|
|
A353715
|
|
a(n) = b(n)+b(n+1), where b is A353709.
|
|
7
|
|
|
1, 3, 6, 12, 11, 19, 28, 44, 49, 23, 46, 104, 69, 15, 58, 113, 79, 142, 161, 51, 86, 77, 43, 54, 92, 107, 167, 156, 90, 102, 61, 155, 226, 109, 157, 242, 354, 277, 63, 234, 449, 279, 126, 233, 387, 286, 125, 481, 410, 63, 357, 456, 143, 87, 240, 171, 95, 372, 419, 207, 348, 433, 231, 334, 313, 183, 462, 840, 531, 63, 492, 961, 543, 254, 992, 783, 127
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Created in an attempt to show that every number appears in A353709. For example, if one could show that the present sequence had a subsequence which was divisible by ever-increasing powers of 2, the desired result would follow. See A353724, A353725, A353726, A353727 for more about this topic.
|
|
LINKS
|
|
|
MAPLE
|
g:= proc() false end: t:= 2:
b:= proc(n) option remember; global t; local k; if n<2 then n
else for k from t while g(k) or Bits[And](k, b(n-2))>0
or Bits[And](k, b(n-1))>0 do od; g(k):=true;
while g(t) do t:=t+1 od; k fi
end:
a:= n-> b(n)+b(n+1):
|
|
MATHEMATICA
|
g[_] = False ; t = 2;
b[n_] := b[n] = Module[{k}, If[n < 2, n,
For[k = t, g[k] || BitAnd[k, b[n-2]] > 0 ||
BitAnd[k, b[n-1]] > 0, k++]; g[k] = True;
While[g[t], t = t+1]; k]];
a[n_] := b[n] + b[n+1];
|
|
PROG
|
(Python)
from itertools import count, islice
def A353715_gen(): # generator of terms
s, a, b, c, ab = {0, 1}, 0, 1, 2, 1
yield 1
while True:
for n in count(c):
if not (n & ab or n in s):
yield b+n
a, b = b, n
ab = a|b
s.add(n)
while c in s:
c += 1
break
|
|
CROSSREFS
|
|
|
KEYWORD
|
base,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|