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.

Previous Showing 21-30 of 33 results. Next

A191595 Order of smallest n-regular graph of girth 5.

Original entry on oeis.org

5, 10, 19, 30, 40, 50
Offset: 2

Views

Author

N. J. A. Sloane, Jun 07 2011

Keywords

Comments

Current upper bounds for a(8)..a(20) are 80, 96, 124, 154, 203, 230, 288, 312, 336, 448, 480, 512, 576. - Corrected from "Lower" to "Upper" and updated, from Table 4 of the Dynamic cage survey, by Jason Kimberley, Dec 29 2012
Current lower bounds for a(8)..a(20) are 67, 86, 103, 124, 147, 174, 199, 230, 259, 294, 327, 364, 403. - from Table 4 of the Dynamic cage survey via Jason Kimberley, Dec 31 2012

Crossrefs

Orders of cages: A054760 (n,k), A000066 (3,n), A037233 (4,n), A218553 (5,n), A218554 (6,n), A218555 (7,n), this sequence (n,5).
Moore lower bound on the orders of (k,g) cages: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306(k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10),A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Nov 02 2011

Formula

a(n) >= A002522(n) with equality if and only if n = 2, 3, 7 or possibly 57. - Jason Kimberley, Nov 02 2011

Extensions

a(2) = 5 prepended by Jason Kimberley, Jan 02 2013

A218553 Order of (5,n) cage, i.e., minimal order of 5-regular graph of girth n.

Original entry on oeis.org

6, 10, 30, 42
Offset: 3

Views

Author

Arkadiusz Wesolowski, Nov 02 2012

Keywords

Comments

a(7) <= 152, a(8) = 170, a(12) = 2730. - From Royle's page via Jason Kimberley, Dec 21 2012

Crossrefs

Orders of cages: A054760 (n,k), A000066 (3,n), A037233 (4,n), this sequence (5,n), A218554 (6,n), A218555 (7,n), A191595 (n,5).

Formula

a(n) >= A061547(n+1).

Extensions

a(7) deleted by Jason Kimberley, Dec 21 2012

A152716 Triangle T(n,k) read by rows: T(n,k) = 4^min(k, n-k) = 4^A004197(n,k).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 4, 4, 1, 1, 4, 16, 4, 1, 1, 4, 16, 16, 4, 1, 1, 4, 16, 64, 16, 4, 1, 1, 4, 16, 64, 64, 16, 4, 1, 1, 4, 16, 64, 256, 64, 16, 4, 1, 1, 4, 16, 64, 256, 256, 64, 16, 4, 1, 1, 4, 16, 64, 256, 1024, 256, 64, 16, 4, 1
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Dec 11 2008

Keywords

Comments

Row sums are: {1, 2, 6, 10, 26, 42, 106, 170, 426, 682, 1706,...} = A061547(n+2).

Examples

			{1},
{1, 1},
{1, 4, 1},
{1, 4, 4, 1},
{1, 4, 16, 4, 1},
{1, 4, 16, 16, 4, 1},
{1, 4, 16, 64, 16, 4, 1},
{1, 4, 16, 64, 64, 16, 4, 1},
{1, 4, 16, 64, 256, 64, 16, 4, 1},
{1, 4, 16, 64, 256, 256, 64, 16, 4, 1},
{1, 4, 16, 64, 256, 1024, 256, 64, 16, 4, 1}
		

Crossrefs

Programs

  • Mathematica
    Clear[a, k, m]; k = 4; a[0] = {1}; a[1] = {1, 1};
    a[n_] := a[n] = Join[{1}, k*a[n - 2], {1}];
    Table[a[n], {n, 0, 10}];
    Flatten[%]

Formula

T(n,k) = 4^min(k, n-k). - Philippe Deléham, Feb 25 2014
T(n,k) = A144464(n,k)^2. - Philippe Deléham, Feb 26 2014

Extensions

Better name by Philippe Deléham, Feb 25 2014

A068976 a(n) = Sum_{d | n} d/core(d) where core(x) is the smallest number such that x*core(x) is a square.

