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%I #24 Jul 03 2023 11:30:04
%S 0,1,2,2,5,10,3,9,17,27,4,14,26,40,56,5,20,37,56,77,100,6,27,50,75,
%T 102,131,162,7,35,65,97,131,167,205,245,8,44,82,122,164,208,254,302,
%U 352,9,54,101,150,201,254,309,366,425,486,10,65,122,181,242,305,370,437,506,577,650,11
%N A linear mapping a(n) = x + d*n of pairs of nonnegative integers (x,d), where the pairs are enumerated by antidiagonals.
%C The sequence solves the following riddle, which has been communicated by _Klaus Nagel_: A flea starts to jump on the nonnegative integers at time = 0 at an unknown location x >= 0 making jumps of unknown, but constant distance d >= 0 at every subsequent time step. By which strategy can the flea be captured with 100% certainty in a finite number of trials? The solution is to hit a(n) at time = n. This works for all enumerations of pairs (x,d) of integers, because eventually any combination of starting location x and jump width d will be addressed.
%H Rainer Rosenthal, <a href="/A304584/b304584.txt">Table of n, a(n) for n = 0..10000</a>
%e d:
%e 5 | 20
%e 4 | 14 19
%e 3 | 9 13 18
%e 2 | 5 8 12 17
%e 1 | 2 4 7 11 16
%e 0 | 0 1 3 6 10 15
%e |________________________
%e x: 0 1 2 3 4 5
%e .
%e a(13) = 1 + 13*3 = 40 because the 13th position in the enumeration corresponds to x=1 and d=3.
%p pos2pair:=proc(n) local w,k,e;w:=floor(sqrt(2*n));if w*(w+1)>2*n then k:=w-1;else k:=w;fi;e:=n-k*(k+1)/2;return [k-e,e];end:WhereFlea:=proc(n) local x,d,pair; pair:=pos2pair(n);x:=pair[1];d:=pair[2];return x+d*n;end:
%p seq(WhereFlea(n),n=0..66);# _Rainer Rosenthal_, May 23 2018
%Y Cf. A304585, A304586, A304587, A305260.
%K nonn
%O 0,3
%A _Hugo Pfoertner_, May 15 2018
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