A091512 - OEIS

A091512

a(n) is the largest integer m such that 2^m divides (2*n)^n, i.e., the exponent of 2 in (2*n)^n.

1, 4, 3, 12, 5, 12, 7, 32, 9, 20, 11, 36, 13, 28, 15, 80, 17, 36, 19, 60, 21, 44, 23, 96, 25, 52, 27, 84, 29, 60, 31, 192, 33, 68, 35, 108, 37, 76, 39, 160, 41, 84, 43, 132, 45, 92, 47, 240, 49, 100, 51, 156, 53, 108, 55, 224, 57, 116, 59, 180, 61, 124, 63

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OFFSET

1,2

COMMENTS

n times one more than the trailing 0's in the binary representation of n. - Ralf Stephan, Aug 22 2013

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for sequences related to binary expansion of n.

FORMULA

a(n) = A007814(A000312(n)) = n*A001511(n) = A069895(n)/2.

G.f.: Sum_{k>=0} 2^k*x^2^k/(1-x^2^k)^2.

Recurrence: a(0) = 0, a(2*n) = 2*a(n) + 2*n, a(2*n+1) = 2*n+1.

Dirichlet g.f.: zeta(s-1)*2^s/(2^s-2). - Ralf Stephan, Jun 17 2007

Mobius transform of A162728, where x/(1-x)^2 = Sum_{n>=1} A162728(n)*x^n/(1+x^n). - Paul D. Hanna, Jul 12 2009

a(n) = A162728(2*n)/phi(2*n), where x/(1-x)^2 = Sum_{n>=1} A162728(n)*x^n/(1+x^n). - Paul D. Hanna, Jul 12 2009

a((2*n-1)*2^p) = (2*n-1)*(p+1)*2^p, p >= 0. Observe that a(2^p) = A001787(p+1). - Johannes W. Meijer, Feb 08 2013

Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Oct 22 2022

a(n) = Sum_{d divides n} d*A048298(n/d); that is, a(n) is the Dirichlet product of A048298(n) and A000027(n). - Peter Bala, Jan 02 2024

MAPLE

nmax:=63: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (2*n-1)*(p+1)*2^p od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 08 2013

# second Maple program:

a:= n-> n*padic[ordp](2*n, 2):

seq(a(n), n=1..63); # Alois P. Heinz, Jan 02 2024

MATHEMATICA

Table[ Part[ Flatten[ FactorInteger[(2 n)^n]], 2], {n, 1, 124}]

Table[IntegerExponent[(2n)^n, 2], {n, 70}] (* Harvey P. Dale, Sep 11 2015 *)

PROG

(PARI) a(n)=n*(valuation(n, 2)+1)

(PARI) a(n)=if(n<1, 0, if(n%2==0, 2*a(n/2)+n, n))

(Magma) [n*(Valuation(n, 2)+1): n in [1..80]]; // Vincenzo Librandi, May 16 2013

(Python)

def A091512(n): return n*(n&-n).bit_length() # Chai Wah Wu, Jul 11 2022