Original entry on oeis.org

1, 2, 2, 6, 2, 4, 2, 10, 11, 4, 2, 12, 2, 4, 4, 26, 2, 22, 2, 12, 4, 4, 2, 20, 27, 4, 20, 12, 2, 8, 2, 42, 4, 4, 4, 66, 2, 4, 4, 20, 2, 8, 2, 12, 22, 4, 2, 52, 51, 54, 4, 12, 2, 40, 4, 20, 4, 4, 2, 24, 2, 4, 22, 106, 4, 8, 2, 12, 4, 8, 2, 110, 2, 4, 54, 12, 4, 8, 2, 52, 101, 4, 2, 24, 4, 4, 4
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2002

Keywords

Comments

More generally, a(n,m) = Sum_{d divides n} gcd(d,n/d)^m is multiplicative with a(p^e,m) = (p^(m*e/2)*(p^m+1)-2)/(p^m-1) if e is even else 2*(p^(m*(e+1)/2)-1)/(p^m-1). - Vladeta Jovovic, May 30 2003

Crossrefs

Programs

  • Maple
    R:= proc(n) uses numtheory; local K,k;
      K:= select(k -> (n mod k^2 = 0), divisors(n));
      add(k^2*2^nops(factorset(n/k^2)),k=K);
    end proc:
    seq(R(n),n=1..100); # Robert Israel, Oct 18 2015
  • Mathematica
    a[n_]:=Total[GCD[#, n/#]^2 & /@ Divisors[n]]; Table[a[n], {n, 1, 87}] (* Jean-François Alcover, Jul 26 2011 *)
    f[p_, e_] := If[OddQ[e], 2*(p^(e+1)-1)/(p^2-1), (p^(e+2)+p^e-2)/(p^2-1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 03 2020 *)

Formula

a(n) = Sum_{d divides n} gcd(d, n/d)^2. Multiplicative with a(p^e) = (p^(e+2)+p^e-2)/(p^2-1) if e is even else 2*(p^(e+1)-1)/(p^2-1). - Vladeta Jovovic, May 30 2003
Dirichlet g.f.: zeta^2(s)*zeta(2s-2)/zeta(2s). Dirichlet convolution of A034444 and the sequence n*A010052(n). - R. J. Mathar, Apr 18 2011
Inverse Mobius transform of A008833. - R. J. Mathar, Oct 31 2011
a(n) = Sum_{d divides n} (-1)^A001222(d) * A000010(d) * A000203(n/d) = Sum_{k^2 divides n} k^2 * 2^A001221(n/k^2). - Robert Israel, Oct 18 2015
Sum_{k=1..n} a(k) ~ Zeta(3/2)^2 * n^(3/2) / (3*Zeta(3)) - (3*n*(log(n) - 1 + 2*gamma + 2*log(2*Pi) - 12*Zeta'(2)/Pi^2))/Pi^2, where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 05 2019
a(2^k) = (3/2)*2^k + (1/6)*(-2)^k - 2/3 = A061547(k+2). - Amiram Eldar, Sep 03 2020

A176965 a(n) = 2^(n-1) - (2^n*(-1)^n + 2)/3.

Original entry on oeis.org

1, 0, 6, 2, 26, 10, 106, 42, 426, 170, 1706, 682, 6826, 2730, 27306, 10922, 109226, 43690, 436906, 174762, 1747626, 699050, 6990506, 2796202, 27962026, 11184810, 111848106, 44739242, 447392426, 178956970, 1789569706, 715827882, 7158278826
Offset: 1

Views

Author

Roger L. Bagula, Apr 29 2010

Keywords

Comments

The ratio a(n+1)/a(n) approaches 10 for even n and 2/5 for odd n as n->infinity.

