A082392 -id:A082392 - cp's OEIS frontend

A129629 Nonzero bisection of Moebius transform of A082392.

1, 1, 3, 7, 14, 31, 63, 123, 255, 511, 1015, 2047, 4092, 8176, 16383, 32767, 65503, 131061, 262143, 524223, 1048575, 2097151, 4194162, 8388607, 16777208, 33554175, 67108863, 134217693, 268434943, 536870911, 1073741823

Offset: 1

Ralf Stephan, May 31 2007

nonn

Possibly identical to A011947.

A129629_upto(N=100)={ my( d=2*N+3, b=Vec( sum( k=0, exponent(d), (x^2^k)/(1-2*x^2^(k+1)),O(x^d)))); vector( N,i, sumdiv(i*2-1,k, moebius((i*2-1)/k)*b[k])) } \\  M. F. Hasler, May 03 2008, updated May 27 2025

a(n+1) = 2^n-A383182(n). - M. F. Hasler, May 03 2008; sequence number corrected by M. F. Hasler, May 27 2025

A220466 a((2n-1)2^p) = 4^p(n-1) + 2^(p-1)(1+2^p), p >= 0 and n >= 1.

Original entry on oeis.org

1, 3, 2, 10, 3, 7, 4, 36, 5, 11, 6, 26, 7, 15, 8, 136, 9, 19, 10, 42, 11, 23, 12, 100, 13, 27, 14, 58, 15, 31, 16, 528, 17, 35, 18, 74, 19, 39, 20, 164, 21, 43, 22, 90, 23, 47, 24, 392, 25, 51, 26, 106, 27, 55, 28, 228, 29, 59, 30, 122, 31, 63, 32, 2080, 33, 67, 34, 138, 35

Offset: 1

Johannes W. Meijer, Dec 24 2012

The a(n) appeared in the analysis of A220002, a sequence related to the Catalan numbers.
The first Maple program makes use of a program by Peter Luschny for the calculation of the a(n) values. The second Maple program shows that this sequence has a beautiful internal structure, see the first formula, while the third Maple program makes optimal use of this internal structure for the fast calculation of a(n) values for large n.
The cross references lead to sequences that have the same internal structure as this sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ralf Stephan, Some divide-and-conquer sequences with simple ordinary generating functions, The OEIS, Jan 01 2004.

Cf. A000027 (the natural numbers), A000120 (1's-counting sequence), A000265 (remove 2's from n), A001316 (Gould's sequence), A001511 (the ruler function), A003484 (Hurwitz-Radon numbers), A003602 (a fractal sequence), A006519 (highest power of 2 dividing n), A007814 (binary carry sequence), A010060 (Thue-Morse sequence), A014577 (dragon curve), A014707 (dragon curve), A025480 (nim-values), A026741, A035263 (first Feigenbaum symbolic sequence), A037227, A038712, A048460, A048896, A051176, A053381 (smooth nowhere-zero vector fields), A055975 (Gray code related), A059134, A060789, A060819, A065916, A082392, A085296, A086799, A088837, A089265, A090739, A091512, A091519, A096268, A100892, A103391, A105321 (a fractal sequence), A109168 (a continued fraction), A117973, A129760, A151930, A153733, A160467, A162728, A181988, A182241, A191488 (a companion to Gould's sequence), A193365, A220466 (this sequence).

-- Following Ralf Stephan's recurrence:
import Data.List (transpose)
a220466 n = a006519_list !! (n-1)
a220466_list = 1 : concat
   (transpose [zipWith (-) (map (* 4) a220466_list) a006519_list, [2..]])
-- Reinhard Zumkeller, Aug 31 2014

# First Maple program
a := n -> 2^padic[ordp](n, 2)*(n+1)/2 : seq(a(n), n=1..69); # Peter Luschny, Dec 24 2012
# Second Maple program
nmax:=69: 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) := 4^p*(n-1)  + 2^(p-1)*(1+2^p) od: od: seq(a(n), n=1..nmax);
# Third Maple program
nmax:=69: for p from 0 to ceil(simplify(log[2](nmax))) do n:=2^p: n1:=1: while n <= nmax do a(n) := 4^p*(n1-1)+2^(p-1)*(1+2^p): n:=n+2^(p+1): n1:= n1+1: od: od:  seq(a(n), n=1..nmax);

