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 1-4 of 4 results.

A001610 a(n) = a(n-1) + a(n-2) + 1, with a(0) = 0 and a(1) = 2.

Original entry on oeis.org

0, 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, 9348, 15126, 24475, 39602, 64078, 103681, 167760, 271442, 439203, 710646, 1149850, 1860497, 3010348, 4870846, 7881195, 12752042, 20633238, 33385281, 54018520
Offset: 0

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Author

Keywords

Comments

For prime p, p divides a(p-1). - T. D. Noe, Apr 11 2009 [This result follows immediately from the fact that A032190(n) = (1/n)*Sum_{d|n} a(d-1)*phi(n/d). - Petros Hadjicostas, Sep 11 2017]
Generalization. If a(0,x)=0, a(1,x)=2 and, for n>=2, a(n,x)=a(n-1,x)+x*a(n-2,x)+1, then we obtain a sequence of polynomials Q_n(x)=a(n,x) of degree floor((n-1)/2), such that p is prime iff all coefficients of Q_(p-1)(x) are multiple of p (sf. A174625). Thus a(n) is the sum of coefficients of Q_(n-1)(x). - Vladimir Shevelev, Apr 23 2010
Odd composite numbers n such that n divides a(n-1) are in A005845. - Zak Seidov, May 04 2010; comment edited by N. J. A. Sloane, Aug 10 2010
a(n) is the number of ways to modify a circular arrangement of n objects by swapping one or more adjacent pairs. E.g., for 1234, new arrangements are 2134, 2143, 1324, 4321, 1243, 4231 (taking 4 and 1 to be adjacent) and a(4) = 6. - Toby Gottfried, Aug 21 2011
For n>2, a(n) equals the number of Markov equivalence classes with skeleton the cycle on n+1 nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
From Gus Wiseman, Feb 12 2019: (Start)
For n > 0, also the number of nonempty subsets of {1, ..., n + 1} containing no two cyclically successive elements (cyclically successive means 1 succeeds n + 1). For example, the a(5) = 17 stable subsets are:
{1}, {2}, {3}, {4}, {5}, {6},
{1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {2,6}, {3,5}, {3,6}, {4,6},
{1,3,5}, {2,4,6}.
(End)
Also the rank of the n-Lucas cube graph. - Eric W. Weisstein, Aug 01 2023

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,n+1)[2] -1); # G. C. Greubel, Jul 12 2019
  • Haskell
    a001610 n = a001610_list !! n
    a001610_list =
       0 : 2 : map (+ 1) (zipWith (+) a001610_list (tail a001610_list))
    -- Reinhard Zumkeller, Aug 21 2011
    
  • Magma
    I:=[0,2]; [n le 2 select I[n] else Self(n-1)+Self(n-2)+1: n in [1..40]]; // Vincenzo Librandi, Mar 20 2015
    
  • Magma
    [Lucas(n+1) -1: n in [0..40]]; // G. C. Greubel, Jul 12 2019
    
  • Mathematica
    t = {0, 2}; Do[AppendTo[t, t[[-1]] + t[[-2]] + 1], {n, 2, 40}]; t
    RecurrenceTable[{a[n] == a[n - 1] +a[n - 2] +1, a[0] == 0, a[1] == 2}, a, {n, 0, 40}] (* Robert G. Wilson v, Apr 13 2013 *)
    CoefficientList[Series[x (2 - x)/((1 - x - x^2) (1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
    Table[Fibonacci[n] + Fibonacci[n + 2] - 1, {n, 0, 40}] (* Eric W. Weisstein, Feb 13 2018 *)
    LinearRecurrence[{2, 0, -1}, {2, 3, 6}, 20] (* Eric W. Weisstein, Feb 13 2018 *)
    Table[LucasL[n] - 1, {n, 20}] (* Eric W. Weisstein, Aug 01 2023 *)
    LucasL[Range[20]] - 1 (* Eric W. Weisstein, Aug 01 2023 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -1,0,2]^n*[0;2;3])[1,1] \\ Charles R Greathouse IV, Sep 08 2016
    
  • PARI
    vector(40, n, f=fibonacci; f(n+1)+f(n-1)-1) \\ G. C. Greubel, Jul 12 2019
    
  • Sage
    [lucas_number2(n+1,1,-1) -1 for n in (0..40)] # G. C. Greubel, Jul 12 2019
    

Formula

a(n) = A000204(n)-1 = A000032(n+1)-1 = A000071(n+1) + A000045(n).
G.f.: x*(2-x)/((1-x-x^2)*(1-x)) = (2*x-x^2)/(1-2*x+x^3). [Simon Plouffe in his 1992 dissertation]
a(n) = F(n) + F(n+2) - 1 where F(n) is the n-th Fibonacci number. - Zerinvary Lajos, Jan 31 2008
a(n) = A014217(n+1) - A000035(n+1). - Paul Curtz, Sep 21 2008
a(n) = Sum_{i=1..floor((n+1)/2)} ((n+1)/i)*C(n-i,i-1). In more general case of polynomials Q_n(x)=a(n,x) (see our comment) we have Q_n(x) = Sum_{i=1..floor((n+1)/2)}((n+1)/i)*C(n-i,i-1)*x^(i-1). - Vladimir Shevelev, Apr 23 2010
a(n) = Sum_{k=0..n-1} Lucas(k), where Lucas(n) = A000032(n). - Gary Detlefs, Dec 07 2010
a(0)=0, a(1)=2, a(2)=3; for n>=3, a(n) = 2*a(n-1) - a(n-3). - George F. Johnson, Jan 28 2013
For n > 1, a(n) = A048162(n+1) + 3. - Toby Gottfried, Apr 13 2013
For n > 0, a(n) = A169985(n + 1) - 1. - Gus Wiseman, Feb 12 2019

A324015 Number of nonempty subsets of {1, ..., n} containing no two cyclically successive elements.

