The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A318958 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A318958

A(n, k) is a square array read in the decreasing antidiagonals, for n >= 0 and k >= 0.
(history; published version)

#31 by Peter Luschny at Mon Jul 22 03:24:52 EDT 2019

STATUS

proposed

approved

#30 by Jean-François Alcover at Mon Jul 22 03:12:19 EDT 2019

STATUS

editing

proposed

#29 by Jean-François Alcover at Mon Jul 22 03:11:58 EDT 2019

DATA

0, 0, 0, 0, -1, -1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 2, 3, 2, 2, 0, 1, 3, 3, 4, 3, 3, 0, 3, 4, 6, 6, 7, 6, 6, 0, 2, 5, 6, 8, 8, 9, 8, 8, 0, 4, 6, 9, 10, 12, 12, 13, 12, 12, 0, 3, 7, 9, 12, 13, 15, 15, 16, 15, 15, 0, 5, 8, 12, 14, 17, 18, 20, 20, 21, 20, 20

MATHEMATICA

h[n_] := If[n < 3, {0, 0, -1}[[n + 1]], Quotient[n^2 - 4 n + 3, 4]];

A[n_, k_] := A[n, k] = If[k == 0, h[n], If[k == 1, h[n+1], A[n, k-2] + n]];

Table[A[n - k, k], {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jul 22 2019, after Peter Luschny *)

STATUS

approved

editing

Discussion

Mon Jul 22

03:12

Jean-François Alcover: Completed last row of data.

#28 by Peter Luschny at Sun Sep 09 06:06:16 EDT 2018

STATUS

proposed

approved

#27 by Paul Curtz at Sun Sep 09 05:55:21 EDT 2018

STATUS

editing

proposed

Discussion

Sun Sep 09

06:04

Peter Luschny: OK, if there still are flaws we can come back later.

#26 by Paul Curtz at Sun Sep 09 05:48:13 EDT 2018

FORMULA

A(n,k) - A(n,k-1) = 0, -1, 1, 0, 2, 1, ... = [1] which variable is running, n or k?

A(n,k) - A(n,k-3) = 0, 0, 3, 3, 6, 6, ... = [3] ?

A(n,k) - A(n,k-4) = 0, 2, 4, 6, 8, 10, ... = [4] ?

A(n,k) - A(n,k-5) = 0, 1, 5, 6, 10, 11, 15, 16, ... = A008851(n+1)).

Note that [1] + [4] = A008851(n+1).

CROSSREFS

Cf. A008851.

Discussion

Sun Sep 09

05:52

Paul Curtz: Last edit removed. Thank you.

#25 by Peter Luschny at Sun Sep 09 05:20:57 EDT 2018

FORMULA

Let h(n) = 0, 0, -1, A198442(1), A198442(2), A198442(3), ... Then A(n, 0) = h(n), A(n, 1) = h(n+1) and A(n, k) = A(n, k-2) + n otherwise [2].

Before: A(n,k) = - A(n,k-1) + (= 0, -1, 1, 0, 2, 1, ... = [1]). which variable is running, n or k?

After: A(n,k) = - A(n,k-3) + (= 0, 0, 3, 3, 6, 6, ... = [3]). ?

A(n,k) = - A(n,k-4) + (= 0, 2, 4, 6, 8, 10, ... = [4]). ?

A(n,k) = - A(n,k-5) + (= 0, 1, 5, 6, 10, 11, 15, 16, ... = A008851(n+1)).

Note that [1] + [4] = A008851(n+1). - _ Paul Curtz _, Sep 09 2018

STATUS

proposed

editing

Discussion

Sun Sep 09

05:28

Peter Luschny: I do not understand comments like 'before', 'after', so I deleted them. Please do not touch the definition. The definition (the first sentence in the formula section) is valid for all n and k. Instead of using references like [1], [2] be more precise and write them in the form A(n,k). I think that this will show that not all the formulas which you added in your last edit are true for all n and k. Anyway, I have no more time to spend on this sequence.

#24 by Paul Curtz at Sun Sep 09 03:42:26 EDT 2018

STATUS

editing

proposed

#23 by Paul Curtz at Sun Sep 09 03:40:27 EDT 2018

FORMULA

Before: A(n,k) = A(n,k-1) + (0, -1, 1, 0, 2, 1, ... = [1] ).

After: A(n,k) = A(n,k-3) + (0, 0, 3, 3, 6, 6, ... = [3] ).

A(n,k) = A(n,k-4) + (0, 2, 4, 6, 8, 10, ... = [4] ).

#22 by Paul Curtz at Sun Sep 09 03:36:09 EDT 2018

FORMULA

Let h(n) = 0, 0, -1, A198442(1), A198442(2), A198442(3), ... Then A(n, 0) = h(n), A(n, 1) = h(n+1) and A(n, k) = A(n, k-2) + n otherwise [2].

Before: A(n,k) = A(n,k-1) +(0, -1, 1, 0, 2, 1, ... = [1] ).

After: A(n,k) = A(n,k-3) + (0, 0, 3, 3, 6, 6, ... = [3] ).

A(n,k) = A(n,k-4) + (0, 2, 4, 6, 8, 10, ... = [4] ).

A(n,k) = A(n,k-5) + (0, 1, 5, 6, 10, 11, 15, 16, ... = A008851(n+1)).

Note that [1] + [4] = A008851(n+1). - _ Paul Curtz _, Sep 09 2018

CROSSREFS

Cf. A008851.

STATUS

proposed

editing