A217833 The largest number not exceeding n^2, such that there are no terms of the sequence in the interval (a(n-1)/2, a(n)/2), with a(0)=0, a(1)=1.

0, 1, 2, 4, 8, 16, 32, 49, 64, 81, 98, 121, 128, 162, 196, 225, 242, 256, 324, 361, 392, 441, 450, 484, 512, 625, 648, 722, 784, 841, 882, 900, 968, 1024, 1156, 1225, 1250, 1296, 1444, 1521, 1568, 1681, 1682, 1764, 1800, 1936, 2048, 2209, 2304, 2312, 2450

Offset: 0

Author

Vladimir Shevelev, Oct 12 2012

Keywords

• nonn

Comments

Every term has the form s*2^k, where s>=0 is a square and k>=0.

Examples

			Let us find a(6), knowing the previous terms. Since a(5) = 16 and a(4)<=16/2<a(5). Then a(6) = 2*a(5) = 32, since 32<6^2 = 36. Further, since a(5)<=a(6)/2<a(6), then a(7) = 7^2 = 49, since 49<2*a(6) = 64.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A217689.

Programs

Maple

a:= proc(n) option remember; local i, j, k, t;
      if n<2 then n
    else i, j, k, t:= 0, n-1, iquo(n-1, 2), a(n-1)/2;
         while k<>i do if a(k)<=t then i:=k else j:=k fi;
                       k:= iquo(i+j,2) od;
         min(n^2, 2*a(k+1))
      fi
    end:
seq (a(n), n=0..100);  # Alois P. Heinz, Nov 03 2012

Mathematica

a[n_] := a[n] = Module[{i, j, k, t}, If[n < 2, n,
     {i, j, k, t} = {0, n-1, Quotient[n-1, 2], a[n-1]/2};
     While[k != i, If[a[k] <= t, i = k, j = k]; k = Quotient[i+j, 2]];
     Min[n^2, 2*a[k+1]]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, May 20 2022, after Alois P. Heinz *)

Formula

a(n) = min(2*a(k+1), n^2) for n>=2 and a(k) <= a(n-1)/2 < a(k+1).

Extensions

More terms from Alois P. Heinz, Nov 02 2012