A209398 -id:A209398 - cp's OEIS frontend

A006498 a(n) = a(n-1) + a(n-3) + a(n-4), a(0) = a(1) = a(2) = 1, a(3) = 2.

1, 1, 1, 2, 4, 6, 9, 15, 25, 40, 64, 104, 169, 273, 441, 714, 1156, 1870, 3025, 4895, 7921, 12816, 20736, 33552, 54289, 87841, 142129, 229970, 372100, 602070, 974169, 1576239, 2550409, 4126648, 6677056, 10803704, 17480761, 28284465, 45765225, 74049690, 119814916

Offset: 0

N. J. A. Sloane

Number of compositions of n into 1's, 3's and 4's. - Len Smiley, May 08 2001
The sum of any two alternating terms (terms separated by one term) produces a number from the Fibonacci sequence. (e.g. 4+9=13, 9+25=34, 6+15=21, etc.) Taking square roots starting from the first term and every other term after, we get the Fibonacci sequence. - Sreyas Srinivasan (sreyas_srinivasan(AT)hotmail.com), May 02 2002
(1 + x + 2*x^2 + x^3)/(1 - x - x^3 - x^4) = 1 + 2*x + 4*x^2 + 6*x^3 + 9*x^4 + 15*x^5 + 25*x^6 + 40*x^7 + ... is the g.f. for the number of binary strings of length where neither 101 nor 111 occur. [Lozansky and Rousseau] Or, equivalently, where neither 000 nor 010 occur.
Equivalently, a(n+2) is the number of length-n binary strings with no two set bits with distance 2; see fxtbook link. - Joerg Arndt, Jul 10 2011
a(n) is the number of words written with the letters "a" and "b", with the following restriction: any "a" must be followed by at least two letters, the second of which is a "b". - Bruno Petazzoni (bpetazzoni(AT)ac-creteil.fr), Oct 31 2005. [This is also equivalent to the previous two conditions.]
Let a(0) = 1, then a(n) = partial products of Product_{n>2} (F(n)/F(n-1))^2 = 1*1*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(8/5)*(8/5)*.... E.g., a(7) = 15 = 1*1*1*2*2*(3/2)*(3/2)*(5/3). - Gary W. Adamson, Dec 13 2009
Number of permutations satisfying -k <= p(i) - i <= r and p(i)-i not in I, i=1..n, with k=1, r=3, I={1}. - Vladimir Baltic, Mar 07 2012
The 2-dimensional version, which counts sets of pairs no two of which are separated by graph-distance 2, is A273461. - Gus Wiseman, Nov 27 2019
a(n+1) is the number of multus bitstrings of length n with no runs of 4 ones. - Steven Finch, Mar 25 2020

			G.f. = 1 + x + x^2 + 2*x^3 + 4*x^4 + 6*x^5 + 9*x^6 + 15*x^7 + 25*x^8 + 40*x^9 + ...
From _Gus Wiseman_, Nov 27 2019: (Start)
The a(2) = 1 through a(7) = 15 subsets with no two elements differing by 2:
  {}  {}   {}     {}     {}     {}
      {1}  {1}    {1}    {1}    {1}
           {2}    {2}    {2}    {2}
           {1,2}  {3}    {3}    {3}
                  {1,2}  {4}    {4}
                  {2,3}  {1,2}  {5}
                         {1,4}  {1,2}
                         {2,3}  {1,4}
                         {3,4}  {1,5}
                                {2,3}
                                {2,5}
                                {3,4}
                                {4,5}
                                {1,2,5}
                                {1,4,5}
(End)

