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.

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A005709 a(n) = a(n-1) + a(n-7), with a(i) = 1 for i = 0..6.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 17, 22, 28, 35, 43, 53, 66, 83, 105, 133, 168, 211, 264, 330, 413, 518, 651, 819, 1030, 1294, 1624, 2037, 2555, 3206, 4025, 5055, 6349, 7973, 10010, 12565, 15771, 19796, 24851
Offset: 0

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Author

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = Sum_{i=0..n/m} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
For n >= 7, a(n-7) is the number of compositions of n in which each part is >=7. - Milan Janjic, Jun 28 2010
Number of compositions of n into parts 1 and 7. - Joerg Arndt, Jun 24 2011
a(n+6) is the number of binary words of length n having at least 6 zeros between every two successive ones. - Milan Janjic, Feb 09 2015
Number of tilings of a 7 X n rectangle with 7 X 1 heptominoes. - M. Poyraz Torcuk, Feb 26 2022

References

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

Crossrefs

Programs

  • Maple
    A005709 := proc(n) option remember; if n <=6 then 1; else A005709(n-1)+A005709(n-7); fi; end;
    with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 6)}, unlabeled]: seq(count(SeqSetU, size=j), j=7..55); # Zerinvary Lajos, Oct 10 2006
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 6)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=6..54); # Zerinvary Lajos, Mar 26 2008
    M:= Matrix(7, (i,j)-> if j=1 and member(i,[1,7]) then 1 elif (i=j-1) then 1 else 0 fi); a:= n-> (M^(n))[1,1]; seq(a(n), n=0..50); # Alois P. Heinz, Jul 27 2008
  • Mathematica
    f[ n_Integer ] := f[ n ]=If[ n>7, f[ n-1 ]+f[ n-7 ], 1 ]
    Table[Sum[Binomial[n-6*i, i], {i, 0, n/7}], {n, 0, 45}] (* Adi Dani, Jun 25 2011 *)
    LinearRecurrence[{1, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
  • PARI
    x='x+O('x^66); Vec(1/(1-(x+x^7))) /* Joerg Arndt, Jun 25 2011 */

Formula

G.f.: 1/(1-x-x^7). - Simon Plouffe in his 1992 dissertation.
For positive integers n and k such that k <= n <= 7*k, and 6 divides n-k, define c(n,k) = binomial(k,(n-k)/6), and c(n,k)=0, otherwise. Then, for n >= 1, a(n) = Sum_{k=1..n} c(n,k). - Milan Janjic, Dec 09 2011
Apparently a(n) = hypergeometric([1/7-n/7, 2/7-n/7, 3/7-n/7, 4/7-n/7, 5/7-n/7, 6/7-n/7, -n/7], [1/6-n/6, 1/3-n/6, 1/2-n/6, 2/3-n/6, 5/6-n/6, -n/6], -7^7/6^6) for n >= 36. - Peter Luschny, Sep 19 2014

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000

A005708 a(n) = a(n-1) + a(n-6), with a(i) = 1 for i = 0..5.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196, 251, 322, 414, 533, 686, 882, 1133, 1455, 1869, 2402, 3088, 3970, 5103, 6558, 8427, 10829, 13917, 17887, 22990, 29548, 37975, 48804, 62721, 80608, 103598, 133146, 171121, 219925, 282646
Offset: 0

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Author

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = sum_{i=0..n/m} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
For n>=6, a(n-6) = number of compositions of n in which each part is >=6. - Milan Janjic, Jun 28 2010
Number of compositions of n into parts 1 and 6. - Joerg Arndt, Jun 24 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=6, 2*a(n-6) equals the number of 2-colored compositions of n with all parts >=6, such that no adjacent parts have the same color. - Milan Janjic, Nov 27 2011
a(n+5) equals the number of binary words of length n having at least 5 zeros between every two successive ones. - Milan Janjic, Feb 07 2015
Number of tilings of a 6 X n rectangle with 6 X 1 hexominoes. - M. Poyraz Torcuk, Mar 26 2022

References

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

Crossrefs

Programs

  • Maple
    with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 5)}, unlabeled]: seq(count(SeqSetU, size=j), j=6..59); # Zerinvary Lajos, Oct 10 2006
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 5)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=5..58); # Zerinvary Lajos, Mar 26 2008
    M := Matrix(6, (i,j)-> if j=1 and member(i,[1,6]) then 1 elif (i=j-1) then 1 else 0 fi); a:= n-> (M^(n))[1,1]; seq(a(n), n=0..60); # Alois P. Heinz, Jul 27 2008
  • Mathematica
    LinearRecurrence[{1, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
  • PARI
    x='x+O('x^66); Vec(x/(1-(x+x^6))) /* Joerg Arndt, Jun 25 2011 */

