A103341 Numbers k such that floor(k*sqrt(2)) is a power of 2.

1, 2, 3, 6, 12, 23, 91, 2897, 5793, 23171, 46341, 92682, 185364, 370728, 1482911, 2965821, 5931642, 23726567, 47453133, 94906266, 379625063, 759250125, 1518500250, 3037000500, 6074001000, 12148002000, 24296004000, 48592008000

Offset: 1

Author

Benoit Cloitre, May 13 2007

Keywords

• nonn

Comments

Sequence is infinite.

If floor(sqrt(2)*2^k) + 1 < sqrt(2)*2^k + sqrt(2)/2, then floor(sqrt(2)*2^k) + 1 is in this sequence. - Jinyuan Wang, Nov 04 2018

References

• Jean-Marie De Koninck and Armel Mercier, 1001 problèmes en théorie classique des nombres, ellipses, 2004, pp. 117, 374-375.

Links

• Robert Israel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001951 (floor(n*sqrt(2))).

Programs

Magma

[n: n in [1..2*10^7] | 2^Ilog(2, s) eq s where s is Floor(n*Sqrt(2))]; // Vincenzo Librandi, Nov 06 2018

Maple

N:= 100: # to get a(1)..a(N)
count:= 0:
for k from 0 while count < N do
  a:= ceil(2^(k-1)*sqrt(2));
  b:= floor((2^(k-1)+1/2)*sqrt(2));
  if a=b then
     count:= count+1;
     A[count]:= a;
  fi
od:
seq(A[n],n=1..N); # Robert Israel, Jul 19 2016

Mathematica

f[k_] := Reduce[n > 0 && (2^k)^2<= 2*n^2 <  (2^k + 1)^2, n, Integers]; n /. ToRules /@ Select[Table[f[k], {k, 0, 40}], # =!= False & ] (* Jean-François Alcover, Sep 13 2011 *)

PARI

for(k=0,50,n=ceil(2^k/sqrt(2));if(floor(n*sqrt(2))==2^k,print1(n,","))) \\ Robert Gerbicz, Jun 09 2007

PARI

isok(n) = my(b=sqrtint(2*n^2)); (b==1) || (b==2) || (isprimepower(b, &p) && (p==2)); \\ Michel Marcus, Mar 12 2019

Extensions

More terms from Robert Gerbicz, Jun 09 2007