A220074 Triangle read by rows giving coefficients T(n,k) of [x^(n-k)] in Sum_{i=0..n} (x-1)^i, 0 <= n <= k.

1, 1, 0, 1, -1, 1, 1, -2, 2, 0, 1, -3, 4, -2, 1, 1, -4, 7, -6, 3, 0, 1, -5, 11, -13, 9, -3, 1, 1, -6, 16, -24, 22, -12, 4, 0, 1, -7, 22, -40, 46, -34, 16, -4, 1, 1, -8, 29, -62, 86, -80, 50, -20, 5, 0, 1, -9, 37, -91, 148, -166, 130, -70, 25, -5, 1

Offset: 0

Author

Mokhtar Mohamed, Dec 03 2012

Keywords

• sign
• tabl
• easy

Comments

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)

Examples

			Triangle begins:
  1;
  1,   0;
  1,  -1,   1;
  1,  -2,   2,    0;
  1,  -3,   4,   -2,    1;
  1,  -4,   7,   -6,    3,    0;
  1,  -5,  11,  -13,    9,   -3,    1;
  1,  -6,  16,  -24,   22,  -12,    4,    0;
  1,  -7,  22,  -40,   46,  -34,   16,   -4,   1;
  1,  -8,  29,  -62,   86,  -80,   50,  -20,   5,   0;
  1,  -9,  37,  -91,  148, -166,  130,  -70,  25,  -5, 1;
  1, -10,  46, -128,  239, -314,  296, -200,  95, -30, 6, 0;
  ...
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened
• Dmitry Efimov, Hafnian of two-parameter matrices, arXiv:2101.09722 [math.CO], 2021.
• Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
• Ângela Mestre, José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
• OEIS Wiki, Autosequence

Crossrefs

Similar to the triangles A080242, A108561, A112555, A071920.

Cf. A000124 (column 2), A003600 (column 3), A223718 (column 4, conjectured), A257890 (column 5).

Cf. A026641, A054651, A059259.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Sum([0..k], j-> (-1)^j*Binomial(n-k+j, j))))); # G. C. Greubel, Feb 18 2019

Magma

[[(&+[(-1)^j*Binomial(n-k+j, j): j in [0..k]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 18 2019

Maple

A059259A := proc(n,k)
    1/(1+y)/(1-x-y) ;
    coeftayl(%,x=0,n) ;
    coeftayl(%,y=0,k) ;
end proc:
A059259 := proc(n,k)
    A059259A(n-k,k) ;
end proc:
A220074 := proc(i,j)
    (-1)^j*A059259(i,j) ;
end proc: # R. J. Mathar, May 14 2014

Mathematica

Table[Sum[(-1)^i*Binomial[n-k+i,i], {i, 0, k}], {n, 0, 12}, {k, 0, n} ]//Flatten (* Michael De Vlieger, Jan 27 2016 *)

PARI

{T(n,k) = sum(j=0,k, (-1)^j*binomial(n-k+j,j))};
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 18 2019

Sage

[[sum((-1)^j*binomial(n-k+j,j) for j in (0..k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 18 2019

Formula

Sum_{k=0..n} T(n,k) = 1.

T(n,k) = Sum_{i=0..k} (-1)^i*binomial(n-k+i, i).

T(2*n,n) = (-1)^n*A026641(n).

T(n,k) = (-1)^k*A059259(n,k).

T(n,0) = 1, T(n,n) = (1+(-1)^n)/2, and T(n,k) = T(n-1,k) - T(n-1,k-1) for 0 < k < n. - Mathew Englander, May 24 2014

Extensions

Definition and comments clarified by Li-yao Xia, May 15 2014