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
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
Taking the component-wise sums of the rows by pairs give the triangle in A178303.
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Jan 17 2020
STATUS
approved