A055325 - OEIS

%I #13 May 17 2021 16:03:25

%S 1,-1,1,3,-4,1,-23,33,-11,1,425,-620,220,-26,1,-18129,26525,-9520,

%T 1180,-57,1,1721419,-2519664,905765,-113050,5649,-120,1,-353654167,

%U 517670461,-186123259,23248085,-1166221,25347,-247,1,153923102577

%N Matrix inverse of Euler's triangle A008292.

%H Robert Israel, <a href="/A055325/b055325.txt">Table of n, a(n) for n = 1..3321</a> (rows 1 to 81, flattened)

%e Triangle starts:

%e [1]         1;

%e [2]        -1,         1;

%e [3]         3,        -4,          1;

%e [4]       -23,        33,        -11,        1;

%e [5]       425,      -620,        220,      -26,        1;

%e [6]    -18129,     26525,      -9520,     1180,      -57,     1;

%e [7]   1721419,  -2519664,     905765,  -113050,     5649,  -120,    1;

%e [8]-353654167, 517670461, -186123259, 23248085, -1166221, 25347, -247, 1;

%p A008292:= proc(n, k) option remember;

%p   if k < 1 or k > n then 0

%p   elif k = 1 or k = n then 1

%p   else (k*procname(n-1, k)+(n-k+1)*procname(n-1, k-1))

%p   fi

%p end proc:

%p T:= Matrix(10,10,(i,j) -> A008292(i,j)):

%p R:= T^(-1):

%p seq(seq(R[i,j],j=1..i),i=1..10); # _Robert Israel_, May 25 2018

%t m = 10 (*rows*);

%t t[n_, k_] := Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];

%t M = Array[t, {m, m}] // Inverse;

%t Table[M[[i, j]], {i, 1, m}, {j, 1, i}] // Flatten (* _Jean-François Alcover_, Mar 05 2019 *)

%Y Cf. A008292, A162498.

%K sign,tabl

%O 1,4

%A _Christian G. Bower_, May 12 2000