A071911 Numbers m such that Stern's diatomic A002487(m) is divisible by 3.

0, 5, 7, 10, 14, 20, 28, 33, 35, 40, 45, 47, 49, 51, 56, 61, 63, 66, 70, 73, 75, 80, 85, 87, 90, 94, 98, 102, 105, 107, 112, 117, 119, 122, 126, 132, 140, 146, 150, 153, 155, 160, 165, 167, 170, 174, 180, 188, 196, 204, 210, 214, 217, 219, 224, 229, 231, 234, 238, 244, 252

Offset: 0

Author

N. J. A. Sloane, Jun 13 2002

Keywords

• nonn

Comments

From Kevin Ryde, Jan 09 2021: (Start)

Dijkstra gives bit pattern {0}1{?0{1}0|?1{0}1}?1{0} for the terms of this sequence, where | is alternative, {x} is zero or more of x, and ? is a single 0 or 1. Dijkstra formed this from a finite state automaton (states as pairs of Stern diatomic values mod 3 = A071412). The minimized automaton is as follows. State A is the start and the sole accepting state.

+---+

1 +----> | B | <-----+ 0

| +---+ |

0 +-- +===+ | +---+ --+ 1

+-> | A | |0,1 | D | <-+

start +===+ v +---+

^ +---+ ^

1 +----- | C | ------+ 0

+---+

a(n) can be calculated from n by a usual unranking in this automaton, using the number of strings of a given length k accepted from each state. Reznick's A122946(k) is the number of strings accepted starting from B. A089977(k-1) is the number accepted starting from C.

Dijkstra shows that for any m, a bit reversal (A030101) or an internal bit complement (A122155) of m are no change to the resulting Diatomic(m) value. So here if m is a term then so are A030101(m) and A122155(m). In the bit pattern, a reversal or complement between (but not including) the outermost 1's is no change.

(End)

Links

• Edsger W. Dijkstra, An exercise for Dr. R. M. Burstall, 1976. Reprinted in Edsger W. Dijkstra, Selected Writings on Computing, Springer-Verlag, 1982, pages 215-216.
• Edsger W. Dijkstra, More about the function ``fusc'', 1976. Reprinted in Edsger W. Dijkstra, Selected Writings on Computing, Springer-Verlag, 1982, pages 230-232.
• Bruce Reznick, Regularity Properties of the Stern Enumeration of the Rationals, Journal of Integer Sequences, volume 11, 2008, article 08.4.1. Also arXiv:math/0610601 [math.NT] 2006, and author's copy. Section 5 theorem 18 onward.

Crossrefs

Cf. A071412 (diatomic mod 3), A122946 (count by bit length), A089977 (half that).

Programs

PARI

{ my(M=Mod('x, 'x^2+'x+2),
            f=[2,1, 0,-'x-1, -2,1, 0,'x-1],
            table=[1,3, 5,5, 7,1, 3,7]);
a(n) = n<<=2; my(k=if(n,logint(n,2)+1), p=M^k, s=1);
  while(k>=0,
    my(t = n + (3<>2; }  \\ Kevin Ryde, Jan 09 2021

Python

def aupto(nn):
  ok = [1] + [0 for i in range(nn)]
  for m in range(nn+1):
    if ok[m]:  # from formula
      for i in [2*m, 8*m-5, 8*m+5, 8*m-7, 8*m+7]:
        if 0 <= i <= nn: ok[i] = 1
  return [m for m in range(nn+1) if ok[m]]
print(aupto(252)) # Michael S. Branicky, Jan 09 2021

Python

from itertools import count, islice
from functools import reduce
def inA071911(n): return not (n and sum(reduce(lambda x,y:(x[0],(x[0]+x[1])%3) if int(y) else ((x[0]+x[1])%3,x[1]),bin(n)[-1:2:-1],(1,0)))%3)
def A071911_gen(startvalue=0): # generator of terms >= startvalue
    return filter(inA071911, count(max(startvalue,0)))
A071911_list = list(islice(A071911_gen(),20)) # Chai Wah Wu, May 18 2023

Formula

If m is in the sequence, then 2*m, 8*m +- 5, and 8*m +- 7 (when nonnegative) are in the sequence. Starting from m=0, this rule generates the sequence. [Reznick section 5 theorem 18] - Kevin Ryde, Jan 09 2021