cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 12 results. Next

A020651 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 6, 5, 6, 4, 9, 5, 9, 3, 10, 7, 10, 4, 11, 7, 11, 2, 9, 7, 9, 5, 12, 7, 12, 3, 11, 8, 11, 5, 13, 8, 13, 1, 7, 6, 7, 5, 11, 6, 11, 4, 13, 9, 13, 5, 14, 9, 14, 3, 13, 10, 13, 7, 17, 10, 17, 4, 15, 11, 15, 7, 18, 11
Offset: 1

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Author

Keywords

Comments

If we insert an initial 1, this is the sequence of numerators in left-hand half of Kepler's tree of fractions. Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j). See A086592 for denominators. See A294442 for the Kepler tree itself.
Level n of the tree consists of 2^n nodes: 1/2; 1/3, 2/3; 1/4, 3/4, 2/5, 3/5; 1 /5, 4/5, 3/7, 4/7, 2/7, 5/7, 3/8, 5/8; ... Fibonacci numbers occur at the right edge this tree, i.e., a(A000225(n)) = A000045(n+1). The fractions are given in their reduced form, thus gcd(A020650(n), A020651(n)) = 1 and gcd(A020651(n), A086592(n)) = 1 for all n. - Antti Karttunen, May 26 2004
A generalization which includes the "rabbit tree" (A226080) and "all rationals tree" (A226130) follows. Suppose that a,b,c,d,e,f,g,h are complex numbers. Let S be the set of numbers defined by these rules: (1) 1 is in S; (2) if x is in S and cx+d is not 0, then U(x) = (ax+b)/(cx+d) is in S; (3) if x is in S and gx+h is not 0, then D(x) = (ex+f)/(gx+h) is in S. If an infinite path in the resulting tree has convergent nodes, then there is some node after which the path is "updown zigzag" ((UoD)o(UoD)o ...) or "downup zigzag" (DoU)o(DoU)o ...). If ag+ch is not 0, then the updown zigzag limit is invariant of x and equals [ae + cf - bg - dh + sqrt(X)]/(2(ag + ch)), where X = (ae + cf - bg - dh)^2 + 4(be + df + ag + ch). If ce + dg is not 0, then the downup zigzag limit is invariant of x and equals [ae + bg - cf - dh + sqrt(Y)]/(2(ce + dg)), where Y = (ae + bg - cf - dh)^2 + 4(af + bh)(ce + dg)) = X. Thus, for the tree A020651, the updown zigzag limit is -1 + sqrt(2) and the downup zigzag limit, sqrt(2). - Clark Kimberling, Nov 10 2013
From Yosu Yurramendi, Jul 13 2014: (Start)
If the terms (n > 0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
1,
1,2,
1,3,2,3,
1,4,3,4,2,5,3,5,
1,5,4,5,3,7,4,7,2, 7,5, 7,3, 8,5, 8,
1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column is an arithmetic sequence. The differences of the arithmetic sequences, except the first on the left, give the sequence A093873 (Numerators in Kepler's tree of harmonic fractions) (a(2^(m+1)-1-k) - a(2^m-1-k) = A093873(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
If the rows are written in a right-aligned fashion:
1,
1, 2,
1, 3,2, 3,
1, 4,3, 4,2, 5,3, 5,
1,5,4,5,3, 7,4, 7,2, 7,5, 7,3, 8,5, 8,
1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then each column k is a Fibonacci sequence. (End)
For m >= 0, a(2^m) = 1 and a(3*2^m) = 2. For n >= 0, a(A070875(n)) = 3 (for m >= 0, a(5*2^m) = 3 and a(7*2^m) = 3). - Yosu Yurramendi, Jun 02 2016

Examples

			1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, ...
		

Crossrefs

See A294442 and A093873/A093875 for two different versions of the Kepler tree.

