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%I #22 Oct 18 2022 19:14:02
%S 1,0,3,1,5,10,0,10,35,35,1,14,91,189,126,0,21,189,651,924,462,1,27,
%T 351,1749,4026,4290,1716,0,36,594,4026,13299,22737,19305,6435,1,44,
%U 946,8294,36751,89375,120835,85085,24310,0,55,1430,15730,89375,289003,551837,615043,369512,92378,1,65,2080,27950,197275,811733,2047123,3203837,3031678,1587222,352716
%N Triangle T(n,k) (n >= k >= 0) read by rows: T(n,0) = (1+(-1)^n)/2; for k>=1, set T(0,k) = 0, S(n,k) = binomial(n,k)*binomial(n+k+1,k), and for n>=1, T(n,k) = S(n,k)-T(n-1,k).
%C The scanned pages of Ser are essentially illegible, and the book is out of print and hard to locate.
%C For Table IV on page 93, it is simplest to ignore the minus signs. The present triangle then matches all the given terms in that triangle, so it seems best to define the triangle by the recurrences given here, and to conjecture (strongly) that this is the same as Ser's triangle.
%D J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93.
%H G. C. Greubel, <a href="/A331432/b331432.txt">Rows n = 0..50 of the triangle, flattened</a>
%H J. Ser, <a href="/A002720/a002720_4.pdf">Les Calculs Formels des Séries de Factorielles</a>, Gauthier-Villars, Paris, 1933 [Local copy].
%H J. Ser, <a href="/A002720/a002720.pdf">Les Calculs Formels des Séries de Factorielles</a> (Annotated scans of some selected pages)
%F T(n, k) = binomial(n,k)*binomial(n+k+1,k) - T(n-1, k), with T(n, 0) = (1 + (-1)^n)/2.
%F T(n, 0) = A000035(n+1).
%F T(n, 1) = A176222(n).
%F T(n, 2) = A331429(n).
%F T(n, n-2) = A002739(n).
%F T(n, n-1) = A002737(n).
%F T(n, n) = A001700(n).
%e Triangle begins:
%e 1;
%e 0, 3;
%e 1, 5, 10;
%e 0, 10, 35, 35;
%e 1, 14, 91, 189, 126;
%e 0, 21, 189, 651, 924, 462;
%e 1, 27, 351, 1749, 4026, 4290, 1716;
%e 0, 36, 594, 4026, 13299, 22737, 19305, 6435;
%e 1, 44, 946, 8294, 36751, 89375, 120835, 85085, 24310;
%e 0, 55, 1430, 15730, 89375, 289003, 551837, 615043, 369512, 92378;
%e 1, 65, 2080, 27950, 197275, 811733, 2047123, 3203837, 3031678, 1587222, 352716;
%p SS := (n,k)->binomial(n,k)*binomial(n+k+1,k);
%p T4:=proc(n,k) local i; global SS; option remember;
%p if k=0 then return((1+(-1)^n)/2); fi;
%p if n=0 then 0 else SS(n,k)-T4(n-1,k); fi; end;
%p rho:=n->[seq(T4(n,k),k=0..n)];
%p for n from 0 to 14 do lprint(rho(n)); od:
%t T[n_, k_]:= T[n, k]= If[n<0, 0, If[k==0, (1 + (-1)^n)/2, Binomial[n, k]*Binomial[n+k+1, k] - T[n-1, k]]];
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Mar 21 2022 *)
%o (Sage)
%o def T(n,k): # A331432
%o if (n<0): return 0
%o elif (k==0): return ((n+1)%2)
%o else: return binomial(n,k)*binomial(n+k+1,k) - T(n-1,k)
%o flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Mar 21 2022
%Y Columns 1 and 2 are A176222 and A331429; the last three diagonals are A002739, A002737, A001700.
%Y Taking the component-wise sums of the rows by pairs give the triangle in A178303.
%Y Ser's tables I and III are A331430 and A331431 (both are still mysterious).
%K nonn,tabl
%O 0,3
%A _N. J. A. Sloane_, Jan 17 2020
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