The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A051714 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A051714

Numerators of table a(n,k) read by antidiagonals: a(0,k) = 1/(k+1), a(n+1,k) = (k+1)*(a(n,k) - a(n,k+1)), n >= 0, k >= 0.
(history; published version)

#52 by Michael De Vlieger at Sun Apr 23 07:29:23 EDT 2023

STATUS

reviewed

approved

#51 by Michel Marcus at Sun Apr 23 02:24:17 EDT 2023

STATUS

proposed

reviewed

#50 by G. C. Greubel at Sun Apr 23 00:56:09 EDT 2023

STATUS

editing

proposed

#49 by G. C. Greubel at Sun Apr 23 00:18:38 EDT 2023

NAME

Numerators of table a(n,k) read by antidiagonals: a(0,k) = 1/(k+1), a(n+1,k) = (k+1)*(a(n,k) - a(n,k+1)), n >= 0, k >= 0.

EXAMPLE

1 1/2 1/3 1/4 1/5 1/6 1/7 ...

0 1/30 1/20 2/35 5/84 ...

Antidiagonals of numerator(a(n,k)):

1;

1, 1;

1, 1, 1;

1, 1, 1, 0;

1, 1, 3, 1, -1;

1, 1, 2, 1, -1, 0;

1, 1, 5, 2, -3, -1, 1;

1, 1, 3, 5, -1, -1, 1, 0;

1, 1, 7, 5, 0, -4, 1, 1, -1;

1, 1, 4, 7, 1, -1, -1, 1, -1, 0;

1, 1, 9, 28, 49, -29, -5, 8, 1, -5, 5;

MATHEMATICA

nmax = 12; a[0, k_] := 1/(k+1); a[n_, k_] := a[n, k] = (k+1)(a[n-1, k]-a[n-1, k+1]); Numerator[ Flatten[ Table[ a[n-k, k], {n, 0, nmax}, {k, n, 0, -1}]]] (* Jean-François Alcover, Nov 28 2011 *)

PROG

(Magma)

function a(n, k)

if n eq 0 then return 1/(k+1);

else return (k+1)*(a(n-1, k) - a(n-1, k+1));

end if;

end function;

A051714:= func< n, k | Numerator(a(n, k)) >;

[A051714(k, n-k): k in [0..n], n in [0..15]]; // G. C. Greubel, Apr 22 2023

(SageMath)

def a(n, k):

if (n==0): return 1/(k+1)

else: return (k+1)*(a(n-1, k) - a(n-1, k+1))

def A051714(n, k): return numerator(a(n, k))

flatten([[A051714(k, n-k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Apr 22 2023

CROSSREFS

Rows 2, 3, 4 give : A026741/A045896, A051712/A051713, A051722/A051723, columns 0, 1, 2, 3 give A000367/A002445, A051716/A051717, A051718/A051719, A051720/A051721.

Columns 0, 1, 2, 3 give: A000367/A002445, A051716/A051717, A051718/A051719, A051720/A051721.

Cf. A027642, A164555.

STATUS

approved

editing

#48 by Alois P. Heinz at Wed Feb 08 10:45:43 EST 2023

STATUS

proposed

approved

#47 by Alois P. Heinz at Mon Feb 06 15:03:22 EST 2023

STATUS

editing

proposed

Discussion

Mon Feb 06

23:03

Fabián Pereyra: It is correct Alois, thanks for fixing it. The numerator of the e.g.f. is x+log(1-t).

#46 by Alois P. Heinz at Mon Feb 06 15:01:56 EST 2023

FORMULA

E.g.f.: A(x,t) = (x+log(1-t))/(1-t-exp(-x)) = (1+(1/2)*x+(1/6)*x^2/2!-(1/30)*x^4/4!+...)*1 + (1/2+(1/3)*x+(1/6)*x^2/2!+...)*t + (1/3+(1/4)*x+(3/20)*x^2/2!+...)*t^2 + .... (End)

Discussion

Mon Feb 06

15:02

Alois P. Heinz: one ")" inserted to get (x+log(1-t))/(1-t-exp(-x)) ... which now seems to be correct ...

#45 by Alois P. Heinz at Mon Feb 06 14:56:11 EST 2023

STATUS

proposed

editing

Discussion

Mon Feb 06

14:57

Alois P. Heinz: (x+log(1-t)/(1-t-exp(-x)) is missing a ")" somewhere ...

#44 by Fabián Pereyra at Sat Jan 14 10:02:35 EST 2023

STATUS

editing

proposed

Discussion

Sat Jan 14

10:12

Jon E. Schoenfield: Thanks!

#43 by Fabián Pereyra at Sat Jan 14 10:01:17 EST 2023

FORMULA

a(n,k) = numerator(Sum_{j,=0..n} (-1)^(n-j)*j!*Stirling2(n,j)/(j+k+1)).

E.g.f.: A(x,t) = (x+log(1-t)/(1-t-exp(-x)) = (1+(1/2)*x+(1/6)*x^2/2!-(1/30)*x^4/4!+...)*1 + (1/2+(1/3)*x+(1/6)*x^2/2!+...)*t + (1/3+(1/4)*x+(3/20)*x^2/2!+...)*t^2 + .... (End)

Discussion

Sat Jan 14

10:02

Fabián Pereyra: Notation fixed.