A127080 Infinite square array read by antidiagonals: Q(m, 0) = 1, Q(m, 1) = 1; Q(m, 2k) = (m - 2k + 1)*Q(m+1, 2k-1) - (2k-1)*Q(m+2,2k-2), m*Q(m, 2k+1) = (m - 2k)*Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1).

1, 1, 1, 1, 1, -2, 1, 1, -1, -5, 1, 1, 0, -4, 12, 1, 1, 1, -3, 3, 43, 1, 1, 2, -2, -4, 28, -120, 1, 1, 3, -1, -9, 15, -15, -531, 1, 1, 4, 0, -12, 4, 48, -288, 1680, 1, 1, 5, 1, -13, -5, 75, -105, 105, 8601, 1, 1, 6, 2, -12, -12, 72, 24, -624, 3984, -30240, 1, 1, 7, 3, -9, -17, 45, 105, -735, 945, -945, -172965

Offset

0,6

Comments

Comment from N. J. A. Sloane, Jan 29 2020: (Start)

It looks like there was a missing 2 in the definition, which I have now corrected. The old definition was:

(Wrong!) Infinite square array read by antidiagonals: Q(m, 0) = 1, Q(m, 1) = 1; Q(m, 2k) = (m - 2k + 1)*Q(m+1, 2k-1) - (2k-1)*Q(m+2, k-2), m*Q(m, 2k+1) = (m - 2k)*Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1). (Wrong!) (End)

References

V. van der Noort and N. J. A. Sloane, Paper in preparation, 2007.

Links

G. C. Greubel, Antidiagonals n = 0..100, flattened

Formula

E.g.f.: Sum_{k >= 0} Q(m,2k) x^k/k! = (1+4x)^((m-1)/2)/(1+2x)^(m/2), Sum_{k >= 0} Q(m,2k+1) x^k/k! = (1+4x)^((m-2)/2)/(1+2x)^((m+1)/2).

Example

Array begins:
     1,    1,    1,    1,    1,   1,   1,    1,    1,    1, ... (A000012)
     1,    1,    1,    1,    1,   1,   1,    1,    1,    1, ... (A000012)
    -2,   -1,    0,    1,    2,   3,   4,    5,    6,    7, ... (A023444)
    -5,   -4,   -3,   -2,   -1,   0,   1,    2,    3,    4, ... (A023447)
    12,    3,   -4,   -9,  -12, -13, -12,   -9,   -4,    3, ... (A127146)
    43,   28,   15,    4,   -5, -12, -17,  -20,  -21,  -20, ... (A127147)
  -120,  -15,   48,   75,   72,  45,   0,  -57, -120, -183, ... (A127148)
  -531, -288, -105,   24,  105, 144, 147,  120,   69,    0, ...
  1680,  105, -624, -735, -432, 105, 720, 1281, 1680, 1833, ...

Maple

f:= proc(k) option remember;
      if `mod`(k, 2)=0 then k!/(k/2)!
    else 2^(k-1)*((k-1)/2)!*add(binomial(2*j, j)/8^j, j=0..((k-1)/2))
      fi; end;
Q:= proc(n, k) option remember;
      if n=0 then (-1)^binomial(k, 2)*f(k)
    elif k<2 then 1
    elif `mod`(k, 2)=0 then (n-k+1)*Q(n+1, k-1) - (k-1)*Q(n+2, k-2)
    else ( (n-k+1)*Q(n+1, k-1) - (k-1)*(n+1)*Q(n+2, k-2) )/n
      fi; end;
seq(seq(Q(n-k, k), k=0..n), n=0..12); # G. C. Greubel, Jan 30 2020

Mathematica

Q[0, k_]:= Q[0, k]= (-1)^Binomial[k, 2]*If[EvenQ[k], k!/(k/2)!, 2^(k-1)*((k-1)/2)!* Sum[Binomial[2*j, j]/8^j, {j, 0, (k-1)/2}] ];
Q[n_, k_]:= Q[n, k]= If[k<2, 1, If[EvenQ[k], (n-k+1)*Q[n+1, k-1] - (k-1)*Q[n+2, k-2], ((n -k+1)*Q[n+1, k-1] - (k-1)*(n+1)*Q[n+2, k-2])/n]];
Table[Q[n-k, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jan 30 2020 *)

Prog

(Sage)
@CachedFunction
def f(k):
    if (mod(k, 2)==0): return factorial(k)/factorial(k/2)
    else: return 2^(k-1)*factorial((k-1)/2)*sum(binomial(2*j, j)/8^j for j in (0..(k-1)/2))
def Q(n, k):
    if (n==0): return (-1)^binomial(k, 2)*f(k)
    elif (k<2): return 1
    elif (mod(k, 2)==0): return (n-k+1)*Q(n+1, k-1) - (k-1)*Q(n+2, k-2)
    else: return ( (n-k+1)*Q(n+1, k-1) - (k-1)*(n+1)*Q(n+2, k-2) )/n
[[Q(n-k, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 30 2020

Crossrefs

See A105937 for another version.

Columns give A127137, A127138, A127144, A127145;

Rows give A127146, A127147, A127148.

Sequence in context: A330942 A141471 A331572 * A216645 A216635 A213945

Adjacent sequences: A127077 A127078 A127079 * A127081 A127082 A127083

Keyword

sign,tabl

Author

N. J. A. Sloane, Mar 24 2007

Extensions

More terms added by G. C. Greubel, Jan 30 2020

Status

approved