A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.
1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1
Offset: 1
A050488 a(n) = 3*(2^n-1) - 2*n.
0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707
Offset: 0
Comments
Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008
Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012
The partial sums of A033484. - J. M. Bergot, Oct 03 2012
Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012
Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013
Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015
Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
- Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
- Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
- Index entries for linear recurrences with constant coefficients, signature (4,-5,2).
Crossrefs
Programs
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GAP
List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018
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Haskell
a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list) -- Reinhard Zumkeller, Jul 24 2015
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Magma
[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018
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Maple
seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018
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Mathematica
Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)
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PARI
a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015
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Python
for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018
Formula
Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n+1) = A156920(n+1,1).
a(n+1) = A156919(n+1,1)/2^n.
a(n+1) = A142963(n+2,1)/2.
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z)).
(End)
a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012
E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023
Comments
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