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%I #32 Apr 12 2023 10:52:32
%S 1,1,1,1,1,1,1,2,1,1,2,1,1,2,2,1,1,2,2,1,1,3,2,3,1,1,3,2,3,1,1,3,3,3,
%T 3,1,1,3,3,3,3,1,1,4,3,6,3,4,1,1,4,3,6,3,4,1,1,4,4,6,6,4,4,1,1,4,4,6,
%U 6,4,4,1,1,5,4,10,6,10,4,5,1,1,5,4,10,6
%N Starting with a(1) = a(2) = 1, proceed in a square spiral, computing each term as the sum of diagonally adjacent prior terms.
%H Peter Kagey, <a href="/A334745/b334745.txt">Table of n, a(n) for n = 1..10000</a>
%H Peter Kagey, <a href="/A334745/a334745.png">Bitmap illustrating the parity of the first 2^22 terms</a>. (Even and odd numbers are represented with black and white pixels respectively.)
%F Conjecture: a(2n-1) = A247976(n).
%e Spiral begins:
%e ... 3---3---3---3---1
%e |
%e 1---1---2---2---1 1
%e | | |
%e 2 1---1---1 1 3
%e | | | | |
%e 2 1 1---1 2 2
%e | | | |
%e 1 1---2---1---1 3
%e | |
%e 1---3---2---3---1---1
%e The last illustrated term above is a(35) = 3 = 2 + 1 because diagonally down-right is 2 and diagonally down-left is 1.
%Y Cf. A141481, A278180, A334741, A334742.
%Y The x- and y-coordinates at n-th step are A174344 and A274923 respectively.
%K nonn
%O 1,8
%A _Alec Jones_ and _Peter Kagey_, May 09 2020
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