A224515 a(n) = least k such that sqrt(k^2 XOR (k+1)^2) = 2*n+1, a(n) = -1 if there is no such k.

0, 4, 3, 24, 23, 44, 43, 112, 111, 180, 76, 264, 248, 348, 164, 480, 479, 411, 611, 327, 183, 115, 139, 943, 1103, 747, 787, 1111, 1447, 323, 699, 1984, 1983, 1851, 2243, 2008, 1576, 1388, 1684, 1072, 976, 1268, 499, 3383, 3271, 4124, 4068, 3679, 4511, 4315, 3804, 4999

Offset

0,2

Comments

Conjectures:

1. a(n) >= 0.

2. Least k is also the only such k.

If both conjectures are true, then the sequence is a permutation of A221643.

Links

Charles R Greathouse IV, Table of n, a(n) for n = 0..1000

Mathematica

a[n_] := For[k=0, k <= 3*n^2+1, k++, If[ Sqrt[ BitXor[k^2, (k+1)^2]] == 2*n+1, Return[k]]] /. Null -> -1; a /@ Range[0, 51] (* Jean-François Alcover, Jun 05 2013 *)

Prog

(Python)
import math
needTerms = n = 1024
i = 0
terms = [-1] * n
while n:
  s = (i*i) ^ ((i+1)*(i+1))
  r = int(math.sqrt(s))
  if s == r*r:
    if (r&1)==0:  break
    r = (r-1)/2
    if r < needTerms:
        if terms[r] >= 0:  break
        terms[r] = i
        n -= 1
  i += 1
if n:  print 'Error'
else:
  for i in range(needTerms):
    t = terms[i]
    print str(t)+', ',  #math.sqrt((t*t) ^ ((t+1)*(t+1)))
(PARI) a(n)=my(k=sqrtint(2*n^2), t); while(!issquare(bitxor(k^2, (k+1)^2), &t)||t!=2*n+1, k++); k \\ Charles R Greathouse IV, Jun 05 2013

Crossrefs

Cf. A221643.

Sequence in context: A287717 A286641 A288360 * A286328 A189742 A350173

Adjacent sequences: A224512 A224513 A224514 * A224516 A224517 A224518

Keyword

nonn,base,less

Author

Alex Ratushnyak, Apr 08 2013

Status

approved