Programs

  • Haskell
    import Data.List (transpose); import Data.Ratio (denominator)
    a020651_list = map denominator ks where
       ks = 1 : concat (transpose [map (+ 1) ks, map (recip . (+ 1)) ks])
    -- Reinhard Zumkeller, Feb 22 2014
    
  • Maple
    A020651 := n -> `if`((n < 2),n,`if`(type(n,even), A020651(n/2), A020650(n-1)));
  • Mathematica
    f[1] = 1; f[n_?EvenQ] := f[n] = f[n/2]+1; f[n_?OddQ] := f[n] = 1/(f[(n-1)/2]+1); a[n_] := Denominator[f[n]]; Table[a[n], {n, 1, 94}] (* Jean-François Alcover, Nov 22 2011 *)
  • R
    N <- 25 # arbitrary
    a <- c(1,1,2)
    for(n in 1:N){
      a[4*n]   <- a[2*n]
      a[4*n+1] <- a[2*n] + a[2*n+1]
      a[4*n+2] <-          a[2*n+1]
      a[4*n+3] <- a[2*n] + a[2*n+1]
    }
    a
    # Yosu Yurramendi, Jul 13 2014

Formula

a(1) = 1, a(2n) = a(n), a(2n+1) = A020650(2n). - Antti Karttunen, May 26 2004
a(2n) = A020650(2n+1). - Yosu Yurramendi, Jul 17 2014
a(2^m + k) = A093873(2^(m+1) + k) = A093873(2^(m+1) + 2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, May 18 2016
a(2^m + 2^r + k) = A093873(2^r + k)*(m-(r-1)) + A093873(k), m >= 0, r <= m-1, 0 <= k < 2^r. For k=0 A093873(0) = 0 is needed. - Yosu Yurramendi, Jul 30 2016
a((2n+1)*2^m) = A086592(n), m >= 0, n > 0. For n = 0 A086592(0) = 1 is needed. - Yosu Yurramendi, Feb 14 2017
a(4n+2) = a(4n+1) - a(4n) = a(2n+1) = a(4n+1) - a(n), n > 0. - Yosu Yurramendi, May 08 2018
a(1) = 1, a(n+1) = 2*floor(1/a(n))+1-1/a(n). - Jan Malý, Jul 30 2019
a(n) = A002487(A231551(n)), n > 0. - Yosu Yurramendi, Jul 15 2021

Extensions

Entry revised by N. J. A. Sloane, May 24 2004

A093873 Numerators in Kepler's tree of harmonic fractions.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 1, 4, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 6, 5, 6, 4, 9, 5, 9, 3, 10, 7, 10, 4, 11, 7, 11, 2, 9, 7, 9, 5, 12, 7, 12, 3, 11, 8, 11, 5, 13, 8, 13, 1, 6
Offset: 1

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Author

Keywords

Comments

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).

Examples

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8
		

Crossrefs

The denominators are in A093875. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.
See A294442 for another version of Kepler's tree of fractions.

Programs

  • Haskell
    {-# LANGUAGE ViewPatterns #-}
    import Data.Ratio((%), numerator, denominator)
    rat :: Rational -> (Integer,Integer)
    rat r = (numerator r, denominator r)
    data Harmony = Harmony Harmony Rational Harmony
    rows :: Harmony -> [[Rational]]
    rows (Harmony hL r hR) = [r] : zipWith (++) (rows hL) (rows hR)
    kepler :: Rational -> Harmony
    kepler r = Harmony (kepler (i%(i+j))) r (kepler (j%(i+j)))
               where (rat -> (i,j)) = r
    -- Full tree of Kepler's harmonic fractions:
    k = rows $ kepler 1 :: [[Rational]] -- as list of lists
    h = concat k :: [Rational] -- flattened
    a093873 n = numerator $ h !! (n - 1)
    a093875 n = denominator $ h !! (n - 1)
    a011782 n = numerator $ (map sum k) !! n -- denominator == 1
    -- length (k !! n) == 2^n
    -- numerator $ (map last k) !! n == fibonacci (n + 1)
    -- denominator $ (map last k) !! n == fibonacci (n + 2)
    -- numerator $ (map maximum k) !! n == n
    -- denominator $ (map maximum k) !! n == n + 1
    -- eop.
    -- Reinhard Zumkeller, Oct 17 2010
  • Maple
    M:= 8: # to get a(1) .. a(2^M-1)
    gen[1]:= [1];
    for n from 2 to M do
      gen[n]:= map(t -> (numer(t)/(numer(t)+denom(t)),
          denom(t)/(numer(t)+denom(t))), gen[n-1]);
    od:
    seq(op(map(numer,gen[i])),i=1..M): # Robert Israel, Jan 11 2016
  • Mathematica
    num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093873 = Table[num[n], {n, 1, 97}] (* Jean-François Alcover, Dec 16 2011 *)