Formula

G.f.: 1/(1-x-x^6). - Simon Plouffe in his 1992 dissertation
a(n) = term (1,1) in the 6 X 6 matrix [1,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; 1,0,0,0,0,0]^n. - Alois P. Heinz, Jul 27 2008
For positive integers n and k such that k <= n <= 6*k and 5 divides n-k, define c(n,k) = binomial(k,(n-k)/5), and c(n,k)=0, otherwise. Then, for n>= 1, a(n) = sum_{k=1..n} c(n,k). - Milan Janjic, Dec 09 2011
Apparently a(n) = hypergeometric([1/6-n/6, 1/3-n/6, 1/2-n/6, 2/3-n/6, 5/6-n/6, -n/6], [1/5-n/5, 2/5-n/5, 3/5- n/5, 4/5-n/5, -n/5], -6^6/5^5) for n>=25. - Peter Luschny, Sep 19 2014

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000

A274142 Number of integers in n-th generation of tree T(1/2) defined in Comments.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 5, 8, 11, 17, 25, 37, 54, 81, 119, 177, 261, 388, 574, 851, 1260, 1868, 2767, 4101, 6077, 9006, 13347, 19781, 29315, 43448, 64392, 95436, 141444, 209636, 310705, 460501, 682519, 1011581, 1499295, 2222155, 3293534, 4881472, 7235018, 10723311, 15893460, 23556367, 34913897, 51747400
Offset: 0

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Author

Clark Kimberling, Jun 11 2016

Keywords

Comments

Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x.
Guide to related sequences:
r sequence
-1/2 A274147
-1/3 A274148
-1/4 A274149
-2/3 A274150
-3/4 A274151
-3/2 A274154
-5/2 A274155
2^(1/2) A000045 (Fibonacci numbers)
2^(1/3) A000930
2^(1/4) A003269
2^(-1/2) A274156
3^(-1/2) A274157
2^(-1/3) A274158
3^(-1/3) A274159
(-1+3i)/2 A274168

Examples

			If r = 1/2, then g(3) = {3,2r,r+1, r^2}, in which the integers are 3 and 1, so that a(3) = 2.
		

Crossrefs

Programs

  • Mathematica
    z = 18; t = Join[{{0}}, Expand[NestList[DeleteDuplicates[Flatten[Map[{# + 1, x*#} &, #], 1]] &, {1}, z]]];
    u = Table[t[[k]] /. x -> 1/2, {k, 1, z}];
    Table[Count[Map[IntegerQ, u[[k]]], True], {k, 1, z}]
    (* second program: *)
    T[0] = {0}; T[n_] := T[n] = Complement[Join[T[n-1]+1, x*T[n-1]], T[n-1]]; Reap[For[n = 0, n <= 25, n++, cnt = Count[T[n] /. x -> 1/2, Integer]; Print[n, " ", cnt]; Sow[cnt]]][[2, 1]] (* _Jean-François Alcover, Jun 14 2016 *)

Extensions

More terms from Jean-François Alcover, Jun 14 2016
More terms from Kenny Lau, Jul 04 2016

A145153 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where sequence a_k of column k is the expansion of x/((1 - x - x^4)*(1 - x)^(k - 1)).

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 1, 3, 3, 1, 1, 0, 1, 4, 6, 4, 2, 1, 0, 1, 5, 10, 10, 6, 3, 1, 0, 1, 6, 15, 20, 16, 9, 4, 1, 0, 1, 7, 21, 35, 36, 25, 13, 5, 2, 0, 1, 8, 28, 56, 71, 61, 38, 18, 7, 3, 0, 1, 9, 36, 84, 127, 132, 99, 56, 25, 10, 4, 0, 1, 10, 45, 120, 211, 259, 231, 155, 81, 35, 14, 5
Offset: 0

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Author

Alois P. Heinz, Oct 03 2008

Keywords

Comments

Each row sequence a_n (for n > 0) is produced by a polynomial of degree n-1, whose (rational) coefficients are given in row n of A145140/A145141. The coefficients *(n-1)! are given in A145142.
Each column sequence a_k is produced by a recursion, whose coefficients are given by row k of A145152.