Crossrefs

Merger of A020988 (even n) and A020989 (odd n).

Programs

  • GAP
    List([1..30], n-> (3*2^(n-1) -(-2)^n -2)/3); # G. C. Greubel, Dec 28 2019
  • Magma
    [(3*2^(n-1) -(-2)^n -2)/3: n in [1..30]]; // G. C. Greubel, Dec 28 2019
    
  • Maple
    seq( (3*2^(n-1) -(-2)^n -2)/3, n=1..30); # G. C. Greubel, Dec 28 2019
  • Mathematica
    a[n_]:= a[n]= 2^(n-1)*If[n==1, 1, a[n-1]/2 +(-1)^(n-1)*Sqrt[(5 +4*(-1)^(n-1) )]/2]; Table[a[n], {n,30}]
    LinearRecurrence[{1,4,-4}, {1,0,6}, 30] (* G. C. Greubel, Dec 28 2019 *)
  • PARI
    vector(30, n, (3*2^(n-1) -(-2)^n -2)/3 ) \\ G. C. Greubel, Dec 28 2019
    
  • Sage
    [(3*2^(n-1) -(-2)^n -2)/3 for n in (1..30)] # G. C. Greubel, Dec 28 2019
    

Formula

From R. J. Mathar, Apr 30 2010: (Start)
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3).
G.f.: x*(1 - x + 2*x^2)/( (1-x)*(1+2*x)*(1-2*x) ). (End)
a(n) = A087231(n), n > 2. - R. J. Mathar, May 03 2010
a(2n-1) = A061547(2n), a(2n) = A061547(2n-1), n > 0. - Yosu Yurramendi, Dec 23 2016
a(n+1) = 2*A096773(n), n > 0. - Yosu Yurramendi, Dec 30 2016
a(2n-1) = A020989(n-1), a(2n) = A020988(n-1), n > 0. - Yosu Yurramendi, Jan 03 2017
a(2n-1) = (A083597(n-1) + A000302(n-1))/2, a(2n) = (A083597(n-1) - A000302(n-1))/2, n > 0. - Yosu Yurramendi, Mar 04 2017
a(n+2) = 4*a(n) + 2, a(1) = 1, a(2) = 0, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n) = (-16 + (9 - (-1)^n) * 2^(n - (-1)^n))/24. - Loren M. Pearson, Dec 28 2019
E.g.f.: (3*exp(2*x) - 4*exp(x) + 3 - 2*exp(-2*x))/6. - G. C. Greubel, Dec 28 2019
a(n) = (2^n*5^(n mod 2) - 4)/6. - Heinz Ebert, Jun 29 2021

A177993 Triangle read by rows, A177990 * A007318.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 2, 4, 3, 1, 3, 8, 9, 5, 1, 3, 9, 13, 11, 5, 1, 4, 15, 28, 31, 20, 7, 1, 4, 16, 34, 46, 40, 22, 7, 1, 5, 24, 62, 102, 110, 78, 35, 9, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 6, 35, 115, 250, 376, 400, 301, 157, 54, 11, 1, 6, 36, 125, 295, 496, 610, 553, 367, 174, 56, 11, 1
Offset: 0

Views

Author

Gary W. Adamson, May 16 2010

Keywords

Examples

			First few rows of the triangle =
1;
1, 1;
2, 3, 1;
2, 4, 3, 1;
3, 8, 9, 5, 1;
3, 9, 13, 11, 5, 1;
4, 15, 28, 31, 20, 7, 1;
4, 16, 34, 46, 40, 22, 7, 1;
5, 24, 62, 102, 110, 78, 35, 9, 1;
5, 25, 70, 130, 166, 148, 91, 37, 9, 1;
6, 35, 115, 250, 376, 400, 301, 157, 54, 11, 1;
6, 36, 125, 295, 496, 610, 553, 367, 174, 56, 11, 1;
7, 48, 191, 515, 991, 1402, 1477, 1159, 669, 276, 77, 13, 1;
7, 49, 203, 581, 1211, 1897, 2269, 2083, 1461, 771, 297, 709, 13, 1;
...
		