Formula

a(n) = a([n/2])*(1 - n mod 2) + A093875([n/2])*(n mod 2).
a(A029744(n-1)) = 1; a(A070875(n-1)) = 2; a(A123760(n-1)) = 3. - Reinhard Zumkeller, Oct 13 2006
A011782(k) = SUM(a(n)/A093875(n): 2^k<=n<2^(k+1)), k>=0. [Reinhard Zumkeller, Oct 17 2010]
a(1) = 1. For all n>0 a(2n) = a(n), a(2n+1) = A093875(n). - Yosu Yurramendi, Jan 09 2016
a(4n+3) = a(4n+1), a(4n+2) = a(4n+1) - a(4n), a(4n+1) = A071585(n). - Yosu Yurramendi, Jan 11 2016
G.f. G(x) satisfies G(x) = x + (1+x) G(x^2) + Sum_{k>=2} x (1+x^(2^(k-1))) G(x^(2^k)). - Robert Israel, Jan 11 2016
a(2^(m+1)+k) = a(2^(m+1)+2^m+k) = A020651(2^m+k), m>=0, 0<=k<2^m. - Yosu Yurramendi, May 18 2016
a(k) = A020651(2^(m+1)+k) - A020651(2^m+k), m>0, 0Yosu Yurramendi, Jun 01 2016
a(2^(m+1)+k) - a(2^m+k) = a(k) , m >=0, 0 <= k < 2^m. For k=0 a(0)=0 is needed. - Yosu Yurramendi, Jul 22 2016
a(2^(m+2)-1-k) = a(2^(m+1)-1-k) + a(2^m-1-k), m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Jul 22 2016
a(2^m-1-(2^r -1)) = A000045(m-r), m >= 1, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
a(2^m+2^r) = m-r, , m >= 1, 0 <= r <= m-1 ; a(2^m+2^r+2^(r-1)) = m-(r-1), m >= 2, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
A093875(2n) - a(2n) = A093875(n), n > 0; A093875(2n+1) - a(2n+1) = a(n), n > 0. - Yosu Yurramendi, Jul 23 2016

A164090 a(n) = 2*a(n-2) for n > 2; a(1) = 2, a(2) = 3.

Original entry on oeis.org

2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152, 3145728
Offset: 1

Views

Author

Klaus Brockhaus, Aug 09 2009

Keywords

Comments

Interleaving of A000079 without initial 1 and A007283.
Agrees from a(2) onward with A145751 for all terms listed there (up to 65536). Apparently equal to 2, 3 followed by A090989. Equals 2 followed by A163978.
Binomial transform is A000129 without first two terms, second binomial transform is A020727, third binomial transform is A164033, fourth binomial transform is A164034, fifth binomial transform is A164035.
Number of achiral necklaces or bracelets with n beads using up to 2 colors. For n=5, the eight achiral necklaces or bracelets are AAAAA, AAAAB, AAABB, AABAB, AABBB, ABABB, ABBBB, and BBBBB. - Robert A. Russell, Sep 22 2018

Crossrefs

Programs

  • Magma
    [ n le 2 select n+1 else 2*Self(n-2): n in [1..42] ];
    
  • Mathematica
    a[n_] := If[EvenQ[n], 3*2^(n/2 - 1), 2^((n + 1)/2)]; Array[a, 42] (* Jean-François Alcover, Oct 12 2017 *)
    RecurrenceTable[{a[1]==2,a[2]==3,a[n]==2a[n-2]},a,{n,50}] (* or *) LinearRecurrence[{0,2},{2,3},50] (* Harvey P. Dale, Mar 01 2018 *)
  • PARI
    a(n) = if(n%2,2,3) * 2^((n-1)\2); \\ Andrew Howroyd, Oct 07 2017