Examples

			Square array A(n,k) begins:
  0, 0, 0,  0,  0,  0,   0, ...
  1, 1, 1,  1,  1,  1,   1, ...
  0, 1, 2,  3,  4,  5,   6, ...
  0, 1, 3,  6, 10, 15,  21, ...
  0, 1, 4, 10, 20, 35,  56, ...
  1, 2, 6, 16, 36, 71, 127, ...
		

Crossrefs

Columns 0-9 give: A017898(n-1) for n>0, A003269, A098578, A145131, A145132, A145133, A145134, A145135, A145136, A145137.
Main diagonal gives: A145138.
Antidiaginal sums give: A145139.
Numerators/denominators of polynomials for rows give: A145140/A145141.

Programs

  • Maple
    A:= proc(n, k) coeftayl (x/ (1-x-x^4)/ (1-x)^(k-1), x=0, n) end:
    seq(seq(A(n, d-n), n=0..d), d=0..13);
  • Mathematica
    a[n_, k_] := SeriesCoefficient[x/(1 - x - x^4)/(1 - x)^(k - 1), {x, 0, n}]; Table[a[n - k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 05 2013 *)

Formula

G.f. of column k: x/((1-x-x^4)*(1-x)^(k-1)).

A005710 a(n) = a(n-1) + a(n-8), with a(i) = 1 for i = 0..7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 18, 23, 29, 36, 44, 53, 64, 78, 96, 119, 148, 184, 228, 281, 345, 423, 519, 638, 786, 970, 1198, 1479, 1824, 2247, 2766, 3404, 4190, 5160, 6358, 7837, 9661, 11908, 14674, 18078, 22268, 27428, 33786, 41623
Offset: 0

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Author

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = Sum_{i=0..n/m} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
For n >= 8, a(n-8) = number of compositions of n in which each part is >= 8. - Milan Janjic, Jun 28 2010
Number of compositions of n into parts 1 and 8. - Joerg Arndt, Jun 24 2011
a(n+7) equals the number of binary words of length n having at least 7 zeros between every two successive ones. - Milan Janjic, Feb 09 2015

References

  • P. Chinn and S. Heubach, (1, k)-compositions, Congr. Numer. 164 (2003), 183-194.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A005710:=-1/(-1+z+z**8); # Simon Plouffe in his 1992 dissertation.
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 7)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=7..62); # Zerinvary Lajos, Mar 26 2008
    M := Matrix(8, (i,j)-> if j=1 and member(i,[1,8]) then 1 elif (i=j-1) then 1 else 0 fi); a := n -> (M^(n))[1,1]; seq(a(n), n=0..55); # Alois P. Heinz, Jul 27 2008
  • Mathematica
    LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
    CoefficientList[Series[1/(1-x-x^8),{x,0,60}],x] (* Harvey P. Dale, Jun 14 2016 *)
  • PARI
    x='x+O('x^66); Vec(x/(1-(x+x^8))) /* Joerg Arndt, Jun 25 2011 */

Formula

G.f.: 1/(1-x-x^8).
For positive integers n and k such that k <= n <= 8*k, and 7 divides n-k, define c(n,k) = binomial(k,(n-k)/7), and c(n,k) = 0, otherwise. Then, for n >= 1, a(n-1) = Sum_{k=1..n} c(n,k). - Milan Janjic, Dec 09 2011
Apparently a(n) = hypergeometric([1/8-n/8, 1/4-n/8, 3/8-n/8, 1/2-n/8, 5/8-n/8, 3/4-n/8, 7/8-n/8, -n/8], [1/7-n/7, 2/7-n/7, 3/7-n/7, 4/7-n/7, 5/7-n/7, 6/7-n/7, -n/7], -8^8/7^7) for n >= 49. - Peter Luschny, Sep 19 2014

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000

A001609 a(1) = a(2) = 1, a(3) = 4; thereafter a(n) = a(n-1) + a(n-3).

Original entry on oeis.org

1, 1, 4, 5, 6, 10, 15, 21, 31, 46, 67, 98, 144, 211, 309, 453, 664, 973, 1426, 2090, 3063, 4489, 6579, 9642, 14131, 20710, 30352, 44483, 65193, 95545, 140028, 205221, 300766, 440794, 646015, 946781, 1387575, 2033590, 2980371, 4367946, 6401536, 9381907
Offset: 1

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Author

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m + 1. The generating function is (x + m*x^m)/(1 - x - x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.
The sequence defined by {a(n) - 1} plays a role for the computation of A065414, A146486, A146487, and A146488 equivalent to the role of A001610 for A005596, A146482, A146483 and A146484, see the variable a_{2,n} in arXiv:0903.2514. - R. J. Mathar, Mar 28 2009
Except for n = 2, a(n) is the number of digits in n-th term of A049064. This can be derived form the T. Sillke link below. - Jianing Song, Apr 28 2019

Examples

			G.f. = x + x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 10*x^6 + 15*x^7 + 21*x^8 + ...
		

