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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
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