Formula

a(n) = A029744(n+1).
a(n) = A052955(n-1) + 1.
a(n) = A027383(n-2) + 2 for n > 1.
a(n) = A060482(n-1) + 3 for n > 3.
a(n) = A070875(n) - A070875(n-1).
a(n) = (7 - (-1)^n)*2^((1/4)*(2*n - 1 + (-1)^n))/4.
G.f.: x*(2+3*x)/(1-2*x^2).
a(n) = A063759(n-1), n>1. - R. J. Mathar, Aug 17 2009
Sum_{n>=1} 1/a(n) = 5/3. - Amiram Eldar, Mar 28 2022

A228405 Pellian Array, A(n, k) with numbers m such that 2*m^2 +- 2^k is a square, and their corresponding square roots, read downward by diagonals.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 2, 2, 3, 5, 0, 2, 4, 7, 12, 4, 4, 6, 10, 17, 29, 0, 4, 8, 14, 24, 41, 70, 8, 8, 12, 20, 34, 58, 99, 169, 0, 8, 16, 28, 48, 82, 140, 239, 408, 16, 16, 24, 40, 68, 116, 198, 338, 577, 985, 0, 16, 32, 56, 96, 164, 280, 478, 816, 1393, 2378
Offset: 0

Views

Author

Richard R. Forberg, Aug 21 2013

Keywords

Comments

The left column, A(n,0), is A000129(n), Pell Numbers.
The top row, A(0,k), is A077957(k) plus an initial 0, which is the inverse binomial transform of A000129.
These may be considered initializing values, or results, depending the perspective taken, since there are several ways to generate the array. See Formula section for details.
The columns of the array hold all values, in sequential order, of numbers m such that 2m^2 + 2^k or 2m^2 - 2^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for k+1.
For example A(n,0) are numbers such that 2m^2 +- 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 2m^2 +- 2 are squares, with those square roots in A(n,2), etc.
A(n, k)/A(n,k-2) = 2; A(n,k)/A(n,k-1) converges to sqrt(2) for large n.
A(n,k)/A(n-1,k) converges to 1 + sqrt(2) for large n.
The other columns of this array hold current OEIS sequences as follows:
A(n,1) = A001333(n); A(n,2) = A163271(n); A(n,3) = A002203(n);
Bisections of these column-oriented sequences also appear in the OEIS, corresponding to the even and odd rows of the array, which in turn correspond to the two different recursive square root equations in the formula section below.
Farey fraction denominators interleave columns 0 and 1, and the corresponding numerators interleave columns 1 and 2, for approximating sqrt(2). See A002965 and A119016, respectively.
The other rows of this array hold current OEIS sequences as follows:
A(1,k) = A016116(k); A(2,k) = A029744(k) less the initial 1;
A(3,k) = A070875(k); A(4,k) = A091523(k) less the initial 8.
The Pell Numbers (A000219) are the only initializing set of numbers where the two recursive square root equations (see below) produce exclusively integer values, for all iterations of k. For any other initial values only even iterations (at k = 2, 4, ...) produce integers.
The numbers in this array, especially the first three columns, are also integer square roots of these expressions: floor(m^2/2), floor(m^2/2 + 1), floor (m^2/2 - 1). See A227972 for specific arrangements and relationships. Also: ceiling(m^2/2), ceiling(m^2/2 + 1), ceiling (m^2/2 -1), m^2+1, m^2-1, m^2*(m^2-1)/2, m^2*(m^2-1)/2, in various different arrangements. Many of these involve: A000129(2n)/2 = A001109(n).
A001109 also appears when multiplying adjacent columns: A(n,k) * A(n,k+1) = (k+1) * A001109(n), for all k.