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

  • Magma
    I:=[1,1,4]; [n le 3 select I[n] else Self(n-1)+Self(n-3): n in [1..45]]; // Vincenzo Librandi, Jun 28 2015
  • Maple
    A001609:=-(1+3*z**2)/(-1+z+z**3); # Simon Plouffe in his 1992 dissertation
    f:= gfun:-rectoproc({a(n) = a(n-1) + a(n-3), a(1)=1,a(2)=1,a(3)=4},a(n),remember):
    map(f, [$1..100]); # Robert Israel, Jun 29 2015
  • Mathematica
    Table[Tr[MatrixPower[{{0, 0, 1}, {1, 0, 0}, {0, 1, 1}}, n]], {n, 1, 60}] (* Artur Jasinski, Jan 10 2007 *)
    Table[ HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -(n/3)}, {1/2 - n/2, 1 - n/2}, -(27/4)], {n, 20}] (* Alexander R. Povolotsky, Nov 21 2008 *)
    a[1] = a[2] = 1; a[3] = 4; m = 3; a[n_] := 1 + n*Sum [Binomial [n - 1 - (m - 1)*i, i - 1]/i, {i, n/m}] A001609 = Table[a[n], {n, 100}] (* Zak Seidov, Nov 21 2008 *)
    LinearRecurrence[{1, 0, 1}, {1, 1, 4}, 50] (* Vincenzo Librandi, Jun 28 2015 *)
  • PARI
    {a(n) = if( n<1, n=-n; polcoeff( (3 + x^2) / (1 + x^2 - x^3) + x * O(x^n), n), polcoeff( x * (1 + 3*x^2) / (1 - x - x^3) + x * O(x^n), n))}; /* Michael Somos, Aug 15 2016 */
    

Formula

G.f.: x*(1 + 3*x^2)/(1 - x - x^3).
a(n) = trace of successive powers of matrix ({{0,0,1},{1,0,0},{0,1,1}})^n. - Artur Jasinski, Jan 10 2007
a(n) = A000930(n) + 3*A000930(n-2). - R. J. Mathar, Nov 16 2007
Logarithmic derivative of Narayana's cows sequence A000930. - Paul D. Hanna, Oct 28 2012
a(n) = w1^n + w2^n + w3^n, where w1,w2,w3 are the roots of the cubic: (-1 - x^2 + x^3), see A092526. - Gerry Martens, Jun 27 2015

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000
More terms from Michael Somos, Oct 03 2002
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A017817 a(n) = a(n-3) + a(n-4), with a(0)=1, a(1)=a(2)=0, a(3)=1.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 1, 2, 1, 1, 3, 3, 2, 4, 6, 5, 6, 10, 11, 11, 16, 21, 22, 27, 37, 43, 49, 64, 80, 92, 113, 144, 172, 205, 257, 316, 377, 462, 573, 693, 839, 1035, 1266, 1532, 1874, 2301, 2798, 3406, 4175, 5099, 6204, 7581, 9274, 11303, 13785, 16855, 20577, 25088
Offset: 0

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Author

Keywords

Comments

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={0,1}. - Vladimir Baltic, Mar 07 2012
Number of compositions (ordered partitions) of n into parts 3 and 4.
For n>=2, a(n-2) is the number of ways to tile the 1xn board with dominoes and squares (ie. monominoes) such that there are either one or two squares between dominoes, no squares at either end of the board, and there is at least one domino. - Enrique Navarrete, Sep 01 2024
For n>=3, a(n-3) is the number of ways to tile the 1xn board with triominoes (ie. size 1x3) and squares (ie. size 1x1) such that there are either none or one squares between triominoes, no squares at either end of the board, and there is at least one triomino. - Enrique Navarrete, Sep 07 2024

Crossrefs

A003269(n) = a(-4-n)(-1)^n.

