A033989 Write 0,1,2,... in a clockwise spiral on a square lattice, writing each digit at a separate lattice point, starting with 0 at the origin and 1 at x=0, y=-1; sequence gives the numbers on the negative x-axis.
0, 3, 1, 1, 3, 2, 7, 9, 1, 1, 6, 9, 4, 7, 9, 1, 2, 1, 2, 1, 6, 7, 4, 3, 6, 1, 2, 9, 5, 1, 1, 0, 9, 3, 1, 3, 6, 6, 1, 8, 6, 9, 2, 5, 0, 2, 2, 4, 6, 6, 2, 5, 6, 0, 3, 8, 9, 5, 3, 3, 6, 9, 4, 0, 5, 4, 4, 9, 8, 0, 5, 0, 4, 5, 5, 3, 3, 1, 6, 8, 5, 8, 6, 5, 1, 4, 7, 4, 9, 1, 8, 5, 1, 9, 9, 8, 6, 6, 9, 1, 1, 6, 4, 8, 1
Offset: 0
Examples
2---3---2---4---2---5---2 | | 2 1---3---1---4---1 6 | | | | 2 2 4---5---6 5 2 | | | | | | 1 1 3 0 7 1 7 | | | | | | | 2 1 2---1 8 6 2 | | | | | 0 1---0---1---9 1 8 | | | 2---9---1---8---1---7 2 We begin with the 0 and wrap the numbers 1 2 3 4 ... around it. Then the sequence is obtained by reading leftwards, starting from the initial 0. - _Andrew Woods_, May 20 2012
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Haskell
a033989 0 = 0 a033989 n = a033307 $ a185950 n -- Reinhard Zumkeller, Aug 14 2013
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Mathematica
nmax = 1000; A033307 = Flatten[IntegerDigits /@ Range[0, nmax^2+10*nmax]]; a[n_] := If[n==0, 0, A033307[[4n^2-n+1]]]; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Apr 23 2017, after Andrew Woods *)
Formula
a(n) = A033307(4*n^2-n-1) for n > 0. - Andrew Woods, May 20 2012
Extensions
More terms from Andrew J. Gacek (andrew(AT)dgi.net)
Edited by Charles R Greathouse IV, Nov 01 2009