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%I #34 Jan 17 2024 09:11:55
%S 2,3,4,5,7,8,9,10,11,13,14,15,16,17,18,19,21,22,23,24,25,26,27,28,29,
%T 30,31,32,33,34,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,
%U 54,55,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,73,74,75,76,77,78,79,80,81,82,83,84,85
%N A bisection of A064736.
%C The terms of A064736 lie on two (curved) lines; this is one of them.
%C Sequence is: a(1) = 2, a(2) = 3. m is in the sequence if and only if there is no i such that a(i) * a(i+1) = m, where i are indices of terms in the sequence so far. By definition, this is the complement of A286090. - _David A. Corneth_, May 25 2017
%C Apparently the same as A121229 shifted by one place. - _R. J. Mathar_, May 25 2017
%H Ray Chandler, <a href="/A286291/b286291.txt">Table of n, a(n) for n = 1..10000</a>
%H Ray Chandler, <a href="/A286291/a286291_1M.gz">Table of n, a(n) for n = 1..1000000</a> (large gzipped file)
%e See comments: 4 is in the sequence, since the terms so far, 2 and 3, don't multiply to 4. Same for 5. Sequence so far is: 2, 3, 4, 5. 6 isn't in the sequence. 7 is. Carrying on we get 2, 3, 4, 5, 7, 8, 9, 10, 11. 12 isn't in the sequence. Further in the sequence, 30 is in the sequence though it's of the form k*(k+1) for k = 5. But 6 isn't in the sequence. And indeed, 5 and 7 are consecutive terms so 5*7 = 35 isn't in the sequence. - _David A. Corneth_, May 25 2017
%o (PARI) upto(n) = {my(l=List([2,3]), i = 1, p = 6, op = 3);
%o while(1, if(op>=n, return(l)); for(j=op + 1, p-1, listput(l, j)); i++; op = p; p = l[i]*l[i+1])}
%o is(n) = !is_A286290(n)
%o is_A286290(n) = if(n < 6, return(n==1)); if(issquare(n+1, &n), is(n), if(sqrtint(4*n+1)^2 == 4*n+1, s = sqrtint(4*n+1); !(is(s\2) || is(s\2+1)), return(0))) \\ _David A. Corneth_, May 25 2017
%Y Cf. A064736, A286290, A286292, A286293, A121229.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, May 23 2017
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