cp's OEIS Frontend

Stirling transform of (-1)^n*a(n-1) = [0, 1, -1, 5, -14, 94, ...] is A000629(n-2) = [0, 1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004

Stirling transform of a(n) = [1, 1, 5, 14, 94, ...] is A052882(n) = [1, 2, 9, 52, 375, ...]. - Michael Somos, Mar 04 2004

a(n) is the number of n-permutations that have a cycle with length greater than n/2. - Geoffrey Critzer, May 28 2009

From Jens Voß, May 07 2010: (Start)

a(4n) is divisible by 6*n + 1 for all n >= 1; the quotient of a(4*n) and 6*n+1 is A177188(n).

a(4*n+3) is divisible by 6*n + 5 for all n >= 0; the quotient of a(4*n+3) and 6*n + 5 is A177174(n). (End)

There is a strong relation between this triangle and triangle A048994 which deals with the binomial (x,n), this triangle being dealing with the summation of this binomial.

Apparently A054651 with reversed rows. - Mathew Englander, May 17 2014

If the triangle is viewed as a square array S(m, k) = T(m+k, k), 0 <= m, 0 <= k, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows have g.f. 1/((1+x)^n*(1-x^2)) (substitute x for -x in g.f. for A059259).

By column, S(m, k) is the coefficient of [x^m] in the generating function Sum_{i=0..k} (-1)^i/(1-x)^(i+1).

This is a rational generating function down column k with a power of (1-x) in the denominator; therefore column k is a polynomial in m respectively n. - Mathew Englander, May 14 2014

Column k multiplied by k! seems to correspond to row k of A054651, considered as a polynomial and then evaluated on the negative integers. For example, row 5 of A054651 represents the polynomial x^5 - 5*x^4 + 25*x^3 + 5*x^2 + 94*x + 120. Evaluating that for x = -1, x = -2, x = -3, ... gives (0, -360, -1440, -4080, -9600, -19920, -37680, ...) which is 5! times column 5 of this triangle. - Mathew Englander, May 23 2014

This triangle provides a solution to a question in the mathematics of gambling. For 0 < p < 1 and positive integers N and G with N < G, suppose you begin with N dollars and make repeated wagers, each time winning 1 dollar with probability p and losing 1 dollar with probability 1-p. You continue betting 1 dollar at a time until you have either G dollars (your Goal) or 0 (bankrupt). What is the probability of reaching your Goal before going bankrupt, as a function of p, N, and G? (This is a type of one-dimensional random walk.) Answer: Let Q_m_(x) be the polynomial whose coefficients are given by row m-1 of the triangle (e.g., Q_6_(x) = 1 - 4x + 7x^2 - 6x^3 + 3x^4). Then, the probability of reaching G dollars before going bankrupt is p^(G-N)*Q_N_(p)/Q_G_(p). - Mathew Englander, May 23 2014

From Paul Curtz, Mar 17 2017: (Start)

Consider the triangle Ja(n+1,k) (here, but generally Ja(n,k)) composed of the triangle a(n) prepended with a column of 0's, i.e.,

0;

0, 1;

0, 1, 0;

0, 1, -1, 1;

0, 1, -2, 2, 0;

0, 1, -3, 4, -2, 1;

0, 1, -4, 7, -6, 3, 0;

0, 1, -5, 11, -13, 9, -3, 1;

... .

The row sums are 0, 1, 1, ... = A057427(n), the most elementary autosequence of the first kind (a sequence of the first kind has 0's as main diagonal of its array of successive differences).

The row sums of the absolute values are A001045(n).

Ja applied to a sequence written in its reluctant form yields an autosequence of the first kind. Example: the reluctant form of A001045(n) is 0, 0, 1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 1, 3, 5, ... = Jl.

Jl multiplied by Ja gives the triangle Jal:

0;

0, 1;

0, 1, 0;

0, 1, -1, 3;

0, 1, -2, 6, 0;

0, 1, -3, 12, -10, 11;

0, 1, -4, 21, -30, 33, 0;

0, 1, -5, 33, -65, 99, -63, 43;

... .

The row sums are A001045(n). (End)