A130595 Triangle read by rows: lower triangular matrix which is inverse to Pascal's triangle (A007318) regarded as a lower triangular matrix.

1, -1, 1, 1, -2, 1, -1, 3, -3, 1, 1, -4, 6, -4, 1, -1, 5, -10, 10, -5, 1, 1, -6, 15, -20, 15, -6, 1, -1, 7, -21, 35, -35, 21, -7, 1, 1, -8, 28, -56, 70, -56, 28, -8, 1, -1, 9, -36, 84, -126, 126, -84, 36, -9, 1, 1, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1, -1, 11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1

Offset: 0

Author

Philippe Deléham, Jun 17 2007

Keywords

• sign
• nice
• tabl

Comments

Triangle T(n,k), read by rows, given by [-1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Coefficients of the polynomials generated by the e.g.f. exp(x*t)*exp(-t). - Peter Luschny, Jul 13 2009

Riordan array (1/(1+x), x/(1+x)). - Philippe Deléham, Nov 29 2009

Multiplication of a sequence (written as column vector) by this matrix (to the left) yields the inverse Binomial transform of the sequence. - M. F. Hasler, Nov 01 2014

From Tom Copeland, Nov 16 2016: (Start)

This signed Pascal matrix IP and the Pascal matrix P contain the coefficients of a prototypical pair of Appell polynomial sequences that are inverse under umbral composition with e.g.f.s e^((x-1)*t) = e^(-t) e^(xt) = f(t) e^(xt) and e^((x+1)t) = e^t e^(xt) = g(t) e^(xt) and row polynomials q_n(x) = (x-1)^n and p_n(x) = (x+1)^n, respectively. The inverse property for an Appell pair is reflected in IP*P = identity matrix, f(t) = 1/g(t), the umbral relation p_n(q.(x)) = x^n = q_n(p.(x)), and their respective raising operators R_(Ip) = x - h(D) and R_P = x + h(D) differing only in the sign of the differential term (h(D) = 1, in this case). The lowering operator for an Appell sequence is L = D = d/dx with L p_n(x) = n*p_(n-1)(x), and the raising operator is defined by R p_n(x) = p_(n+1)(x).

The related signed Pascal matrix M with M(n,k) = (-1)^n IP(n,k) = (-1)^k P(n,k) has the e.g.f. e^((1-x)t) = e^t e^(-xt), and w_n(x) = (1-x)^n is not an Appell sequence, but it is a Sheffer sequence with lowering and raising operators L = -D and R = 1 - x, and M = M^(-1) since w_n(w.(x)) = (1-w.(x))^n = sum_{k = 0,..,n} binomial(n,k) (-1)^k w_k(x) = (1-(1-x))^n = x^n.

Umbral composition of a pair of Sheffer polynomial sequences, of which Appell sequences are a special class, is equivalent to the multiplication of their respective coefficient matrices.

(End)

Examples

			Triangle begins with T(0,0):
   1;
  -1,    1;
   1,   -2,    1;
  -1,    3,   -3,    1;
   1,   -4,    6,   -4,    1;
  -1,    5,  -10,   10,   -5,    1;
   1,   -6,   15,  -20,   15,   -6,    1;
  -1,    7,  -21,   35,  -35,   21,   -7,    1;
   1,   -8,   28,  -56,   70,  -56,   28,   -8,    1;
  -1,    9,  -36,   84, -126,  126,  -84,   36,   -9,    1;
  ...
As polynomials:
  + 1;
  - 1 + 1 x;
  + 1 - 2 x + 1 x^2;
  - 1 + 3 x - 3 x^2 + 1 x^3;
  + 1 - 4 x + 6 x^2 - 4 x^3 + 1 x^4;
		

Links

• Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
• Shishuo Fu, Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
• Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974. [_N. J. A. Sloane_, Nov 26 2011]
• Wikipedia, Relation between binomial coefficients and harmonic numbers.
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Cf. A007318, A084938.

Sums include: A000007 (row sums), A019590, A039834 (diagonal sums), A049347 (alternating sign diagonal sums), A063524, A085750, A122803 (alternating sign sums).

Programs

Haskell

a130595 n = a130595_list !! n
a130595_list = concat $ iterate ([-1,1] *) [1]
instance Num a => Num [a] where
   fromInteger k = [fromInteger k]
   (p:ps) + (q:qs) = p + q : ps + qs
   ps + qs         = ps ++ qs
   (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
    *                = []
-- Reinhard Zumkeller, Apr 02 2011

Haskell

a130595 n k = a130595_tabl !! n !! k
a130595_row n = a130595_tabl !! n
a130595_tabl = iterate (\row -> zipWith (-) ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Apr 13 2013

Magma

[(-1)^(n+k)*Binomial(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jun 22 2024

Maple

A130595 := proc(n,k)
        (-1)^(n+k)*binomial(n,k) ;
end proc: # R. J. Mathar, Feb 13 2013

Mathematica

nmax = 11; t[n_, k_] := (-1)^(n-k)*Binomial[n, k]; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}] ] (* Jean-François Alcover, Dec 01 2011 *)
Table[Binomial[-1-k, n-k],{n,0,11},{k,0,n}]//Flatten (* Robert A. Russell, Jan 16 2020 *)

PARI

A130595(n,k)=(-1)^(n+k)*binomial(n,k) \\ M. F. Hasler, Nov 01 2014

SageMath

flatten([[(-1)^(n+k)*binomial(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jun 22 2024

Formula

T(n,k) = (-1)^(n-k)*binomial(n,k) = (-1)^(n-k)*A007318(n,k).

T(n,k) = T(n-1,k-1) - T(n-1,k). - Philippe Deléham, Oct 10 2011

G.f.: 1/(1+x-x*y). - R. J. Mathar, Aug 11 2015 [corrected by Anders Claesson, Nov 28 2015]

Conjecture from Dale Gerdemann, Nov 28 2015:

T(n,k) = (n-k+1)*T(n-1,k-1) + (k-1)*T(n-1,k).

Proof from Anders Claesson, Nov 29 2015:

It follows from T(n,k) = T(n-1,k-1) - T(n-1,k) and n*T(n-1,k-1) = k*T(n,k) that: (n-k+1)*T(n-1,k-1) + (k-1)*T(n-1,k) = n*T(n-1,k-1) - (k-1)*T(n-1,k-1) + (k-1)*T(n-1,k) = n*T(n-1,k-1) - (k-1)*(T(n-1,k-1) - T(n-1,k)) = n*T(n-1,k-1) - (k-1)*T(n,k) = n*T(n-1,k-1) - k*T(n,k) + T(n,k) = T(n,k). QED

(-1)^(n+1) Sum_{k=1..n} T(n,k)/k = Sum_{k=1..n} 1/k = H(n) where H(n) is the n-th harmonic number. For a proof see link "Relation between binomial coefficients and harmonic numbers". - Wolfgang Hintze, Oct 22 2016

T(n,k) = binomial(-1-k,n-k). - Robert A. Russell, Jan 16 2020

From G. C. Greubel, Jun 22 2024: (Start)

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

Sum_{k=0..n} T(n, k) = A000007(n).

Sum_{k=0..n} (-1)^k*T(n, k) = A122803(n).

Sum_{k=0..floor(n/2)} T(n-k, k) = A039834(n+1).

Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A049347(n).

Sum_{k=0..n} k*T(n, k) = A063524(n).

Sum_{k=0..n} (-1)^k*k*T(n, k) = A085750(n+1).

Sum_{k=0..n} (k+1)*T(n, k) = A019590(n). (End)

Extensions

Edited by N. J. A. Sloane, Nov 27 2011