Crossrefs

Row sums are A061547(n+1).
Cf. A061547.

Programs

  • PARI
    T(n,k) = {binomial(n,k) + sum(j=0, n\2-1, binomial(2*j+1,k))} \\ Andrew Howroyd, Apr 13 2021

Formula

As infinite lower triangular matrices, A177990 * Pascal's triangle, (A007318).
T(n,k) = binomial(n,k) + Sum_{j=0..floor(n/2)-1} binomial(2*j+1,k). - Andrew Howroyd, Apr 13 2021

Extensions

Terms a(55) and beyond from Andrew Howroyd, Apr 13 2021

A183158 T(n,k) is the number of partial isometries of an n-chain of fix k (fix of alpha is the number of fixed points of alpha).

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 12, 6, 3, 1, 38, 10, 6, 4, 1, 90, 26, 10, 10, 5, 1, 220, 42, 15, 20, 15, 6, 1, 460, 106, 21, 35, 35, 21, 7, 1, 1018, 170, 28, 56, 70, 56, 28, 8, 1, 2022, 426, 36, 84, 126, 126, 84, 36, 9, 1, 4304, 682, 45, 120, 210, 252, 210, 120, 45, 10, 1
Offset: 0

Views

Author

Abdullahi Umar, Dec 28 2010

Keywords

Examples

			T (4,2) = 6 because there are exactly 6 partial isometries (on a 4-chain) of fix 2, namely: (1,2)-->(1,2); (2,3)-->(2,3); (3,4)-->(3,4); (1,3)-->(1,3); (2,4)-->(2,4); (1,4)-->(1,4) - the mappings are coordinate-wise.
...1.
...1....1.
...4....2....1.
..12....6....3....1.
..38...10....6....4....1.
..90...26...10...10....5....1.
.220...42...15...20...15....6....1.
.460..106...21...35...35...21....7....1.
1018..170...28...56...70...56...28....8....1.
2022..426...36...84..126..126...84...36....9....1.
4304..682...45..120..210..252..210..120...45...10....1.
		

Crossrefs

Cf. A183156 (row sums).

Programs

  • Maple
    A183159 := proc(n) nh := floor(n/2) ; if type(n,'even') then 13*4^nh-12*nh^2-18*nh-10; else 25*4^nh-12*nh^2-30*nh-22; end if; %/3 ; end proc:
    A061547 := proc(n) 3*2^n/8 +(-2)^n/24 - 2/3; end proc:
    A183158 := proc(n,k) if k = 0 then A183159(n) ; elif k = 1 then A061547(n+1) ; else binomial(n,k) ; end if; end proc: # R. J. Mathar, Jan 06 2011
  • Mathematica
    T[n_, 0] := (51*2^n + (-2)^n - 40)/12 - n*(n + 3);
    T[n_, 1] := (9*2^n + (-1)^(n+1)*2^n - 8)/12;
    T[n_, k_] := Binomial[n, k];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2017 *)

Formula

T(n,0)= A183159(n). T(n,1)=A061547(n+1). T(n,k)=binomial(n,k), k > 1.

A345253 Maximal Fibonacci tree: Arrangement of the positive integers as labels of a complete binary tree.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 13, 11, 14, 16, 21, 12, 15, 17, 22, 18, 23, 26, 34, 19, 24, 27, 35, 29, 37, 42, 55, 20, 25, 28, 36, 30, 38, 43, 56, 31, 39, 44, 57, 47, 60, 68, 89, 32, 40, 45, 58, 48, 61, 69, 90, 50, 63, 71, 92, 76, 97, 110, 144, 33, 41, 46, 59, 49
Offset: 1

