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 A048881 a(n) = A000120(n+1) - 1 = wt(n+1) - 1. 29
 0, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,7 COMMENTS Highest power of 2 dividing n-th Catalan number (A000108). a(n) = 0 iff n = 2^k - 1, k=0,1,... Appears to be number of binary left-rotations (iterations of A006257) to reach fixed point of form 2^k-1. Right-rotation analog is A063250. This would imply that for n >= 0, a(n)=f(n), recursively defined to be 0 for n=0, otherwise as f( ( (1-n)(1-p)(1-s) - (1-n-p-s) ) / 2) + p (s+1) / 2, where p = n mod 2 and s = - signum(n) (f(n<0) is A000120(-n)). - Marc LeBrun, Jul 11 2001 Let f(0) = 01, f(1) = 12, f(2) = 23, f(3) = 34, f(4) = 45, etc. Sequence gives concatenation of 0, f(0), f(f(0)), f(f(f(0))), ... Also f(f(...f(0)...)) converges to A000120. - Philippe Deléham, Aug 14 2003 Highest power of 2 dividing binomial(n,floor(n/2)). - Benoit Cloitre, Oct 20 2003 2^a(n) are numerators in the Maclaurin series for (sin x)^2. - Jacob A. Siehler, Nov 11 2009 a(n) = A000120(A129760(n+1)). - Reinhard Zumkeller, Jun 30 2010 a(n+k) = A240857(n,k), 0<=k<=n; in particular: a(n) = A240857(n,0). - Reinhard Zumkeller, Apr 14 2014 LINKS R. Alter and K. K. Kubota, Prime and Prime Power Divisibility of Catalan Numbers, J. Comb. Th. A 15 (1973) pp. 243-256. E. Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, J. Num. Theory 117 (2006), 191-215. FORMULA Writing n as 2^m+k with -1<=k<2^m-1, then a(n) = A000120(k+1). - Henry Bottomley, Mar 28 2000 a(n) = k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n). a(2*n) = a(n-1)+1, a(2*n+1) = a(n). - Vladeta Jovovic, Oct 10 2002 G.f.: 1/(x-x^2) * {x^2/(1-x) - Sum_{k>=1} x^(2^k)/(1-x^(2^k))}. - Ralf Stephan, Apr 13 2002 C(n, k) is the number of occurrence of k in the n-th group of terms in this sequence read by rows: {0}; {0, 1}; {0, 1, 1, 2}; {0, 1, 1, 2, 1, 2, 2, 3}; {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 }; ... - Philippe Deléham, Jan 01 2004 EXAMPLE From Omar E. Pol, Mar 08 2011: (Start) Sequence can be written in the following form (irregular triangle): 0, 0,1, 0,1,1,2, 0,1,1,2,1,2,2,3, 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4, 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5, Row sums are A001787. (End) MAPLE A048881 := proc(n)     A000120(n+1)-1 ; end proc: seq(A048881(n), n=0..200) ; # R. J. Mathar, Mar 12 2018 MATHEMATICA a[n_] := IntegerExponent[ CatalanNumber[n], 2]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jun 21 2013 *) PROG (PARI) { a(n) = if( n<0, 0, n++; n /= 2^valuation(n, 2); subst( Pol( binary( n ) ), x, 1) - 1 ) } /* Michael Somos, Aug 23 2007 */ (PARI) {a(n) = if( n<0, 0, valuation( (2*n)! / n! / (n+1)!, 2 ) ) } /* Michael Somos, Aug 23 2007 */ (Haskell) a048881 n = a048881_list !! n a048881_list = c [0] where c (x:xs) = x : c (xs ++ [x, x+1]) -- Reinhard Zumkeller, Mar 07 2011 CROSSREFS Cf. A000120, A006257, A007318, A063250. First differences of A078903. Sequence in context: A061358 A025866 A259920 * A026931 A127506 A007968 Adjacent sequences:  A048878 A048879 A048880 * A048882 A048883 A048884 KEYWORD nonn,easy AUTHOR EXTENSIONS Entry revised by N. J. A. Sloane, Jun 07 2009 STATUS approved

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Last modified July 21 02:02 EDT 2019. Contains 325189 sequences. (Running on oeis4.)