A286291 A bisection of A064736.

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, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 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

Offset: 1

Author

N. J. A. Sloane, May 23 2017

Keywords

• nonn

Comments

The terms of A064736 lie on two (curved) lines; this is one of them.

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

Apparently the same as A121229 shifted by one place. - R. J. Mathar, May 25 2017

Examples

			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
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Ray Chandler, Table of n, a(n) for n = 1..1000000 (large gzipped file)

Crossrefs

Cf. A064736, A286290, A286292, A286293, A121229.

Programs

PARI

upto(n) = {my(l=List([2,3]), i = 1, p = 6, op = 3);
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])}
is(n) = !is_A286290(n)
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