OFFSET
1,2
COMMENTS
In a tournament where each entry fee is treated as an "across the board" bet equally divided among all paid-out placings, each placing's pool is shared among players who finish at or above that placing. The sequence forms a triangle read by rows, T(n,k), where n = number of paid-out players and k = placing. Dividing T(n,k) by n*n! gives the fraction of the total pool awarded to that placing.
Each row sums to n*n!.
T(n-1,k)/n! is also the probability that, when sampling elements of a permutation at random and observing their cycles, the second distinct cycle observed has length l_2 = k. (This sums to 1-1/n due to the chance 1/n that all n elements already inhabit the first cycle so the second's absence isn't recorded.) Given that the first cycle has length l_1, the second has chance 1/(n-l_1) of being length l_2 (uniformly for 1 <= l_2 <= n - l_1), so T(n-1,k)/n! = (1/n) * Sum_{l_1=1..n-k} 1/(n-l_1). In general, the chance of the c-th cycle (counting from c=0) having length l is Stirling1(l; n, c+l)*(l-1)!/n!, where Stirling1(r; n, k) is the unsigned r-Stirling number of the first kind. - Natalia L. Skirrow, May 21 2026
REFERENCES
J. H. Conway and R. K. Guy (1996). The Book of Numbers, New York: Springer-Verlag, p. 258.
LINKS
Andrei Z. Broder, The r-Stirling numbers, Discrete Mathematics vol. 49 iss. 3 pp. 241-259, 1984.
Niels Erik Nørlund, Vorlesungen über Differenzenrechnung, Springer, Berlin, 1924, p. 261.
Harold Ruben, A Note on the Trigamma Function. Am. Math. Monthly, Vol. 83, No. 8 (Oct 1976), pp. 622-623 (typesetting of a typewritten 1965 technical report).
Natalia L. Skirrow, more on Faulhabernoulli and Gregory.
FORMULA
T(n,k) = n! * Sum_{j=k..n} (1/j) = n! * (H_n - H_(k-1)), where H_n = A001008(n)/A002805(n) denotes the n-th harmonic number, for 1 <= k <= n.
From Natalia L. Skirrow, Apr 16 2026 (revised May 21 2026): (Start)
D-finite with (k-1)*T(n, k-1) + (n+2-k)*T(n, k) = T(n+1, k).
E.g.f.: y * log((1-x*y)/(1-x)) / ((1-x)*(1-y)).
D.e.g.f.: T(n+k,k) = n!*k!*[x^n * y^k] log(1/(1-x)) * log((1-x)/(1-x-y)). (Doubly-exponential generating function.)
D.e.g.f.: T(n+k+1,k+1) = n!*k!*[x^n * y^k] log(1/(1-x)) / (1-x-y).
T(n,k+1) = (n-k)!*Sum_{j=1..n-k} (n-j)!/((n-j-k)!*j) = (n-k)*(n-1)!*hypergeom([1,1,1+k-n], [2,1-n],1). (See Nørlund link, last equation in Ruben link.) This equivalence can be rewritten as Sum_{1 <= i_0 < i_1 < i_2 < ... < i_k <= n} 1/i_1 = C(n,k)*(H_n - H_k), and is explained elementarily in the Conway & Guy reference.
Sum_{k=1..n} T(n,k) = n!*n = A001563(n).
Sum_{k=1..n} T(n,k)*k = n!*n*(n+3)/4 = A138772(n+1).
Sum_{k=m..n} T(n,k)*C(k,m) = n!*(C(n+1,m+1) + C(n,m)/m)/(m+1) for m >= 1. (Binomial moments.)
The placing of player that a unit of money goes to has expectation (n+3)/4 and variance n^2/9 + 5*n/12 + 17/36.
(End)
EXAMPLE
In a tournament with 5 paid out winners, a participant places 3rd. They receive (1/3+1/4+1/5)*(1/5) = T(5,3)/(5*5!) = 94/600 = 15.67 percent of the pool.
Triangle begins:
1;
3, 1;
11, 5, 2;
50, 26, 14, 6;
274, 154, 94, 54, 24;
1764, 1044, 684, 444, 264, 120;
...
MAPLE
T := (n, k) -> n! * (harmonic(n) - harmonic(k-1)):
seq(seq(T(n, k), k = 1..n), n = 1..9); # Peter Luschny, Apr 17 2026
MATHEMATICA
T[n_, k_]:=n!*Sum[1/j, {j, k, n}]; Table[T[n, k], {n, 1, 9}, {k, 1, n}]//Flatten (* James C. McMahon, Sep 24 2025 *)
PROG
(PARI) row(n) = vector(n, k, n!*sum(j=k, n, 1/j)); \\ Michel Marcus, Sep 24 2025
CROSSREFS
KEYWORD
AUTHOR
Daniel Vatcher, Sep 20 2025
STATUS
approved