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%I #26 Dec 23 2015 14:07:49
%S 0,2,1,6,4,2,12,8,6,3,18,0,12,5,4,22,0,18,7,8,5,30,0,22,0,0,7,6,34,0,
%T 30,0,0,0,12,7,42,0,34,0,0,0,18,0,8,46,0,42,0,0,0,22,0,0,9,54,0,46,0,
%U 0,0,30,0,0,11,10,58,0,54,0,0,0,34,0,0,16,14,11
%N Square array A(row,col): A(row,0) = row and for col >= 1, if A262686(row) is 0, then A(row,col) = 0, otherwise A(row,col) = A(A262686(row),col-1).
%C The square array A(row>=0, col>=0) is read by downwards antidiagonals as: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), A(0,3), A(1,2), A(2,1), A(3,0), ...
%C Each row n lists all the nodes in A263267-tree that one encounters when one starts from node n and always chooses the largest possible child of it (A262686), and then the largest possible child of that child, etc, until a leaf-child (one of the terms of A045765) is encountered, after which the rest of the row contains only zeros.
%H Antti Karttunen, <a href="/A263271/b263271.txt">Table of n, a(n) for n = 0..10439; the first 144 antidiagonals</a>
%F A(row,0) = row and for col >= 1, if A262686(row) is 0, then A(row,col) = 0, otherwise A(row,col) = A(A262686(row),col-1).
%F A(0,0) = 0, A(0,1) = 2; if col = 0, A(row,0) = row; and for col > 0, if A(row,col-1) = 0, then A(row,col) = 0, otherwise A(row,col) = A262686(A(row,col-1)). [Another, perhaps more intuitive recurrence for this array.] - _Antti Karttunen_, Dec 21 2015
%e The top left corner of the array:
%e 0, 2, 6, 12, 18, 22, 30, 34, 42, 46, 54, 58, 66, 0, 0, 0, 0
%e 1, 4, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 2, 6, 12, 18, 22, 30, 34, 42, 46, 54, 58, 66, 0, 0, 0, 0, 0
%e 3, 5, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 4, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 5, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 6, 12, 18, 22, 30, 34, 42, 46, 54, 58, 66, 0, 0, 0, 0, 0, 0
%e 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 9, 11, 16, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 10, 14, 20, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 11, 16, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 12, 18, 22, 30, 34, 42, 46, 54, 58, 66, 0, 0, 0, 0, 0, 0, 0
%e 13, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 14, 20, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e 15, 17, 21, 23, 27, 36, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
%e ...
%o (Scheme)
%o (define (A263271 n) (A263271bi (A002262 n) (A025581 n)))
%o (define (A263271bi row col) (cond ((zero? col) row) ((A262686 row) => (lambda (lad) (if (zero? lad) lad (A263271bi lad (- col 1)))))))
%o ;; An alternative implementation, reflecting the new recurrence:
%o (define (A263271bi row col) (cond ((zero? col) row) ((and (zero? row) (= 1 col)) 2) ((zero? (A263271bi row (- col 1))) 0) (else (A262686 (A263271bi row (- col 1))))))
%Y Column 0: A001477, Column 1: A262686.
%Y Cf. A264971 (number of significant terms on each row, position where the first trailing zero-term occurs).
%Y Cf. A264970.
%Y Cf. also A000005, A045765, A263267.
%Y See also array A265751 constructed in the same way, but obtained by following always the smallest child A082284, instead of the largest child A262686.
%K nonn,tabl
%O 0,2
%A _Antti Karttunen_, Nov 29 2015
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