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Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).
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%I #29 Mar 27 2021 23:52:00

%S 1,1,1,1,1,1,1,1,3,1,1,1,7,5,1,1,1,13,13,11,1,1,1,21,25,55,21,1,1,1,

%T 31,41,181,133,43,1,1,1,43,61,461,481,463,85,1,1,1,57,85,991,1281,

%U 2653,1261,171,1,1,1,73,113,1891,2821,10501,8425,4039,341,1,1,1,91,145,3305

%N Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).

%C Square array of solutions of a family of recurrences.

%C Rows of the array give solutions to the recurrences a(n)=a(n-1)+k(k-1)a(n-2), a(0)=a(1)=1.

%C Subarray of array in A072024. - _Philippe Deléham_, Nov 24 2013

%H Andrew Howroyd, <a href="/A081297/b081297.txt">Table of n, a(n) for n = 0..1274</a>

%F T(k, n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).

%F Rows of the array have g.f. 1/((1+kx)(1-(k+1)x)).

%e Rows begin

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 3, 5, 11, 21, ...

%e 1, 1, 7, 13, 55, 133, ...

%e 1, 1, 13, 25, 181, 481, ...

%e 1, 1, 21, 41, 461, 1281, ...

%t T[n_, k_]:=((n + 1)^(k + 1) - (-n)^(k + 1)) / (2n + 1); Flatten[Table[T[n - k, k], {n, 0, 10}, {k, 0, n}]] (* _Indranil Ghosh_, Mar 27 2017 *)

%o (PARI)

%o for(k=0, 10, for(n=0, 9, print1(((k+1)^(n+1)-(-k)^(n+1))/(2*k+1), ", "); ); print(); ) \\ _Andrew Howroyd_, Mar 26 2017

%o (Python)

%o def T(n, k): return ((n + 1)**(k + 1) - (-n)**(k + 1)) // (2*n + 1)

%o for n in range(11):

%o print([T(n - k, k) for k in range(n + 1)]) # _Indranil Ghosh_, Mar 27 2017

%Y Cf. A059259, A072024.

%Y Rows include A001045, A015441, A053404, A053428, A053430.

%Y Columns include A002061, A001844, A072025.

%Y Diagonals include A081298, A081299, A081300, A081301, A081302.

%K easy,nonn,tabl

%O 0,9

%A _Paul Barry_, Mar 17 2003

%E Name clarified by _Andrew Howroyd_, Mar 27 2017