A036360 a(n) = Sum_{k=1..n} n! * n^(n-k+1) / (n-k)!

1, 12, 153, 2272, 39225, 776736, 17398969, 435538944, 12058401393, 366021568000, 12090393761721, 431832459644928, 16585599200808937, 681703972229640192, 29858718555221585625, 1388451967046195347456, 68316647610168842824161, 3546179063131198669848576, 193670918442059606406896473

Offset

1,2

Comments

This formula is given as a solution to Exercise 1.15a in the Harary and Palmer reference on page 30. However, the formula may not be correct and could be a misprint for Sum_{k=2..n} n! * n^(n-k-1) / (n-k)! which is a formula for A000435(n). - Andrew Howroyd, Feb 06 2024

It appears that a(n) * n^-(n+1) is the mean position of the first duplicate in sequences of n elements randomly drawn with replacement. - Brian P Hawkins, Jan 06 2024

References

F. Harary and E. Palmer, Graphical Enumeration, (1973), p. 30, Exercise 1.15a.

Links

N. J. A. Sloane, Table of n, a(n) for n = 1..100

Formula

a(n) = n^2 * A001865(n). - Gerald McGarvey, Apr 17 2008

a(n) = Sum_{k=1..n} n! * k^2 * n^(n-k) / (n-k)!. - Brian P Hawkins, Jan 06 2024

Maple

f := proc(n) local k; add(n!*n^(n-k+1)/(n-k)!, k=1..n); end;

Mathematica

Table[Sum[n!*n^(n-k+1)/(n-k)!, {k, 1, n}], {n, 19}] (* James C. McMahon, Feb 07 2024 *)

Prog

(Python)
def a(n):
    total_sum = 0
    for k in range(1, n + 1):
        term = (math.factorial(n) / math.factorial(n - k))*(k**2)*(n**(n - k))
        total_sum += term
    return total_sum
# Brian P Hawkins, Jan 06 2024
(PARI) a(n) = sum(k=1, n, n! * n^(n-k+1) / (n-k)!) \\ Andrew Howroyd, Jan 06 2024

Crossrefs

Cf. A000435, A001373.

Sequence in context: A189490 A180808 A004356 * A120657 A015612 A085260

Adjacent sequences: A036357 A036358 A036359 * A036361 A036362 A036363

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

Offset corrected by Brian P Hawkins, Jan 06 2024

Name edited by Andrew Howroyd, Feb 06 2024

Status

approved