Views

Author

J. Parker Shectman, Jun 12 2021

Keywords

Comments

Every positive integer occurs exactly once, so that, as a sequence, a(n) is a permutation of the positive integers.
Descending from the root node 1, generate tree by outer composition of L(n) = n + F(Finv(n)) and R(n) = n + F(Finv(n) + 1), respectively, according to left or right branching, where F(n) = A000045(n) are the Fibonacci numbers and Finv(n) = A130233(n) is the 'lower' Fibonacci inverse. This produces each number by maximal Fibonacci expansion (cf. example below of Method 2, entry A343152, and links).
(Level of tree): The number of terms in this expansion of n is the level of the tree on which n appears, A112310(n-1) + 1 = A200648(n+1). The number of terms in the expansion of a(n) is floor(log_2(n)) + 1 = A113473(n) = A070939(n) = A029837(n+1).
"Maximal Fibonacci expansion" maximizes the sum of coefficients over all Fibonacci numbers (of positive index), allowing both F(1) = 1 and F(2) = 1. Thus, it is just an expansion and not a representation (like "greedy" and "lazy"), as it "breaks the rule" by using two bits that correspond to elements of equal value, rather than using distinct basis elements (link). This reveals connections to the cf. sequences: Binary strings that emerge in lexicographic order from "maximal Fibonacci gaps" (example), binary trees of the positive integers, and I-D arrays "harvested" from the trees. To define the expansion uniquely, always include F(1), so that the expansion of positive integer n equals F(1) for n = 1 and F(1) prepended to the lazy Fibonacci representation of n-1 for n > 1. Hence, a(1) = 1, and for n > 1, a(n) = A095903(n-1) + 1. The "redundant" expansion arranges the positive integers in the single binary tree {T(n,k)}, rather than the two trees at A255773 and A255774 that result from representation (see link).
(Left-to-right order in tree): Each F(t)-sized block (F(t+1), ..., F(t+2) - 1) of successive positive integers ("Fibonacci cohort" t) appears in right-to-left order in the tree as reordered in A343152, where elements of each cohort appear consecutively (see link).
Descending from the root node 1, generate tree by the inner composition of A026351 and A026352, that is, one plus the sequences of lower and upper Wythoff numbers, A000201 and A001950, respectively, according to left or right branching (see example below of Method 1 and links).
Generate tree from (one plus) the number of (initial) zeros on the positive integers for the outer composition of sequences, A060143 and A060144, respectively, according to left or right branching descending from the identity (c.f example below of Method 3 and links).
The lower Wythoff numbers, A000201, appear exclusively in the 1st, 3rd, 5th, ... right clades of the tree, while the upper Wythoff numbers A001950, appear exclusively in the 2nd, 4th, 6th, ... right clades of the tree. Here, the k-th right clade comprises the nodes at positions 2^(k+1) and 2^k + 1, together with all descendants of the latter (link).
(Duality with tree A232560, and related arrays): Consider the labeled binary trees a(n) = A232560(A059893(n)) and A232560(n) = a(A059893(n)). Labels along maximal straight paths that always branch left in a(n) give rows of array A345252, while labels along maximal straight paths that always branch left in A232560 give rows of array A083047.
Sorting the labels from each successive right clade of the binary tree a(n) gives the successive columns of A083047, while sorting labels from each successive right clade of A232560 gives each successive column of A345252. This makes the trees a(n) and A232560 "blade-duals," blade being a contraction of branch-clade (see entry for A345254 and link). A200648(n)+1 gives the level of the tree on which elements of array first-columns A345252(n,1) and A083047(n,1) appear.
(Palindromes and coincidence of elements): Trees a(n) and A232560 coincide when the sequence of left and right branching is a palindrome: a(A329395(n)) = A232560(A329395(n)). As Kimberling notes (cf. A059893), this happens at fixed points of A059893(n) or, equivalently, at n for which A081242(n) is a palindrome.
The inverse permutation of a(n) as a sequence can be read from a "tetrangle" or "irregular triangle" tableau with F(t) (Fibonacci number) entries on each row t, for t = 1, 2, 3, ..., in which an entry on row t is 2*x the entry x immediately above it on row t-1, if such exists, or otherwise 2*x + 1 the entry x in the corresponding position on row t-2 (thus generating new rows as in A243571 but without sorting the numbers into increasing order, linked reference):
1,
2,
3, 4,
5, 6, 8,
7, 9, 10, 12, 16,
11, 13, 17, 14, 18, 20, 24, 32,
...
With the right-justified tableau substituted by a left-justified tableau, the same procedure yields the inverse permutation for the "minimal Fibonacci tree," A048680(A059893(n)), the "cohort-dual" tree of a(n), where "cohort" t is the F(t)-sized block of successive entries in the tableau (see entry for A345252, linked reference).
(Coincidence of elements): a(A020988(n)) = A048680(A059893(A020988(n))) = A099919(n) and a(A020989(n)) = A048680(A059893(A020989(n))) = A049651(n). Collectively, a(A061547(n)) = A048680(A059893(A061547(n))) = union(A049651(n), A099919(n)).
With two types of duality, the tree forms a quartet of binary-tree arrangements of the positive integers, together with its blade dual A232560, its cohort dual A048680(A059893), and blade dual A048680 of the latter.
Order in the tree is "memory-less": Let a(n) and a(m) label nodes at positions n and m, respectively. Let d1 and d2 be two descending paths, i.e., sequences branching left or right from a starting node. (Nodal positions for the left and right children of the node at position p are given by 2*p and 2*p + 1, resp., and d1 and d2 are compositions of these.) Then a(d1(n)) < a(d2(n)) if and only if a(d1(m)) < a(d2(m)) (linked reference).

