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%I #83 Sep 08 2022 08:46:04
%S 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,
%T -6,16,-24,22,-12,4,0,1,-7,22,-40,46,-34,16,-4,1,1,-8,29,-62,86,-80,
%U 50,-20,5,0,1,-9,37,-91,148,-166,130,-70,25,-5,1
%N Triangle read by rows giving coefficients T(n,k) of [x^(n-k)] in Sum_{i=0..n} (x-1)^i, 0 <= n <= k.
%C 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).
%C 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).
%C 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
%C 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
%C 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
%C From _Paul Curtz_, Mar 17 2017: (Start)
%C 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.,
%C 0;
%C 0, 1;
%C 0, 1, 0;
%C 0, 1, -1, 1;
%C 0, 1, -2, 2, 0;
%C 0, 1, -3, 4, 2, 1;
%C 0, 1, -4, 7, -6, 3, 0;
%C 0, 1, -5, 11, -13, 9, -3, 1;
%C ... .
%C 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).
%C The row sums of the absolute values are A001045(n).
%C 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.
%C Jl multiplied by Ja gives the triangle Jal:
%C 0;
%C 0, 1;
%C 0, 1, 0;
%C 0, 1, -1, 3;
%C 0, 1, -2, 6, 0;
%C 0, 1, -3, 12, -10, 11;
%C 0, 1, -4, 21, -30, 33, 0;
%C 0, 1, -5, 33, -65, 99, -63, 43;
%C ... .
%C The row sums are A001045(n). (End)
%H G. C. Greubel, <a href="/A220074/b220074.txt">Rows n = 0..100 of triangle, flattened</a>
%H Dmitry Efimov, <a href="https://arxiv.org/abs/2101.09722">Hafnian of two-parameter matrices</a>, arXiv:2101.09722 [math.CO], 2021.
%H Kyu-Hwan Lee, Se-jin Oh, <a href="http://arxiv.org/abs/1601.06685">Catalan triangle numbers and binomial coefficients</a>, arXiv:1601.06685 [math.CO], 2016.
%H Ângela Mestre, José Agapito, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Mestre/mestre2.html">Square Matrices Generated by Sequences of Riordan Arrays</a>, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
%H OEIS Wiki, <a href="https://oeis.org/wiki/Autosequence">Autosequence</a>
%F Sum_{k=0..n} T(n,k) = 1.
%F T(n,k) = Sum_{i=0..k} (-1)^i*binomial(n-k+i, i).
%F T(2*n,n) = (-1)^n*A026641(n).
%F T(n,k) = (-1)^k*A059259(n,k).
%F 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
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, -1, 1;
%e 1, -2, 2, 0;
%e 1, -3, 4, -2, 1;
%e 1, -4, 7, -6, 3, 0;
%e 1, -5, 11, -13, 9, -3, 1;
%e 1, -6, 16, -24, 22, -12, 4, 0;
%e 1, -7, 22, -40, 46, -34, 16, -4, 1;
%e 1, -8, 29, -62, 86, -80, 50, -20, 5, 0;
%e 1, -9, 37, -91, 148, -166, 130, -70, 25, -5, 1;
%e 1, -10, 46, -128, 239, -314, 296, -200, 95, -30, 6, 0;
%e ...
%p A059259A := proc(n,k)
%p 1/(1+y)/(1-x-y) ;
%p coeftayl(%,x=0,n) ;
%p coeftayl(%,y=0,k) ;
%p end proc:
%p A059259 := proc(n,k)
%p A059259A(n-k,k) ;
%p end proc:
%p A220074 := proc(i,j)
%p (-1)^j*A059259(i,j) ;
%p end proc: # _R. J. Mathar_, May 14 2014
%t 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 *)
%o (PARI) {T(n,k) = sum(j=0,k, (-1)^j*binomial(n-k+j,j))};
%o for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Feb 18 2019
%o (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
%o (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
%o (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
%Y Similar to the triangles A080242, A108561, A112555, A071920.
%Y Cf. A000124 (column 2), A003600 (column 3), A223718 (column 4, conjectured), A257890 (column 5).
%Y Cf. A026641, A054651, A059259.
%K sign,tabl,easy
%O 0,8
%A _Mokhtar Mohamed_, Dec 03 2012
%E Definition and comments clarified by _Li-yao Xia_, May 15 2014
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