OFFSET
0,3
COMMENTS
The sequence is a solution to the riddle described in the comments of A304584 without the restriction of x and d to nonnegative numbers.
LINKS
Rainer Rosenthal, Table of n, a(n) for n = 0..10000
EXAMPLE
This is the standard counterclockwise square spiral starting at 0. - N. J. A. Sloane, Oct 17 2019
d:
3 | 36--35--34--33--32--31--30 55
| | | |
2 | 37 16--15--14--13--12 29 54
| | | | | |
1 | 38 17 4---3---2 11 28 53
| | | | | | | |
0 | 39 18 5 0---1 10 27 52
| | | | | | |
-1 | 40 19 6---7---8---9 26 51
| | | | |
-2 | 41 20--21--22--23--24--25 50
| | |
-3 | 42--43--44--45--46--47--48--49
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x: -3 -2 -1 0 1 2 3 4
.
a(9) = 2 + 9*(-1) = -7 because the 9th position in the spiral corresponds to x = 2 and d = -1,
a(14) = 0 + 14*2 = 28 because the 14th position in the spiral corresponds to x = 0 and d = 2,
a(25) = 3 + 25*(-2) = -47 because the 25th position in the spiral corresponds to x = 3 and d = -2.
MAPLE
square2pair:=proc(sq)local w, k; w:=floor(sqrt(sq)); k:=floor(w/2); if modp(sq, 2)=0 then return[-k, k]; else return[k+1, -k]; fi; end:pos2pS:=proc(n)local w, q, Q, e, E, sp; w:=floor(sqrt(n)); q := w^2; Q:=(w+1)^2; e:=n-q; E:=Q-n; if e<E then sp:=square2pair(q); if modp(q, 2)=0 then return[sp[1], sp[2]-e]; else return[sp[1], sp[2]+e]; fi; else sp:=square2pair(Q); if modp(Q, 2)=0 then return[sp[1]+E, sp[2]]; else return[sp[1]-E, sp[2]]; fi; fi; end:WhereFlea:=proc(n) local x, d, pair; pair:=pos2pS(n); x:=pair[1]; d:=pair[2]; return x+d*n; end: seq(WhereFlea(n), n=0..61); # Rainer Rosenthal, May 24 2018
CROSSREFS
KEYWORD
sign,look
AUTHOR
Hugo Pfoertner, May 16 2018
EXTENSIONS
a(1) and a(2) corrected by Rainer Rosenthal, May 24 2018
STATUS
approved