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A136480 Number of trailing equal digits in binary representation of n. 20
1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 5, 5, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 6, 6, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 2, 2, 1, 1, 5, 5, 1, 1, 2, 2, 1, 1, 3, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
a(even) = number of trailing binary zeros;
a(odd) = number of trailing binary ones.
For n>0, power of 2 associated with n^2 + n, e.g. n=4 gives 20, so a(4)=2. - Jon Perry, Sep 12 2014
LINKS
Francis Laclé, 2-adic parity explorations of the 3n+ 1 problem, hal-03201180v2 [cs.DM], 2021.
FORMULA
a(n) = A050603(n-1) for n>0;
a(2*n + n mod 2) = a(n) + 1.
For n>0: a(n) = A007814(n + n mod 2).
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=0..m} a(k) = 2. - Amiram Eldar, Sep 15 2022
a(n) = A007814(A002378(n)), n>0. - R. J. Mathar, Mar 20 2023
MAPLE
A136480 := proc(n)
if n = 0 then
1;
else
A007814(n*(n+1)) ;
end if;
end proc:
seq( A136480(n), n=0..80) ; # R. J. Mathar, Mar 20 2023
MATHEMATICA
Length[Last[Split[IntegerDigits[#, 2]]]]&/@Range[0, 140] (* Harvey P. Dale, Mar 31 2011 *)
PROG
(PARI) a(n)=if (n, valuation(n+n%2, 2), 1) \\ Charles R Greathouse IV, Oct 14 2013
(Haskell)
a136480 0 = 1
a136480 n = a007814 $ n + mod n 2 -- Reinhard Zumkeller, Jul 22 2014
(JavaScript)
for (n=1; n<120; n++) {
m=n*n+n;
c=0;
while (m%2==0) {m/=2; c++; }
document.write(c+", ");
} // Jon Perry, Sep 12 2014
(Python)
def A136480(n): return (~(m:=n+(n&1))& m-1).bit_length() # Chai Wah Wu, Jul 08 2022
CROSSREFS
Sequence in context: A064894 A003638 A094267 * A050603 A286554 A352784
KEYWORD
nonn,base,easy
AUTHOR
Reinhard Zumkeller, Dec 31 2007
STATUS
approved

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Last modified March 18 22:56 EDT 2024. Contains 370952 sequences. (Running on oeis4.)