A289206 Greedy strictly increasing sequence starting at a(1)=1 avoiding both arithmetic and geometric progressions of length 3.

1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88, 90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, 223, 232, 235, 240, 243, 244, 249, 250, 252, 259, 267, 271, 289, 304, 314, 318, 325, 340, 342, 397, 504, 508, 511, 531, 549

Offset: 1

Author

Roderick MacPhee, Jun 28 2017

Keywords

• nonn

Comments

By avoiding arithmetic progressions, at most 2/3 of the numbers up to a(n) are in the sequence. The sequence doesn't contain 3 consecutive powers in arithmetic progression for any base c.

Where a(n)+1 = a(n+1): 1, 3, 5, 7, 9, 12, 17, 21, 23, 26, 30, 32, 37, 39, etc. - Robert G. Wilson v, Jul 02 2017

Examples

			5 is in the sequence because 1,2,5 is neither an arithmetic progression nor a geometric progression.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• Math StackExchange, A sequence that avoids both arithmetic and geometric progression (2014)

Crossrefs

Cf. A000452, A003278, A005836, A224853, A225571.

Programs

PARI

{my(a=[1,2]);
for(x=3,100,
if(#select(r->#select(q->q==2*r,b)==0,b=vecsort(apply(r->x-r,a)))==#a && #select(r->#select(q->q==r^2,b)==0,b=vecsort(apply(r->x/r,a)))==#a,a=concat(a,x)));a
}

PARI

first(n)=my(v=vector(n)); v[1]=1; for(k=2,n, my(avoid=List(),t,last=v[k-1]); for(i=2,k-1, for(j=1,i-1, t=2*v[i]-v[j]; if(t>last, listput(avoid, t)); if(denominator(t=v[i]^2/v[j])==1 && t>last, listput(avoid,t)))); avoid=Set(avoid); for(i=v[k-1]+1,v[k-1]+#avoid+1, if(!setsearch(avoid,i), v[k]=i; break))); v \\ Charles R Greathouse IV, Jun 29 2017

Formula

a(n) >= 3n/2 for n > 2.

Extensions

More terms from Alois P. Heinz, Jun 28 2017