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Square array A(row,col) read by antidiagonals: A(1,col) = A055938(col), and for row > 1, A(row,col) = A011371(A(row-1,col)).
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%I #16 May 04 2015 02:13:38

%S 2,5,1,6,3,0,9,4,1,0,12,7,3,0,0,13,10,4,1,0,0,14,10,8,3,0,0,0,17,11,8,

%T 7,1,0,0,0,20,15,8,7,4,0,0,0,0,21,18,11,7,4,3,0,0,0,0,24,18,16,8,4,3,

%U 1,0,0,0,0,27,22,16,15,7,3,1,0,0,0,0,0,28,23,19,15,11,4,1,0,0,0,0,0,0,29,25,19,16,11,8,3,0,0,0,0,0,0,0

%N Square array A(row,col) read by antidiagonals: A(1,col) = A055938(col), and for row > 1, A(row,col) = A011371(A(row-1,col)).

%C The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

%C Column n gives the trajectory of iterates of A011371, when starting from A055938(n), thus stepping through successive parent-nodes when starting from the n-th leaf of binary beanstalk, until finally reaching the fixed point 0, which is the root of the whole binary tree.

%C The hanging tails of columns (upward from the first encountered zero) converge towards A179016.

%H Antti Karttunen, <a href="/A257264/b257264.txt">Table of n, a(n) for n = 1..10440; the first 144 antidiagonals of the array</a>

%H Paul Tek, <a href="/A179016/a179016.png">Illustration of how natural numbers in range 0 .. 133 are organized as a binary tree in the binary beanstalk</a>

%e The top left corner of the array:

%e 2, 5, 6, 9, 12, 13, 14, 17, 20, 21, 24, 27, 28, 29, 30, 33, 36, 37, 40, 43

%e 1, 3, 4, 7, 10, 10, 11, 15, 18, 18, 22, 23, 25, 25, 26, 31, 34, 34, 38, 39

%e 0, 1, 3, 4, 8, 8, 8, 11, 16, 16, 19, 19, 22, 22, 23, 26, 32, 32, 35, 35

%e 0, 0, 1, 3, 7, 7, 7, 8, 15, 15, 16, 16, 19, 19, 19, 23, 31, 31, 32, 32

%e 0, 0, 0, 1, 4, 4, 4, 7, 11, 11, 15, 15, 16, 16, 16, 19, 26, 26, 31, 31

%e 0, 0, 0, 0, 3, 3, 3, 4, 8, 8, 11, 11, 15, 15, 15, 16, 23, 23, 26, 26

%e 0, 0, 0, 0, 1, 1, 1, 3, 7, 7, 8, 8, 11, 11, 11, 15, 19, 19, 23, 23

%e 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 7, 7, 8, 8, 8, 11, 16, 16, 19, 19

%e 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 4, 4, 7, 7, 7, 8, 15, 15, 16, 16

%e 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 3, 3, 4, 4, 4, 7, 11, 11, 15, 15

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 3, 3, 3, 4, 8, 8, 11, 11

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 3, 7, 7, 8, 8

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 7, 7

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 4, 4

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 3, 3

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1

%e ...

%o (Scheme)

%o (define (A257264 n) (A257264bi (A002260 n) (A004736 n)))

%o (define (A257264bi row col) (if (= 1 row) (A055938 col) (A011371 (A257264bi (- row 1) col))))

%Y Row 1: A055938, Row 2: A257507.

%Y Cf. A011371, A055938.

%Y Cf. also A071542, A179016, A218254, A256993, A256997, A257265.

%K nonn,tabl

%O 1,1

%A _Antti Karttunen_, May 03 2015