A220466 = Module[{n, p}, p = IntegerExponent[#, 2]; n = (#/2^p + 1)/2; 4^p*(n - 1) + 2^(p - 1)*(1 + 2^p)] &; Array[A220466, 50] (* JungHwan Min, Aug 22 2016 *)

a(n)=if(n%2,n\2+1,4*a(n/2)-2^valuation(n/2,2)) \\ Ralf Stephan, Dec 17 2013

a((2*n-1)*2^p) = 4^p*(n-1) + 2^(p-1)*(1+2^p), p >= 0 and n >= 1. Observe that a(2^p) = A007582(p).
a(n) = ((n+1)/2)*(A060818(n)/A060818(n-1))
a(n) = (-1/64)*(q(n+1)/q(n))/(2*n+1) with q(n) = (-1)^(n+1)*2^(4*n-5)*(2*n)!*A060818(n-1) or q(n) = (1/8)*A220002(n-1)*1/(A098597(2*n-1)/A046161(2*n))*1/(A008991(n-1)/A008992(n-1))
Recurrence: a(2n) = 4a(n) - 2^A007814(n), a(2n+1) = n+1. - Ralf Stephan, Dec 17 2013

A337821 For n >= 0, a(4n+1) = 0, a(4n+3) = a(2n+1) + 1, a(2n+2) = a(n+1).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 0, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 1, 2, 0, 3, 4, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 0, 2, 3, 1, 0, 0, 1, 0, 0, 1, 2, 2, 0, 0, 1, 3, 0, 4, 5, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 0, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 1, 2, 0, 3, 4, 1

Offset: 1

Peter Munn, Sep 23 2020

This sequence is the ruler sequence A007814 interleaved with this sequence; specifically, the odd bisection is A007814, the even bisection is the sequence itself.
The 3-adic valuation of the Doudna sequence (A005940).
The 2-adic valuation of Kimberling's paraphrase (A003602) of the binary number system. [Edited Peter Munn, Aug 13 2025.]

			Start of table showing the interleaving with ruler sequence, A007814:
   n  a(n)  A007814    a(n/2)
            ((n+1)/2)
   1   0       0
   2   0                 0
   3   1       1
   4   0                 0
   5   0       0
   6   1                 1
   7   2       2
   8   0                 0
   9   0       0
  10   0                 0
  11   1       1
  12   1                 1
  13   0       0
  14   2                 2
  15   3       3
  16   0                 0
  17   0       0
  18   0                 0
  19   1       1
  20   0                 0
  21   0       0
  22   1                 1
  23   2       2
  24   1                 1

Odd bisection: A007814.
A000265, A003602, A005940, A007949 are used in a formula defining this sequence.
Positions of zeros: A091072.
Sequences with similar interleaving: A089309, A014577, A025480, A034947, A038189, A082392, A099545, A181363, A274139.

a[n_] := IntegerExponent[(n/2^IntegerExponent[n, 2] + 1)/2, 2]; Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)

a(n) = valuation(n>>valuation(n,2)+1, 2) - 1; \\ Kevin Ryde, Apr 06 2024

a(2*n) = a(n).
a(2*n+1) = A007814(n+1).
a(n) = A007949(A005940(n)).
a(n) = A007814(A003602(n)) = A007814((A000265(n)+1) / 2) = A089309(n) - 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Sep 13 2024

A274139 a(n) = 2^A000265(n) = 2^numerator(n/2^n), a sequence related to Oresme numbers.

Original entry on oeis.org

2, 2, 8, 2, 32, 8, 128, 2, 512, 32, 2048, 8, 8192, 128, 32768, 2, 131072, 512, 524288, 32, 2097152, 2048, 8388608, 8, 33554432, 8192, 134217728, 128, 536870912, 32768, 2147483648, 2, 8589934592, 131072, 34359738368, 512, 137438953472, 524288, 549755813888, 32

Offset: 1

Jean-François Alcover and Paul Curtz, Jun 11 2016

nonn

Differences: 0, 6, -6, 30, -24, 120, -126, 510, -480, 2016, -2040, 8184, -8064, 32640, -32766, 131070, -130560, ...

GCD of differences is 6.

A. F. Horadam, Oresme numbers, Fib. Quart., 12 (1974), 267-271.

Cf. A000265, A006519, A007814, A082392.

a[n_] := 2^(n/2^IntegerExponent[n, 2]);
Array[a, 40]

a(n) = 2^(n/2^valuation(n,2)); \\ Michel Marcus, Jun 12 2016

a(n) = 2^denominator(2^n/n).
a(n) = 2^(n/2^valuation(n,2)) = 2^A007814(n).
a(n) = 2*A082392(n)^2 (conjecture).

cp's OEIS Frontend

A129629 Nonzero bisection of Moebius transform of A082392.

Views

Author

Keywords

Comments

Programs

PARI

Formula

A220466 a((2*n-1)*2^p) = 4^p*(n-1) + 2^(p-1)*(1+2^p), p >= 0 and n >= 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Formula

A337821 For n >= 0, a(4n+1) = 0, a(4n+3) = a(2n+1) + 1, a(2n+2) = a(n+1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Formula

A274139 a(n) = 2^A000265(n) = 2^numerator(n/2^n), a sequence related to Oresme numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A220466 a((2n-1)2^p) = 4^p(n-1) + 2^(p-1)(1+2^p), p >= 0 and n >= 1.