E. Lozansky and C. Rousseau, Winning Solutions, Springer, 1996; see pp. 157 and 172.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
Katharine A. Ahrens, Combinatorial Applications of the k-Fibonacci Numbers: A Cryptographically Motivated Analysis, Ph. D. thesis, North Carolina State University (2020).
Michael A. Allen, Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings, arXiv:2409.00624 [math.CO], 2024. See pp. 16, 18.
D. Applegate, M. LeBrun, and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
Said Amrouche and Hacène Belbachir, Unimodality and linear recurrences associated with rays in the Delannoy triangle, Turkish Journal of Mathematics (2019) Vol. 44, 118-130.
Joerg Arndt, Matters Computational (The Fxtbook), section 14.10.1, p.320.
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (2010), 119-135.
V. Baltic, Applications of the finite state automata for counting restricted permutations and variations, Yug. J. Oper. Res. 22 (2012) 183-198, Sec. 3
G. E. Bergum and V. E. Hoggatt, Jr., A combinatorial problem involving recursive sequences and tridiagonal matrices, Fib. Quart., 16 (1978), 113-118.
M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461.
K. Edwards, A Pascal-like triangle related to the tribonacci numbers, Fib. Q., 46/47 (2008/2009), 18-25.
Steven Finch, Cantor-solus and Cantor-multus distributions, arXiv:2003.09458 [math.CO], 2020.
T. Guardia and D. Jiménez, Fiboquadratic Sequences and Extensions of the Cassini Identity Raised From the Study of Rithmomachia, arXiv preprint arXiv:1509.03177 [math.HO], 2015-2016.
John Konvalina, Yi-Hsin Liu, subsets without q-separation and binomial products of Fibonacci numbers, J. Comb. Theo. A 57 (2) (1991) 306-310, T_n.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A060945 (for 1's, 2's and 4's). Essentially the same as A074677.
Diagonal sums of number triangle A059259.
Cf. A000032, A000034, A000045, A001654, A007598, A209398.
Numbers whose binary expansion has no subsequence (1,0,1) are A048716.
Cf. A003242, A005251, A027383, A167606, A261041, A273461, A329860, A329871.

a006498 n = a006498_list !! n
a006498_list = 1 : 1 : 1 : 2 : zipWith (+) (drop 3 a006498_list)
   (zipWith (+) (tail a006498_list) a006498_list)
-- Reinhard Zumkeller, Apr 07 2012

[ n eq 1 select 1 else n eq 2 select 1 else n eq 3 select 1 else n eq 4 select 2 else Self(n-1)+Self(n-3)+ Self(n-4): n in [1..40] ]; // Vincenzo Librandi, Aug 20 2011

LinearRecurrence[{1,0,1,1},{1,1,1,2},50] (* Harvey P. Dale, Jul 13 2011 *)
Table[Fibonacci[Floor[n/2] + 2]^Mod[n, 2]*Fibonacci[Floor[n/2] + 1]^(2 - Mod[n, 2]), {n, 0, 40}] (* David Nacin, Feb 29 2012 *)
a[ n_] := Fibonacci[ Quotient[ n+2, 2]] Fibonacci[ Quotient[ n+3, 2]] (* Michael Somos, Jan 19 2014 *)
Table[Length[Select[Subsets[Range[n]],!MatchQ[#,{_,x_,_,y_,_}/;x+2==y]&]],{n,10}] (* Gus Wiseman, Nov 27 2019 *)

{a(n) = fibonacci( (n+2)\2 ) * fibonacci( (n+3)\2 )} /* Michael Somos, Mar 10 2004 */

PARI
```
Vec(1/(1-x-x^3-x^4)+O(x^66))
```

def a(n, adict={0:1, 1:1, 2:1, 3:2}):
    if n in adict:
        return adict[n]
    adict[n]=a(n-1)+a(n-3)+a(n-4)
    return adict[n] # David Nacin, Mar 07 2012

