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 A070940 Number of digits that must be counted from left to right to reach the last 1 in the binary representation of n. 9
 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 4, 2, 4, 3, 4, 1, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 1, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 2, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7, 6, 7, 3, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Length of longest carry sequence when adding numbers <= n to n in binary representation: a(n)=T(n, A080079(n)) and T(n,k) <= a(n) for 1 <= k <= n, with T defined as in A080080. - Reinhard Zumkeller, Jan 26 2003 a(n+1) is the number of distinct values of gcd(2^n, binomial(n,j)) (or, equivalently, A007814(binomial(n,j))) arising for j=0,...,n-1. Proof using Kummer's Theorem given by Marc Schwartz. - Labos Elemer, Apr 23 2003 E.g., n=10: 10th row of Pascal's triangle = {1,10,45,120,210,252,210,120,45,10,1}, largest powers of 2 dividing binomial coefficients is: {1,2,1,8,2,4,2,8,1,2,1}; including distinct powers of 2, thus a(10)=4. If m=-1+2^k, i.e., m=0,1,3,7,15,31,.. then a(m)=1. This corresponds to "odd rows" of Pascal triangle. - Labos Elemer Smallest x > 0 for which a(x)=n equals 2^n. - Labos Elemer a(n) <= A070939(n), a(n) = A070939(n) iff n is odd, where A070939(n) = floor(log_2(n)) + 1. - Reinhard Zumkeller, Jan 26 2003 Can be regarded as a table with row n having 2^(n-1) columns, with odd columns repeating the previous row, and even columns containing the row number. - Franklin T. Adams-Watters, Nov 08 2011 It appears that a(n) is the greatest number in a periodicity equivalence class defined at A269570; e.g., the 5 classes for n = 35 are (1, 1, 2, 2, 6), (1, 1, 1, 1, 4, 2, 2), (3), (1, 3), (1, 2); in these the greatest number is 6, so that a(35) = 6. - Clark Kimberling, Mar 01 2016 Number of binary digits of the largest odd factor of n. - Andres Cicuttin, May 18 2017 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA a(n) = floor(log_2(n)) - A007814(n) = A070939(n) - A007814(n). a(n) = f(n, 1), f(n, k) = if n=1 then k else f(floor(n/2), k+(if k>1 then 1 else n mod 2)). - Reinhard Zumkeller, Feb 01 2003 G.f.: sum(k>=0, t/(1-t^2) * [1 + sum(l>=1, t^2^l)], t=x^2^k). - Ralf Stephan, Mar 15 2004 a(n) = A070939(A000265(n)). Andres Cicuttin, May 19 2017 EXAMPLE a(10)=3 is the number of digits that must be counted from left to right to reach the last 1 in 1010, the binary representation of 10. The table starts: 1 1 2 1 3 2 3 1 4 3 4 2 4 3 4 MAPLE A070940 := n -> if n mod 2 = 0 then A070939(n)-A001511(n/2) else A070939(n); fi; MATHEMATICA Table[Length[Union[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}]]], {n, 0, 256}] f[n_] := Position[ IntegerDigits[n, 2], 1][[ -1, 1]]; Table[ f[n], {n, 105}] (* Robert G. Wilson v, Dec 01 2004 *) (* By exploiting the "positional" regularity of the sequence *) b = {}; a = {1, 1}; Do[a = Riffle[a, j];   b = AppendTo[b, a[[1 ;; Floor[Length[a]/2]]]] // Flatten, {j, 1, 10}]; Print[b[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *) (* By following the alternative definition "Number of binary digits of the largest integer odd factor of n" *) c = Table[IntegerDigits[n/(2^IntegerExponent[n, 2]), 2] // Length , {n,     2^10 - 1}]; Print[c[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *) PROG (Haskell) a070940 = maximum . a080080_row  -- Reinhard Zumkeller, Apr 22 2013 CROSSREFS Cf. A070939, A001511. Differs from A002487 around 11th term. Cf. A000005, A007318, A000079, A082907, A082908. Bisections give A070941 and this sequence (again). Cf. A002064 (row sums), A199570. Sequence in context: A284266 A038568 A071912 * A020651 A294442 A281392 Adjacent sequences:  A070937 A070938 A070939 * A070941 A070942 A070943 KEYWORD nonn,nice,easy,tabf AUTHOR N. J. A. Sloane, May 18 2002 EXTENSIONS Entry revised by Ralf Stephan, Nov 29 2004 STATUS approved

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Last modified March 19 06:59 EDT 2018. Contains 300836 sequences. (Running on oeis4.)