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Triangle T, read by rows, such that the m-th matrix power satisfies T^m = I + m*(T - I) and consequently the matrix logarithm satisfies log(T) = T - I, where I is the identity matrix.
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%I #33 Jun 05 2021 06:09:34

%S 1,1,1,-1,0,1,1,1,1,1,-1,-2,-2,0,1,1,3,4,2,1,1,-1,-4,-7,-6,-3,0,1,1,5,

%T 11,13,9,3,1,1,-1,-6,-16,-24,-22,-12,-4,0,1,1,7,22,40,46,34,16,4,1,1,

%U -1,-8,-29,-62,-86,-80,-50,-20,-5,0,1,1,9,37,91,148,166,130,70,25,5,1,1,-1,-10,-46,-128,-239,-314,-296,-200,-95,-30,-6,0

%N Triangle T, read by rows, such that the m-th matrix power satisfies T^m = I + m*(T - I) and consequently the matrix logarithm satisfies log(T) = T - I, where I is the identity matrix.

%C Signed version of A108561. Row sums equal A084247. The n-th unsigned row sum = A001045(n) + 1 (Jacobsthal numbers). Central terms of even-indexed rows are a signed version of A072547. Sums of squared terms in rows yields A112556, which equals the first differences of the unsigned central terms.

%C Equals row reversal of triangle A112468 up to sign, where A112468 is the Riordan array (1/(1-x),x/(1+x)). - _Paul D. Hanna_, Jan 20 2006

%C The elements here match A108561 in absolute value, but the signs are crucial to the properties that the matrix A112555 exhibits; the main property being T^m = I + m*(T - I). This property is not satisfied by A108561. - _Paul D. Hanna_, Nov 10 2009

%C Eigensequence of the triangle = A140165. - _Gary W. Adamson_, Jan 30 2009

%C Triangle T(n,k), read by rows, given by [1,-2,0,0,0,0,0,0,0,...] DELTA [1,0,-1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Sep 17 2009

%H Paul D. Hanna, <a href="/A112555/b112555.txt">Table of n, a(n) for n = 0..1080</a>

%F G.f.: 1/(1-x*y) + x/((1-x*y)*(1+x+x*y)).

%F The m-th matrix power T^m has the g.f.: 1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)).

%F Recurrence: T(n, k) = [T^-1](n-1, k) + [T^-1](n-1, k-1), where T^-1 is the matrix inverse of T.

%F Sum_{k=0..n} T(n,k)*x^(n-k) = A165760(n), A165759(n), A165758(n), A165755(n), A165752(n), A165746(n), A165751(n), A165747(n), A000007(n), A000012(n), A084247(n), A165553(n), A165622(n), A165625(n), A165638(n), A165639(n), A165748(n), A165749(n), A165750(n) for x= -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 respectively. - _Philippe Deléham_, Oct 07 2009

%F Sum_{k=0..n} T(n,k)*x^k = A166157(n), A166153(n), A166152(n), A166149(n), A166036(n), A166035(n), A091004(n+1), A077925(n), A000007(n), A165326(n), A084247(n), A165405(n), A165458(n), A165470(n), A165491(n), A165505(n), A165506(n), A165510(n), A165511(n) for x = -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 respectively. - _Philippe Deléham_, Oct 08 2009

%e Triangle T begins:

%e 1;

%e 1, 1;

%e -1, 0, 1;

%e 1, 1, 1, 1;

%e -1, -2, -2, 0, 1;

%e 1, 3, 4, 2, 1, 1;

%e -1, -4, -7, -6, -3, 0, 1;

%e 1, 5, 11, 13, 9, 3, 1, 1;

%e -1, -6, -16, -24, -22, -12, -4, 0, 1;

%e 1, 7, 22, 40, 46, 34, 16, 4, 1, 1;

%e -1, -8, -29, -62, -86, -80, -50, -20, -5, 0, 1;

%e ...

%e Matrix log, log(T) = T - I, begins:

%e 0;

%e 1, 0;

%e -1, 0, 0;

%e 1, 1, 1, 0;

%e -1, -2, -2, 0, 0;

%e 1, 3, 4, 2, 1, 0;

%e -1, -4, -7, -6, -3, 0, 0;

%e ...

%e Matrix inverse, T^-1 = 2*I - T, begins:

%e 1;

%e -1, 1;

%e 1, 0, 1;

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

%e 1, 2, 2, 0, 1;

%e -1, -3, -4, -2, -1, 1;

%e ...

%e where adjacent sums in row n of T^-1 gives row n+1 of T.

%t Clear[t]; t[0, 0] = 1; t[n_, 0] = (-1)^(Mod[n, 2]+1); t[n_, n_] = 1; t[n_, k_] /; k == n-1 := t[n, k] = Mod[n, 2]; t[n_, k_] /; 0 < k < n-1 := t[n, k] = -t[n-1, k] - t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 06 2013 *)

%o (PARI) {T(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k)); polcoeff( polcoeff( (1+2*x+x*y)/((1-x*y)*(1+x+x*y)),n,X),k,Y)}

%o for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))

%o (PARI) {T(n,k)=local(m=1,x=X+X*O(X^n),y=Y+Y*O(Y^k)); polcoeff(polcoeff(1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)),n,X),k,Y)}

%o for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))

%o (Sage)

%o def A112555_row(n):

%o @cached_function

%o def prec(n, k):

%o if k==n: return 1

%o if k==0: return 0

%o return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))

%o return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n+1)]

%o for n in (0..12): print(A112555_row(n)) # _Peter Luschny_, Mar 16 2016

%Y Cf. A108561, A084247, A001045, A072547, A112556.

%Y Cf. A112468 (reversed rows).

%Y Cf. A140165. - _Gary W. Adamson_, Jan 30 2009

%K sign,tabl

%O 0,12

%A _Paul D. Hanna_, Sep 21 2005