A136480 - OEIS

%I #47 Mar 20 2023 05:30:48

%S 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,

%T 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,

%U 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

%N Number of trailing equal digits in binary representation of n.

%C a(even) = number of trailing binary zeros;

%C a(odd) = number of trailing binary ones.

%C 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

%H James Spahlinger, <a href="/A136480/b136480.txt">Table of n, a(n) for n = 0..10000</a>

%H Francis Laclé, <a href="https://hal.archives-ouvertes.fr/hal-03201180v2">2-adic parity explorations of the 3n+ 1 problem</a>, hal-03201180v2 [cs.DM], 2021.

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(n) = A050603(n-1) for n>0;

%F a(2*n + n mod 2) = a(n) + 1.

%F For n>0: a(n) = A007814(n + n mod 2).

%F Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=0..m} a(k) = 2. - _Amiram Eldar_, Sep 15 2022

%F a(n) = A007814(A002378(n)), n>0. - _R. J. Mathar_, Mar 20 2023

%p A136480 := proc(n)

%p     if n = 0 then

%p         1;

%p     else

%p         A007814(n*(n+1)) ;

%p     end if;

%p end proc:

%p seq( A136480(n),n=0..80) ; # _R. J. Mathar_, Mar 20 2023

%t Length[Last[Split[IntegerDigits[#,2]]]]&/@Range[0,140]  (* _Harvey P. Dale_, Mar 31 2011 *)

%o (PARI) a(n)=if (n, valuation(n+n%2,2), 1) \\ _Charles R Greathouse IV_, Oct 14 2013

%o (Haskell)

%o a136480 0 = 1

%o a136480 n = a007814 $ n + mod n 2  -- _Reinhard Zumkeller_, Jul 22 2014

%o (JavaScript)

%o for (n=1;n<120;n++) {

%o m=n*n+n;

%o c=0;

%o while (m%2==0) {m/=2;c++;}

%o document.write(c+", ");

%o }  // _Jon Perry_, Sep 12 2014

%o (Python)

%o def A136480(n): return (~(m:=n+(n&1))& m-1).bit_length() # _Chai Wah Wu_, Jul 08 2022

%Y Cf. A007814, A050603, A094267, A163575, A001511, A039963 (parity).

%K nonn,base,easy

%O 0,4

%A _Reinhard Zumkeller_, Dec 31 2007