Examples

			As a complete binary tree:
                    1
           /                 \
          2                   3
      /       \          /        \
     4         5        6          8
    / \       / \      / \        / \
   7    9    10   13   11   14   16   21
  / \  / \  /  \ /  \ /  \ /  \ /  \ /  \
  ...
By maximal Fibonacci expansion:
                                        F(1)
                      /                                       \
                F(1) + F(2)                               F(1) + F(3)
           /                    \                    /                  \
  F(1) + F(2) + F(3)   F(1) + F(2) + F(4)   F(1) + F(3) + F(4)   F(1) + F(3) + F(5)
  ...
"Fibonacci gaps," or differences between successive indices in maximal Fibonacci expansion above, are A007931(n-1) for n > 1 (see link):
                   *
          /                  \
         1                    2
     /       \           /        \
    11        12        21        22
   /  \      /  \      /  \      /  \
  111  112  121  122  211  212  221  222
  / \  / \  / \  / \  / \  / \  / \  / \
  ...
In examples of the three methods below:
Branch left-right-right down the tree to arrive at nodal position n = 2*(2*(2*1) + 1) + 1 = 11;
Branch right-left-left down the tree to arrive at nodal position n = 2*(2*(2*1 + 1)) = 12.
Tree by inner composition of (one plus) the lower and upper Wythoff sequences, A000201 and A001950 (Method 1):
a(11) = A000201(A001950(A001950(1) + 1) + 1) + 1 = 13.
a(12) = A001950(A000201(A000201(1) + 1) + 1) + 1 = 11.
Tree by (outer) composition of branching functions L(n) = n + F(Finv(n)) and R(n) = n + F(Finv(n) + 1), where F(n) = A000045(n) and Finv(n) = A130233(n) (Method 2):
a(11) = R(R(L(1))) = 13.
a(12) = L(R(R(1))) = 11.
Tree by outer composition of A060143 and A060144 (Wythoff inverse sequences) (Method 3):
a(11) = 13, position of first nonzero in A060144(A060144(A060143(m))) = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ..., for m = 1, 2, 3, ....
a(12) = 11, position of first nonzero in A060143(A060143(A060144(m))) = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ..., for m = 1, 2, 3, ....
		