G.f.: 1 / ((1 + x^2) * (1 - x - x^2)); a(2*n) = F(n+1)^2, a(2*n - 1) = F(n+1)*F(n). a(n) = a(-4-n) * (-1)^n. - Michael Somos, Mar 10 2004
The g.f. -(1+z+2*z^2+z^3)/((z^2+z-1)*(1+z^2)) for the truncated version 1, 2, 4, 6, 9, 15, 25, 40, ... was given in the Simon Plouffe thesis of 1992. [edited by R. J. Mathar, May 13 2008]
From Vladeta Jovovic, May 03 2002: (Start)
a(n) = round((-(1/5)*sqrt(5) - 1/5)*(-2*1/(-sqrt(5)+1))^n/(-sqrt(5)+1) + ((1/5)*sqrt(5) - 1/5)*(-2*1/( sqrt(5)+1))^n/(sqrt(5)+1)).
G.f.: 1/(1-x-x^2)/(1+x^2). (End)
a(n) = (-i)^n*Sum{k=0..n} U(n-2k, i/2) where i^2=-1. - Paul Barry, Nov 15 2003
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*F(n-2k+1). - Paul Barry, Oct 12 2007
F(floor(n/2) + 2)^(n mod 2)*F(floor(n/2) + 1)^(2 - (n mod 2)) where F(n) is the n-th Fibonacci number. - David Nacin, Feb 29 2012
a(2*n - 1) = A001654(n), a(2*n) = A007598(n+1). - Michael Somos, Mar 10 2004
a(n+1)*a(n+3) = a(n)*a(n+2) + a(n+1)*a(n+2) for all n in Z. - Michael Somos, Jan 19 2014
a(n) = round(1/(1/F(n+2) + 2/F(n+3))), where F(n) = A000045, and 0.5 is rounded to 1. - Richard R. Forberg, Aug 04 2014
5*a(n) = (-1)^floor(n/2)*A000034(n+1) + A000032(n+2). - R. J. Mathar, Sep 16 2017
a(n) = Sum_{j=0..floor(n/3)} Sum_{k=0..j} binomial(n-3j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 18 2017
E.g.f.: (2*cos(x) + sin(x) + exp(x/2)*(3*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Mar 12 2024

A209400 Number of subsets of {1,...,n} containing a subset of the form {k,k+1,k+3} for some k.

Original entry on oeis.org

0, 0, 0, 0, 2, 7, 19, 46, 107, 242, 535, 1162, 2490, 5281, 11108, 23206, 48206, 99663, 205218, 421115, 861585, 1758249, 3580075, 7275377, 14759592, 29897683, 60481359, 122206014, 246665382, 497414751, 1002231335, 2017877779, 4060069150, 8164204342

Offset: 0

David Nacin, Mar 07 2012

Also, number of subsets of {1,...,n} containing {a,a+2,a+3} for some a.

Also, number of bitstrings of length n containing 1101 or 1111.

			When n=4 the only subsets containing an {a,a+1,a+3} happen when a=1 with the two subsets {1,2,3,4} and {1,2,4}.  Thus a(4)=2.

David Nacin, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2,1,-1,-2).

Cf. A209398, A209399, A164387.

I:=[0, 0, 0, 0, 2, 7]; [n le 6 select I[n] else 3*Self(n-1) - Self(n-2)-2*Self(n-3)+Self(n-4)-Self(n-5)-2*Self(n-6): n in [0..30]]; // G. C. Greubel, Jan 03 2018

LinearRecurrence[{3, -1, -2, 1, -1, -2}, {0, 0, 0, 0, 2, 7}, 40]
CoefficientList[Series[x^4*(2+x)/(1-3*x+x^2+2*x^3-x^4+x^5+2*x^6), {x,0, 50}], x] (* G. C. Greubel, Jan 03 2018 *)

x='x+O('x^30); concat([0,0,0,0], Vec(x^4*(2+x)/(1-3*x+x^2+2*x^3-x^4+x^5+2*x^6))) \\ G. C. Greubel, Jan 03 2018

#From recurrence
def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:2, 5:7}):
 if n in adict:
  return adict[n]
 adict[n]=3*a(n-1)-a(n-2)-2*a(n-3)+a(n-4)-a(n-5)-2*a(n-6)
 return adict[n]

#Returns the actual list of valid subsets
def contains1101(n):
 patterns=list()
 for start in range (1,n-2):
  s=set()
  for i in range(4):
   if (1,1,0,1)[i]:
    s.add(start+i)
  patterns.append(s)
 s=list()
 for i in range(2,n+1):
  for temptuple in comb(range(1,n+1),i):
   tempset=set(temptuple)
   for sub in patterns:
    if sub <= tempset:
     s.append(tempset)
     break
 return s
#Counts all such sets
def countcontains1101(n):
 return len(contains1101(n))

a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3) + a(n-4) - a(n-5) - 2*a(n-6), a(4)=2, a(5)=7, a(i)=0 for i<4.
G.f.: x^4*(2 + x)/(1 - 3*x + x^2 + 2*x^3 - x^4 + x^5 + 2*x^6) = x^4*(2 + x)/((1 - 2*x)*(1 - x - x^2 - x^4 - x^5)).
a(n) = 2^n - A164387(n).