Examples

			With row # as n. and column # as k, and n, k =>0, the array begins:
0,     1,     0,     2,     0,     4,     0,     8, ...
1,     1,     2,     2,     4,     4,     8,     8, ...
2,     3,     4,     6,     8,    12,    16,    24, ...
5,     7,    10,    14,    20,    28,    40,    56, ...
12,   17,    24,    34,    48,    68,    96,   136, ...
29,   41,    58,    82,   116,   164,   232,   328, ...
70,   99,   140,   198,   280,   396,   560,   792, ...
169,  239,  338,   478,   676,   956,  1352,  1912, ...
408,  577,  816,  1154,  1632,  2308,  3264,  4616, ...
		

Crossrefs

Formula

If using the left column and top row to initialize: A(n,k) = A(n,k-1) + A(n-1,k-1).
If using only the top row to initialize, then each column for k = i is the binomial transform of A(0,k) restricted to k=> i as input to the transform with an appropriate down shift of index. The inverse binomial transform with a similar condition can produce each row from A000129.
If using only the first two rows to initialize then the Pell equation produces each column, as: A(n,k) = 2*A(n-1, k) + A(n-2, k).
If using only the left column (A000219(n) = Pell Numbers) to initialize then the following two equations will produce each row:
A(n,k) = sqrt(2*A(n,k-1) + (-2)^(k-1)) for even rows
A(n,k) = sqrt(2*A(n,k-1) - (-2)^(k-1)) for odd rows.
Interestingly, any portion of the array can also be filled "backwards" given the top row and any column k, using only: A(n,k-1) = A(n-1,k-1) + A(n-1, k), or if given any column and its column number by rearranging the sqrt recursions above.

A070876 Binary expansion is 1xx100...0 where xx = 00 or 11.

Original entry on oeis.org

9, 15, 18, 30, 36, 60, 72, 120, 144, 240, 288, 480, 576, 960, 1152, 1920, 2304, 3840, 4608, 7680, 9216, 15360, 18432, 30720, 36864, 61440, 73728, 122880, 147456, 245760, 294912, 491520, 589824, 983040, 1179648, 1966080, 2359296, 3932160
Offset: 0

Views

Author

N. J. A. Sloane, May 19 2002

Keywords

Crossrefs

Cf. A070875.

Programs

  • Magma
    [n le 2 select 6*n+3 else 2*Self(n-2): n in [1..38]]; // Bruno Berselli, Mar 02 2011
    
  • Mathematica
    a = {}; Do[a = Append[a, FromDigits[ Join[{1, 0, 0, 1}, Table[0, {n}]], 2]]; a = Append[a, FromDigits[ Join[{1, 1, 1, 1}, Table[0, {n}]], 2]], {n, 0, 20}]; a
    CoefficientList[Series[3 (3 + 5 x) / (1 - 2 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 19 2013 *)
    LinearRecurrence[{0,2},{9,15},40] (* Harvey P. Dale, Aug 06 2021 *)
  • PARI
    my(x='x+O('x^99)); Vec(3*(3+5*x)/(1-2*x^2)) \\ Altug Alkan, Sep 20 2018

Formula

From Bruno Berselli, Mar 02 2011: (Start)
G.f.: 3*(3+5*x)/(1-2*x^2).
a(n) = 3*(4-(-1)^n)*2^((2*n+(-1)^n-1)/4). Therefore: a(n) = 9*2^(n/2) for n even, otherwise a(n) = 15*2^((n-1)/2).
a(n) = 2*a(n-2) for n>1. (End)
Sum_{n>=0} 1/a(n) = 16/45. - Amiram Eldar, Mar 28 2022

Extensions

More terms from Robert G. Wilson v, May 20 2002

A123760 Numbers whose binary expansion is 1xy100...0 where x,y = 0 or 1.

