A331432 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).

1, 0, 3, 1, 5, 10, 0, 10, 35, 35, 1, 14, 91, 189, 126, 0, 21, 189, 651, 924, 462, 1, 27, 351, 1749, 4026, 4290, 1716, 0, 36, 594, 4026, 13299, 22737, 19305, 6435, 1, 44, 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

Offset

0,3

Comments

The scanned pages of Ser are essentially illegible, and the book is out of print and hard to locate.

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.

References

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93.

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].

J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)

Formula

T(n, k) = binomial(n,k)*binomial(n+k+1,k) - T(n-1, k), with T(n, 0) = (1 + (-1)^n)/2.

T(n, 0) = A000035(n+1).

T(n, 1) = A176222(n).

T(n, 2) = A331429(n).

T(n, n-2) = A002739(n).

T(n, n-1) = A002737(n).

T(n, n) = A001700(n).

Example

Triangle begins:
  1;
  0,  3;
  1,  5,   10;
  0, 10,   35,    35;
  1, 14,   91,   189,    126;
  0, 21,  189,   651,    924,    462;
  1, 27,  351,  1749,   4026,   4290,    1716;
  0, 36,  594,  4026,  13299,  22737,   19305,    6435;
  1, 44,  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;

Maple

SS := (n, k)->binomial(n, k)*binomial(n+k+1, k);
T4:=proc(n, k) local i; global SS; option remember;
if k=0 then return((1+(-1)^n)/2); fi;
if n=0 then 0 else SS(n, k)-T4(n-1, k); fi; end;
rho:=n->[seq(T4(n, k), k=0..n)];
for n from 0 to 14 do lprint(rho(n)); od:

Mathematica

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]]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 21 2022 *)

Prog

(Sage)
def T(n, k): # A331432
    if (n<0): return 0
    elif (k==0): return ((n+1)%2)
    else: return binomial(n, k)*binomial(n+k+1, k) - T(n-1, k)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 21 2022

Crossrefs

Columns 1 and 2 are A176222 and A331429; the last three diagonals are A002739, A002737, A001700.

Taking the component-wise sums of the rows by pairs give the triangle in A178303.

Ser's tables I and III are A331430 and A331431 (both are still mysterious).

Sequence in context: A202504 A146916 A146255 * A122366 A228781 A103327

Adjacent sequences: A331429 A331430 A331431 * A331433 A331434 A331435

Keyword

nonn,tabl

Author

N. J. A. Sloane, Jan 17 2020

Status

approved