A209399 Number of subsets of {1,...,n} containing two elements whose difference is 3.

Original entry on oeis.org

0, 0, 0, 0, 4, 14, 37, 83, 181, 387, 824, 1728, 3584, 7360, 15032, 30571, 61987, 125339, 252883, 509294, 1024300, 2057848, 4130724, 8285758, 16610841, 33285207, 66673209, 133512759, 267294832, 535025408, 1070755840, 2142652160, 4287149680, 8577285255

Offset: 0

David Nacin, Mar 07 2012

Also, the number of bitstrings of length n containing one of the following: 1001, 1101, 1011, 1111.

			When n=4 any such subset must contain 1 and 4.  There are four such subsets so a(4) = 4.

David Nacin, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,3,-1,-1,-3,1,2).

Cf. A209398, A209400, A006500.

[2^n - Fibonacci(Floor(n/3) + 2)*Fibonacci(Floor((n + 1)/3) + 2)*Fibonacci(Floor((n + 2)/3) + 2): n in [0..30]]; // G. C. Greubel, Jan 03 2018

LinearRecurrence[{3, -1, -3, 3, -1, -1, -3, 1, 2}, {0, 0, 0, 0, 4, 14, 37, 83, 181}, 50]
Table[2^n - Product[Fibonacci[Floor[(n + i)/3] + 2], {i, 0, 2}], {n, 0, 50}]

x='x+O('x^30); concat([0,0,0,0], Vec(x^4*(4+2*x-x^2-2*x^3-x^4)/( (1-2*x)*(1-x-x^2)*(1+x^3-x^6)))) \\ G. C. Greubel, Jan 03 2018

#From recurrence
def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:4, 5:14, 6:37, 7:83, 8:181}):
 if n in adict:
  return adict[n]
 adict[n]=3*a(n-1)-a(n-2)-3*a(n-3)+3*a(n-4)-a(n-5)-a(n-6)-3*a(n-7)+a(n-8)+2*a(n-9)
 return adict[n]

#Returns the actual list of valid subsets
def contains1001(n):
 patterns=list()
 for start in range (1,n-2):
  s=set()
  for i in range(4):
   if (1,0,0,1)[i]:
    s.add(start+i)
  patterns.append(s)
 s=list()
 for i in range(2,n+1):
  for temptuple in comb(range(1,n+1),i):
   tempset=set(temptuple)
   for sub in patterns:
    if sub <= tempset:
     s.append(tempset)
     break
 return s
#Counts all such sets
def countcontains1001(n):
 return len(contains1001(n))

a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + 3*a(n-4) - a(n-5) - a(n-6) - 3*a(n-7) + a(n-8) + 2*a(n-9).
G.f.: (4*x^4 + 2*x^5 - x^6 - 2*x^7 - x^8)/(1 - 3*x + 1*x^2 + 3*x^3 - 3*x^4 + x^5 + x^6 + 3*x^7 - x^8 - 2*x^9) = x^4*(4 + 2*x - x^2 - 2*x^3 - x^4)/((1 - 2*x)*(1 - x - x^2)*(1 + x^3 - x^6)).
a(n) = 2^n - A006500(n).
a(n) = 2^n - Product(i=0 to 2) F(floor((n+i)/3)+2) where F(n) is the n-th Fibonacci number.

cp's OEIS Frontend

A006498 a(n) = a(n-1) + a(n-3) + a(n-4), a(0) = a(1) = a(2) = 1, a(3) = 2.

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A209400 Number of subsets of {1,...,n} containing a subset of the form {k,k+1,k+3} for some k.

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Magma

Mathematica

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A209399 Number of subsets of {1,...,n} containing two elements whose difference is 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Python

Formula