Crossrefs

Programs

  • Mathematica
    (* For binary tree implementations, see supporting file under LINKS *)
    a[n_] := (x = 0; y = 0; BDn = Reverse[IntegerDigits[n, 2]]; imax = Length[BDn] - 1; For[i = 0, i <= imax, i++, {x, y} = {y + 1, x + y}; If[BDn[[i + 1]] == 1, {x, y} = {y, x + y}]]; y);
    (* Adapted from PARI code of Kevin Ryde *)
  • PARI
    a(n) = my(x=0,y=0); for(i=0,logint(n,2), [x,y]=[y+1,x+y]; if(bittest(n,i), [x,y]=[y,x+y])); y; \\ Kevin Ryde, Jun 19 2021

Formula

a(1) = 1 and for n > 1, a(n) = A095903(n-1) + 1.
a(n) = A232560(A059893(n)).

A166753 Partial sums of A166752.

Original entry on oeis.org

1, 2, 5, 6, 17, 18, 61, 62, 233, 234, 917, 918, 3649, 3650, 14573, 14574, 58265, 58266, 233029, 233030, 932081, 932082, 3728285, 3728286, 14913097, 14913098, 59652341, 59652342, 238609313, 238609314, 954437197, 954437198, 3817748729, 3817748730
Offset: 0

Views

Author

Paul Barry, Oct 21 2009

Keywords

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1+x-2*x^2-4*x^3)/((1-x)*(1-5*x^2+4*x^4)) )); // G. C. Greubel, Jun 06 2019
    
  • Mathematica
    LinearRecurrence[{1,5,-5,-4,4}, {1,2,5,6,17}, 40] (* G. C. Greubel, May 24 2016 *)
    Accumulate[LinearRecurrence[{0,5,0,-4},{1,1,3,1},40]] (* Harvey P. Dale, Aug 12 2024 *)
  • PARI
    my(x='x+O('x^40)); Vec((1+x-2*x^2-4*x^3)/((1-x)*(1-5*x^2+4*x^4))) \\ G. C. Greubel, Sep 30 2017
    
  • Sage
    ((1+x-2*x^2-4*x^3)/((1-x)*(1-5*x^2+4*x^4))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jun 06 2019

Formula

G.f.: (1+x-2*x^2-4*x^3)/((1-x)*(1-5*x^2+4*x^4)).
a(n) = a(n+1) + 5*a(n+2) - 5*a(n-3) - 4*a(n-4) + 4*a(n-5).
a(n) = (4/3)*A061547(n+1) - (1/3)*A166754(n).
a(n) = (4/3)*A061547(n+1) - (1/3)*A000975(n) + (4/3)*A011377(n-2).

A266846 Decimal representation of the n-th iteration of the "Rule 70" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 6, 20, 104, 336, 1696, 5440, 27264, 87296, 436736, 1397760, 6989824, 22368256, 111845376, 357908480, 1789558784, 5726601216, 28633071616, 91625881600, 458129670144, 1466015154176, 7330076819456, 23456246661120, 117281237499904, 375299963355136
Offset: 0

Views

Author

Robert Price, Jan 04 2016

Keywords

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Programs

  • Mathematica
    rule=70; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) Table[FromDigits[catri[[k]],2],{k,1,rows}]   (* Decimal Representation of Rows *)

Formula

Conjectures from Colin Barker, Jan 05 2016 and Apr 18 2019: (Start)
a(n) = 2^(n-1)*(-(-2)^n+9*2^n-2)/3 = 2^(n-1)*A061547(n+3).
a(n) = 2*a(n-1)+16*a(n-2)-32*a(n-3) for n>2.
G.f.: (1+4*x-8*x^2) / ((1-2*x)*(1-4*x)*(1+4*x)).
(End)
Previous Showing 21-30 of 33 results. Next