The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A054120 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A054120

Triangular array T(n,k): start with T(n,0)=T(n,n)=1 for n >= 0; recursively, draw vertical lines through T(n-1,k-1) if present and T(n-1,k) if present; then T(n,k) is the sum of T(i,j) that lie on or between the lines and not below T(n,k).
(history; published version)

#27 by R. J. Mathar at Wed Jan 26 10:04:02 EST 2022

STATUS

editing

approved

#26 by R. J. Mathar at Wed Jan 26 10:03:58 EST 2022

CROSSREFS

Row sums: A052945. A054122 (diagonal), A052392 (subdiagonal).

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approved

editing

#25 by R. J. Mathar at Wed Jan 26 10:01:47 EST 2022

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editing

approved

#24 by R. J. Mathar at Wed Jan 26 10:01:42 EST 2022

CROSSREFS

Row sums: A052945. A054122 (diagonal).

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approved

editing

#23 by R. J. Mathar at Wed Jan 26 10:01:02 EST 2022

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editing

approved

#22 by R. J. Mathar at Wed Jan 26 10:00:57 EST 2022

COMMENTS

Consider the array with g.f. (1-u*v)/(1-u-v-2*u*v). The triangle appears to be that symmetric array read by antidiagonals. - R. J. Mathar, Jan 26 2022

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approved

editing

#21 by Bruno Berselli at Mon Jun 15 04:05:30 EDT 2015

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reviewed

approved

#20 by Russell Jay Hendel at Sat Jun 13 22:29:30 EDT 2015

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proposed

reviewed

#19 by Jon E. Schoenfield at Sat Jun 13 02:59:32 EDT 2015

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editing

proposed

Discussion

Sat Jun 13

22:29

Russell Jay Hendel: "We have   T..." I am ok on removing this also.

#18 by Jon E. Schoenfield at Sat Jun 13 02:59:09 EDT 2015

COMMENTS

We illustrate the proof with n=5 and k=3. We first calculate T(3,2). By (D), we have (G) T(3,2)=6=SUM(L(2,1),L(1,1),L(2,2)) with L(2,1)={3,1,0,0,0,...}, L(1,1)={1,0,0,....} and L(2,2)={1,0,0,...}. By another application of (D), we have T(5,3)=39=(SUM(L(4,2),L(3,2),L(4,3)) where L(4,2)={18,3,1,...}={18} UNION L(2,1), L(3,2)={6,1,0,...}={6} UNION L(1,1), and L(4,3) = {1,0,0,...}={1} UNION L(2,2). Combining this last equation with (G), we have T(5,3)=39=18+6+1+SUM(L(2,1),L(1,1),L(2,2))=18+6+1+6=18+2*6+1 as required.

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reviewed

editing

Discussion

Sat Jun 13

02:59

Jon E. Schoenfield: Yes (although I just now saw and removed one I'd overlooked earlier).  :-)