A123562 - OEIS

%I #25 Oct 16 2017 02:04:15

%S 1,1,1,1,-1,1,1,-3,-3,1,1,-5,-3,-5,1,1,-7,1,1,-7,1,1,-9,9,11,9,-9,1,1,

%T -11,21,17,17,21,-11,1,1,-13,37,11,1,11,37,-13,1,1,-15,57,-15,-39,-39,

%U -15,57,-15,1,1,-17,81,-69,-87,-81,-87,-69,81,-17,1

%N Pascal-(1,-3,1) array, read by antidiagonals.

%C Riordan array (1/(1-x), x*(1-3x)/(1-x)).

%H G. C. Greubel, <a href="/A123562/b123562.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%F Sum_{k=0..n} T(n,k) = A088137(n+1).

%F T(n,k) = T(n-1,k-1) + T(n-1,k) - 3*T(n-2,k-1), n>0.

%F From _Paul Barry_, Jan 24 2011: (Start)

%F T(n,k) = Sum_{j=0..n} binomial(n-j,k)*binomial(k,j)*(-3)^j.

%F T(n,k) = [k<=n]*Hypergeometric2F1(-k,k-n,1,-2). (End)

%F E.g.f. for the n-th subdiagonal: exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(-2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 - 4*x + 4*x^2/2) = 1 - 3*x - 3*x^2/2! + x^3/3! + 9*x^4/4! + 21*x^5/5! + .... - _Peter Bala_, Mar 05 2017

%e Triangle begins:

%e   1;

%e   1,   1;

%e   1,  -1,   1;

%e   1,  -3,  -3,   1;

%e   1,  -5,  -3,  -5,   1;

%e   1,  -7,   1,   1,  -7,   1;

%e   1,  -9,   9,  11,   9,  -9,  1;

%e   1, -11,  21,  17,  17,  21, -11,   1;

%e   1, -13,  37,  11,   1,  11,  37, -13,   1;

%t T[n_, k_] := Sum[Binomial[n - j, k]*Binomial[k, j]*(-3)^j, {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Oct 15 2017 *)

%o (PARI) for(n=0,10, for(k=0,n, print1(sum(j=0,n, binomial(n-j,k)*  binomial(k,j)*(-3)^j), ", "))) \\ _G. C. Greubel_, Oct 15 2017

%Y Cf. Pascal (1,m,1) array: A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

%K sign,tabl,easy

%O 0,8

%A _Philippe Deléham_, Nov 12 2006