OFFSET
1,3
COMMENTS
If bit i is 1, then bits i+-2 must be 0. All terms satisfy A048725(n) = 5*n.
It appears that n is in the sequence if and only if C(5n,n) is odd (cf. A003714). - Benoit Cloitre, Mar 09 2003
Yes, as remarked in A048715, "This is easily proved using the well-known result that the multiplicity with which a prime p divides C(n+m,n) is the number of carries when adding n+m in base p." - Jason Kimberley, Dec 21 2011
A116361(a(n)) <= 2. - Reinhard Zumkeller, Feb 04 2006
LINKS
MATHEMATICA
Reap[Do[If[OddQ[Binomial[5n, n]], Sow[n]], {n, 0, 400}]][[2, 1]]
(* Second program: *)
filterQ[n_] := With[{bb = IntegerDigits[n, 2]}, MatchQ[bb, {0}|{1}|{1, 1}|{___, 0, _, 1, ___}|{___ 1, _, 0, ___}] && !MatchQ[bb, {___, 1, _, 1, ___}]];
Select[Range[0, 201], filterQ] (* Jean-François Alcover, Dec 31 2020 *)
PROG
(PARI) is(n)=!bitand(n, n>>2) \\ Charles R Greathouse IV, Oct 03 2016
(PARI) list(lim)=my(v=List(), n, t); while(n<=lim, t=bitand(n, n>>2); if(t, n+=1<<valuation(t, 2), listput(v, n); n++)); Vec(v) \\ Charles R Greathouse IV, Oct 22 2021
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Antti Karttunen, Mar 30 1999
STATUS
approved