Original entry on oeis.org

9, 11, 13, 15, 18, 22, 26, 30, 36, 44, 52, 60, 72, 88, 104, 120, 144, 176, 208, 240, 288, 352, 416, 480, 576, 704, 832, 960, 1152, 1408, 1664, 1920, 2304, 2816, 3328, 3840, 4608, 5632, 6656, 7680, 9216, 11264, 13312, 15360, 18432, 22528, 26624
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 13 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Union[Join @@ (Table[FromDigits[Join[#, Table[0, {n}]], 2], {n, 0, 11}] & /@ {{1, 0, 0, 1}, {1, 0, 1, 1}, {1, 1, 0, 1}, {1, 1, 1, 1}})] (* Amiram Eldar, Mar 28 2022 *)

Formula

a(n) = 2*a(n-4) for n > 4.
A093873(a(n)) = 3.
Sum_{n>=1} 1/a(n) = 4448/6435. - Amiram Eldar, Mar 28 2022

A334873 a(n) = A329697(n) * A334204(n).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 2, 0, 6, 2, 4, 1, 4, 2, 4, 0, 4, 6, 9, 2, 9, 4, 12, 1, 6, 4, 9, 2, 9, 4, 6, 0, 15, 4, 12, 6, 12, 9, 18, 2, 8, 9, 20, 4, 15, 12, 16, 1, 16, 6, 8, 4, 12, 9, 12, 2, 16, 9, 12, 4, 9, 6, 8, 0, 18, 15, 20, 4, 20, 12, 32, 6, 15, 12, 21, 9, 28, 18, 24, 2, 20, 8, 18, 9, 12, 20, 24, 4, 18, 15, 20, 12, 20
Offset: 1

Views

Author

Antti Karttunen, Jun 09 2020

Keywords

Crossrefs

Cf. A000079 (positions of zeros); It is also conjectured that A007283 gives the positions of ones, and that A070875 gives the indices of twos.

Programs

Formula

a(n) = A329697(n) * A334204(n) = A329697(n) * A329697(A163511(n)).

A063757 G.f.: (1+3*x+2*x^2)/((1-x)*(1-2*x^2)).

Original entry on oeis.org

1, 4, 8, 14, 22, 34, 50, 74, 106, 154, 218, 314, 442, 634, 890, 1274, 1786, 2554, 3578, 5114, 7162, 10234, 14330, 20474, 28666, 40954, 57338, 81914, 114682, 163834, 229370, 327674, 458746, 655354, 917498, 1310714, 1835002, 2621434
Offset: 0

Views

Author

N. J. A. Sloane, Aug 14 2001

Keywords

References

  • P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 158.

Programs

  • Mathematica
    CoefficientList[Series[(1+3x+2x^2)/((1-x)(1-2x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{1,2,-2},{1,4,8},41] (* Harvey P. Dale, Jun 05 2012 *)

Formula

a(0)=1, a(1)=4, a(2)=8, a(n)=a(n-1)+2*a(n-2)-2*a(n-3) From Harvey P. Dale, Jun 05 2012
a(n)=2^((n-3)/2)*((5*Sqrt[2]-7)*(-1)^n+7+5*Sqrt[2])-6 From Harvey P. Dale, Jun 05 2012
a(2*n) = 7*2^n - 6 = A048489(n), a(2*n+1) = 10*2^n - 6 = A020714(n+1) - 6, a(n) = A070875(n+1) - 6. - Philippe Deléham, Apr 13 2013

A265027 First differences of A048701 divided by 6.

Original entry on oeis.org

1, 1, 3, 2, 1, 2, 11, 4, 2, 4, 1, 4, 2, 4, 43, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 171, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 1, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 683, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8, 32, 16, 32, 2, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8
Offset: 2

Views

Author

N. J. A. Sloane, Nov 30 2015

Keywords

Comments

Indices n such that a(n) = 1 are equal to row sums of Lucas triangle. In other words, a(A042950(n)) = 1. Additionally, a(A070875(n)) = 2 and a(A123760(n)) = 4. - Altug Alkan, Dec 04 2015

Crossrefs

Programs

  • Mathematica
    Differences@ Select[Range@ 12000, Reverse@ # == # && EvenQ@ Length@ # &@ IntegerDigits[#, 2] &]/6 (* Michael De Vlieger, Dec 04 2015 *)
  • PARI
    a048701(n) = my(f); f = length(binary(n)) - 1; 2^(f+1)*n + sum(i=0, f, bittest(n, i) * 2^(f-i));
    vector(100, n, (a048701(n+1) - a048701(n)) / 6) \\ Altug Alkan, Dec 03 2015

