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Triangle T(n,m) (n >= 1, 0 <= m < n) giving coefficients of (n-1)! P_n, where P_n is the polynomial formula for row n of A213086.
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%I #5 May 24 2016 09:38:36

%S 1,0,1,0,1,1,0,1,3,2,0,2,9,10,3,0,2,25,50,35,8,0,-12,86,270,260,102,

%T 14,0,-120,140,1344,2030,1260,350,36,0,-1248,-1016,7336,15862,13048,

%U 5236,1024,78,0,-9216,-22464,28528,124488,139776,76104,22152,3312,200,0,-90720,-322344,1860,1036990,1514205,1018563,379890,80760,9165,431

%N Triangle T(n,m) (n >= 1, 0 <= m < n) giving coefficients of (n-1)! P_n, where P_n is the polynomial formula for row n of A213086.

%F The first formulas (stripped of factorials) :

%F 1,

%F k,

%F k + k^2,

%F k + 3 k^2 + 2 k^3,

%F 2 k + 9 k^2 + 10 k^3 + 3 k^4,

%F 2 k + 25 k^2 + 50 k^3 + 35 k^4 + 8 k^5,

%F -12 k + 86 k^2 + 270 k^3 + 260 k^4 + 102 k^5 + 14 k^6,

%F -120 k + 140 k^2 + 1344 k^3 + 2030 k^4 + 1260 k^5 + 350 k^6 + 36 k^7,

%F ...

%e Row T(5) = {0, 2, 9, 10, 3}, so P_5(k) = (1/4!)(2k + 9k^2 + 10k^3 + 3k^4), which gives 1, 7, 25, 65, 140, 266, ..., that is A001296 (row 5 of A213086), for k >=1.

%e Triangle begins:

%e {1},

%e {0, 1},

%e {0, 1, 1},

%e {0, 1, 3, 2},

%e {0, 2, 9, 10, 3},

%e {0, 2, 25, 50, 35, 8},

%e {0, -12, 86, 270, 260, 102, 14},

%e ...

%Y Cf. A213086, A213074.

%K sign,tabl

%O 1,9

%A _Jean-François Alcover_, May 24 2016