)*f(m-1).

Mathematica

a[n_]:=Module[{O, I, M}, O=ConstantArray[1, {n, n}]; I=IdentityMatrix[n]; M=LowerTriangularize[O]; Det[nI-M+MatrixPower[M, 2]/n-O.M/n]]; Table[a[i], {i, 3, 19}] (* Robert P. P. McKone, Aug 12 2023 *)
A[n_]:=I (n+1)/Sqrt[3]; B[n_]:=I Sqrt[3]/2; a[n_]:=(n-1+A[n])/2(n-1/2+B[n])^(n-1) + (n-1-A[n])/2(n-1/2-B[n])^(n-1); Simplify[Table[a[n], {n, 3, 19}]] (* Stefano Spezia, Aug 23 2023 *)

Prog

(Python)
def a(n):
    A, B = n-1, n*(n-2)
    for i in range(n-2):
        A, B = B, (2*n-1)*B-(n*n-n+1)*A
    return B

Crossrefs

Cf. A000312.

Sequence in context: A267731 A001024 A012643 * A067222 A154597 A041422

Adjacent sequences: A364854 A364855 A364856 * A364858 A364859 A364860

Keyword

nonn

Author

Daniel Chen, Aug 10 2023

Status

approved