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Column 1 of A223725.

Column 1 of A223663.

Apparently also column 4 of A071920. - R. J. Mathar, May 17 2014

Table starts

..2...4...7...11....16.....22.....29......37......46.......56.......67

..3...9..22...46....86....148....239.....367.....541......771.....1068

..4..16..50..130...296....610...1163....2083....3544.....5776.....9076

..5..25..95..295...791...1897...4166....8518...16414....30086....52834

..6..36.161..581..1792...4900..12174...27966...60172...122464...237590

..7..49.252.1036..3612..11088..30738...78354..186142...416394...884236

..8..64.372.1716..6672..22716..69498..194634..505912..1233584..2845492

..9..81.525.2685.11517..43065.144111..439791.1241383..3276559..8157227

.10.100.715.4015.18832..76714.278707..920491.2803658..7963384.21280337

.11.121.946.5786.29458.129844.508937.1808521.5911763.17978389.51325352

From Charles A. Lane, Aug 22 2013: (Start)

The first column is also the coefficients of a in y''[x] - a*x^n*y[x] + b*en*y[x] = 0 where n = 0. The recursion yields coefficients of a, a*b*en, a*b^2*en^2 etc.

The second column is obtained when n=1, the third column when n=2. The final column is for n=10.

Example: Write a normal recursion for n=4. For convenience set x to 1. Running the recursion yields

1-(b en)/2+(b^2 en^2)/24+1/30 (a-(b^3 en^3)/24)+(-384 a b en+b^4 en^4)/40320+(2064 a b^2 en^2-b^5 en^5)/3628800+(120960 a^2-7104 a b^3 en^3+b^6 en^6)/479001600+(-4682880 a^2 b en+18984 a b^4 en^4-b^7 en^7)/87178291200+(54268416 a^2 b^2 en^2-43008 a b^5 en^5+b^8 en^8)/20922789888000.

The coefficient of a is 24, the coefficient of a b en is 384 and the coefficient of a b^2 en^2 is 2064. Dividing by 4! yields a sequence of 1,16,86... , the same as column 5 without the leading 1. There is a hint of unity among the oscillators. (End)

If one uses a definition of unimodality that involves universal quantifiers on the domain of a function then a(0,m)=1 a priori.

Table starts

...3......9........22..........46............86............148..............239

...9.....81.......484........2116..........7396..........21904............57121

..22....484......5883.......46613........273562........1285547..........5087912

..46...2116.....46613......608855.......5537147.......38566854........218619076

..86...7396....273562.....5537147......74140718......733129227.......5733691150

.148..21904...1285547....38566854.....733129227.....9991366555.....105179942478

.239..57121...5087912...218619076....5733691150...105179942478....1461411174370

.367.134689..17557701..1051051942...37151436091...899048082946...16206659678557

.541.292681..54161878..4413826871..206162721823..6467662339564..148957213473086

.771.594441.152154419.16548850432.1004293423046.40225236084575.1167498177877341

First two terms in row n: n,(n-1)^2; last term: 1.

Period of alternating row sums: (1,1,0).

For a discussion and guide to related arrays, see A208510.

Apparently this is A071920 without the marginal zeros, read by downwards antidiagonals, or T(n,k) = A071922(n,k). - R. J. Mathar, 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)