Formula

a(n) = A265026(n) / 6, for n > 1. - Altug Alkan, Dec 03 2015

A383961 Square array read by upward antidiagonals: T(n,k) is the n-th number whose largest odd divisor is its k-th divisor, n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 15, 16, 11, 10, 20, 18, 32, 13, 12, 21, 50, 36, 64, 17, 14, 27, 81, 45, 30, 128, 19, 22, 28, 88, 63, 42, 105, 256, 23, 24, 33, 98, 75, 54, 135, 60, 512, 29, 25, 35, 104, 99, 66, 165, 84, 120, 1024, 31, 26, 39, 136, 117, 70, 189, 108, 140, 90
Offset: 1

Views

Author

Omar E. Pol, May 16 2025

Keywords

Comments

This is a permutation of the positive integers.
From Peter Munn, May 18 2025: (Start)
Numbers with the same factorization pattern of their sequence of divisors (see A290110) and the same parity appear here in the same column.
For example, each column k > 2 includes the subsequence 2^(k-2) * p for all prime p > 2^(k-2).
(End)

Examples

			The corner 15 X 15 of the square array is as follows:
      1,  3,  6,  15,  18,  36,  30, 105,  60, 120,  90, 315,  816, 1360, 180, ...
      2,  5,  9,  20,  50,  45,  42, 135,  84, 140, 126, 324,  880, 1520, 210, ...
      4,  7, 10,  21,  81,  63,  54, 165, 108, 168, 150, 432,  912, 1632, 252, ...
      8, 11, 12,  27,  88,  75,  66, 189, 132, 220, 198, 440, 1040, 1760, 270, ...
     16, 13, 14,  28,  98,  99,  70, 195, 156, 240, 216, 495, 1056, 1824, 300, ...
     32, 17, 22,  33, 104, 117,  72, 200, 162, 260, 234, 520, 1104, 1840, 330, ...
     64, 19, 24,  35, 136, 147,  78, 231, 204, 308, 264, 525, 1120, 1904, 378, ...
    128, 23, 25,  39, 152, 153, 100, 255, 225, 340, 280, 528, 1144, 2000, 390, ...
    256, 29, 26,  40, 176, 171, 102, 273, 228, 364, 294, 560, 1232, 2080, 396, ...
    512, 31, 34,  44, 184, 175, 110, 285, 276, 380, 306, 585, 1248, 2128, 462, ...
   1024, 37, 38,  51, 208, 207, 114, 297, 348, 405, 312, 616, 1392, 2208, 468, ...
   2048, 41, 46,  52, 232, 243, 130, 345, 372, 460, 336, 624, 1456, 2288, 510, ...
   4096, 43, 48,  55, 242, 245, 138, 351, 400, 476, 342, 675, 1458, 2320, 546, ...
   8192, 47, 49,  56, 248, 261, 144, 357, 441, 480, 350, 680, 1488, 2464, 570, ...
  16384, 53, 58,  57, 296, 272, 154, 375, 444, 500, 408, 693, 1496, 2480, 588, ...
  ...
		

Crossrefs

Column 1 gives A000079.
Column 2 gives A065091.
Column 3 consists of (A001248 U A091629 U A100484)\{4}.
Column 4 consists of numbers >= 15 in (A001749 U A030078 U A046388 U A070875).
Row 1 gives A383402.

Programs

  • Mathematica
    f[n_] := If[OddQ[n], DivisorSigma[0, n], FirstPosition[Divisors[n], n/2^IntegerExponent[n, 2]][[1]]]; seq[m_] := Module[{t = Table[0, {m}, {m}], v = Table[0, {m}], c = 0, k = 1, i, j}, While[c < m*(m + 1)/2, i = f[k]; If[i <= m, j = v[[i]] + 1; If[j <= m - i + 1, t[[i]][[j]] = k; v[[i]]++; c++]]; k++]; Table[t[[j]][[i - j + 1]], {i, 1, m}, {j, 1, i}] // Flatten]; seq[11] (* Amiram Eldar, May 16 2025 *)
Showing 1-10 of 12 results. Next