%I
%S 0,1,0,2,1,3,4,0,5,6,2,7,8,9,10,1,11,12,3,13,14,15,16,4,17,18,0,19,20,
%T 21,22,5,23,24,25,26,27,28,6,29,30,2,31,32,33,34,35,36,7,37,38,39,40,
%U 8,41,42,9,43,44,45,46,10,47,48,49,50,51,52,1,53,54
%N Minimal sequence whose forwards van Eck transform is the sequence of prime numbers.
%C At each step, the minimum available integer is used.
%C From _Rémy Sigrist_, Aug 12 2017: (Start)
%C a(n)=0 iff n belongs to A074271.
%C a(n)=1 iff n > 1 and n belongs to A259408.
%C For any k > 0, A064427(k) = least n such that a(n) = k1.
%C (End)
%H Rémy Sigrist, <a href="/A181642/b181642.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1)=0. Next 0 is at distance 2 (1st prime): a(3)=0.
%e a(2)=1. Next 1 is at distance 3 (2nd prime): a(5)=1.
%e a(3)=0. Next 0 is at distance 5 (3rd prime): a(8)=0.
%e For a(4), we can use neither 0 (distance 1 from previous 0 would lead to an incongruence) nor 1 (distance 1 from subsequent 1 would lead to another incongruence). Therefore we must use 2.
%e Next 2 must be at distance 7 (4th prime): a(11)=2. And so on.
%p P:=proc(q,h) local i,k,n,t,x; x:=array(1..h); for k from 1 to h do x[k]:=1; od; x[1]:=0; i:=0; t:=0;for n from 1 to q do if isprime(n) then i:=i+1; if x[i]>1 then x[i+n]:=x[i]; else t:=t+1; x[i]:=t; x[i+n]:=x[i]; fi; fi; od; seq(x[k],k=1..79); end: P(400,500);
%o (PARI) a = vector(71, i, 1); u = 0; for (n=1, #a, if (a[n]<0, o = n; while (o <= #a, a[o] = u; o += prime(o)); u++); print1 (a[n] ", ")) \\ _Rémy Sigrist_, Aug 12 2017
%Y Cf. A000040, A064427, A074271, A181391, A259408.
%K easy,nonn,changed
%O 1,4
%A _Paolo P. Lava_ & _Giorgio Balzarotti_, Nov 03 2010
%E More terms from _Rémy Sigrist_, Aug 12 2017