Original entry on oeis.org

0, 1, 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, 9348, 15126, 24475, 39602, 64078, 103681, 167760, 271442, 439203, 710646, 1149850, 1860497, 3010348, 4870846, 7881195, 12752042, 20633238, 33385281, 54018520, 87403802
Offset: 0

Views

Author

Gus Wiseman, Feb 12 2019

Keywords

Comments

Cyclically successive means 1 succeeds n.
After a(1) = 1, same as A001610 shifted once to the right. Also, a(n) = A169985(n) - 1.

Examples

			The a(6) = 17 stable subsets:
  {1}, {2}, {3}, {4}, {5}, {6},
  {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {2,6}, {3,5}, {3,6}, {4,6},
  {1,3,5}, {2,4,6}.
		

Crossrefs

Programs

  • Mathematica
    stabsubs[g_]:=Select[Rest[Subsets[Union@@g]],Select[g,Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&];
    Table[Length[stabsubs[Partition[Range[n],2,1,1]]],{n,0,10}]

Formula

For n <= 3, a(n) = n. Otherwise, a(n) = a(n - 1) + a(n - 2) + 1.

A323949 Number of set partitions of {1, ..., n} with no block containing three distinct cyclically successive vertices.

Original entry on oeis.org

1, 1, 2, 4, 10, 36, 145, 631, 3015, 15563, 86144, 508311, 3180930, 21018999, 146111543, 1065040886, 8117566366, 64531949885, 533880211566, 4587373155544, 40865048111424, 376788283806743, 3590485953393739, 35312436594162173, 357995171351223109, 3736806713651177702
Offset: 0

Views

Author

Gus Wiseman, Feb 10 2019

Keywords

Comments

Cyclically successive means 1 is a successor of n.

Examples

			The a(1) = 1 through a(4) = 10 set partitions:
  {{1}}  {{1,2}}    {{1},{2,3}}    {{1,2},{3,4}}
         {{1},{2}}  {{1,2},{3}}    {{1,3},{2,4}}
                    {{1,3},{2}}    {{1,4},{2,3}}
                    {{1},{2},{3}}  {{1},{2},{3,4}}
                                   {{1},{2,3},{4}}
                                   {{1,2},{3},{4}}
                                   {{1},{2,4},{3}}
                                   {{1,3},{2},{4}}
                                   {{1,4},{2},{3}}
                                   {{1},{2},{3},{4}}
		

Crossrefs

Programs

  • Mathematica
    spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
    Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],3,1,1],Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&],Range[n]]],{n,8}]

Extensions

a(12)-a(25) from Alois P. Heinz, Feb 10 2019

A323955 Regular triangle read by rows where T(n, k) is the number of set partitions of {1, ..., n} with no block containing k cyclically successive vertices, n >= 1, 2 <= k <= n + 1.

Original entry on oeis.org

1, 1, 2, 1, 4, 5, 4, 10, 14, 15, 11, 36, 46, 51, 52, 41, 145, 184, 196, 202, 203, 162, 631, 806, 855, 869, 876, 877, 715, 3015, 3847, 4059, 4115, 4131, 4139, 4140, 3425, 15563, 19805, 20813, 21056, 21119, 21137, 21146, 21147, 17722, 86144, 109339, 114469
Offset: 1

Views

Author

Gus Wiseman, Feb 10 2019

Keywords

Comments

Cyclically successive means 1 is a successor of n.

Examples

			Triangle begins:
    1
    1    2
    1    4    5
    4   10   14   15
   11   36   46   51   52
   41  145  184  196  202  203
  162  631  806  855  869  876  877
  715 3015 3847 4059 4115 4131 4139 4140
Row 4 counts the following partitions:
  {{13}{24}}      {{12}{34}}      {{1}{234}}      {{1234}}
  {{1}{24}{3}}    {{13}{24}}      {{12}{34}}      {{1}{234}}
  {{13}{2}{4}}    {{14}{23}}      {{123}{4}}      {{12}{34}}
  {{1}{2}{3}{4}}  {{1}{2}{34}}    {{124}{3}}      {{123}{4}}
                  {{1}{23}{4}}    {{13}{24}}      {{124}{3}}
                  {{12}{3}{4}}    {{134}{2}}      {{13}{24}}
                  {{1}{24}{3}}    {{14}{23}}      {{134}{2}}
                  {{13}{2}{4}}    {{1}{2}{34}}    {{14}{23}}
                  {{14}{2}{3}}    {{1}{23}{4}}    {{1}{2}{34}}
                  {{1}{2}{3}{4}}  {{12}{3}{4}}    {{1}{23}{4}}
                                  {{1}{24}{3}}    {{12}{3}{4}}
                                  {{13}{2}{4}}    {{1}{24}{3}}
                                  {{14}{2}{3}}    {{13}{2}{4}}
                                  {{1}{2}{3}{4}}  {{14}{2}{3}}
                                                  {{1}{2}{3}{4}}
		

Crossrefs

First column (k = 2) is A000296. Second column (k = 3) is A323949. Rightmost terms are A000110. Second to rightmost terms are A058692.

Programs

  • Mathematica
    spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
    Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],k,1,1],Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&],Range[n]]],{n,7},{k,2,n+1}]
Showing 1-4 of 4 results.