Programs

  • GAP
    a:=[1,0,0,1];; for n in [5..60] do a[n]:=a[n-3]+a[n-4]; od; a; # G. C. Greubel, Mar 05 2019
  • Magma
    I:=[1,0,0,1]; [n le 4 select I[n] else Self(n-3) +Self(n-4): n in [1..60]]; // G. C. Greubel, Mar 05 2019
    
  • Mathematica
    LinearRecurrence[{0,0,1,1}, {1,0,0,1}, 60] (* G. C. Greubel, Mar 05 2019 *)
  • PARI
    a(n)=polcoeff(if(n<0,(1+x)/(1+x-x^4),1/(1-x^3-x^4)) +x*O(x^abs(n)), abs(n))
    
  • Sage
    (1/(1-x^3-x^4)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Mar 05 2019
    

Formula

G.f.: 1/(1-x^3-x^4).
a(n)/a(n-1) tends to A060007. - Gary W. Adamson, Oct 22 2006
a(n) = Sum_{k=0..floor(n/3)} binomial(k,n-3*k). - Seiichi Manyama, Mar 06 2019

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 17 1999

A017898 Expansion of (1-x)/(1-x-x^4).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272, 8657, 11949, 16493, 22765, 31422, 43371, 59864, 82629, 114051, 157422, 217286, 299915, 413966, 571388, 788674
Offset: 0

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Author

Keywords

Comments

A Lamé sequence of higher order.
Essentially the same as A003269, which has much more information.
Number of compositions of n into parts >= 4. - Joerg Arndt, Aug 13 2012

Crossrefs

For Lamé sequences of orders 1 through 9 see A000045, A000930, this one, and A017899-A017904.

Programs

  • Maple
    f := proc(r) local t1,i; t1 := []; for i from 1 to r do t1 := [op(t1),0]; od: for i from 1 to r+1 do t1 := [op(t1),1]; od: for i from 2*r+2 to 50 do t1 := [op(t1),t1[i-1]+t1[i-1-r]]; od: t1; end; # set r = order
    a:= n-> (Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 0$2, 1][i] else 0 fi)^n)[4,4]: seq(a(n), n=0..42); # Alois P. Heinz, Aug 04 2008
  • Mathematica
    LinearRecurrence[{1, 0, 0, 1}, {1, 0, 0, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 11 2012 *)
    CoefficientList[Series[(1-x)/(1-x-x^4),{x,0,50}],x] (* Harvey P. Dale, Sep 12 2019 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,0,0,1]^n*[1;0;0;0])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

Formula

a(n) = a(n-1) + a(n-4). - R. J. Mathar, Mar 06 2008
G.f.: 1/(1-sum(k>=4, x^k)). - Joerg Arndt, Aug 13 2012
Apparently a(n) = hypergeometric([1-1/4*n, 5/4-1/4*n, 3/2-1/4*n, 7/4-1/4*n],[4/3-1/3*n, 5/3-1/3*n, 2-1/3*n], -4^4/3^3) for n>=13. - Peter Luschny, Sep 18 2014
a(n) = A003269(n+1)-A003269(n). - R. J. Mathar, Jun 10 2018

A086106 Decimal expansion of positive root of x^4 - x^3 - 1 = 0.

Original entry on oeis.org

1, 3, 8, 0, 2, 7, 7, 5, 6, 9, 0, 9, 7, 6, 1, 4, 1, 1, 5, 6, 7, 3, 3, 0, 1, 6, 9, 1, 8, 2, 2, 7, 3, 1, 8, 7, 7, 8, 1, 6, 6, 2, 6, 7, 0, 1, 5, 5, 8, 7, 6, 3, 0, 2, 5, 4, 1, 1, 7, 7, 1, 3, 3, 1, 2, 1, 1, 2, 4, 9, 5, 7, 4, 1, 1, 8, 6, 4, 1, 5, 2, 6, 1, 8, 7, 8, 6, 4, 5, 6, 8, 2, 4, 9, 0, 3, 5, 5, 0, 9, 3, 7
Offset: 1

Views

Author

Eric W. Weisstein, Jul 09 2003

Keywords

Comments

Also the growth constant of the Fibonacci 3-numbers A003269 [Stakhov et al.]. - R. J. Mathar, Nov 05 2008

Examples

			1.380277569...
The four solutions are the present one, -A230151, and the two complex ones 0.2194474721... - 0.9144736629...*i and its complex conjugate. - _Wolfdieter Lang_, Aug 19 2022
		

Crossrefs

Cf. -A230151 (other real root).
Cf. A060006.

Programs

  • Mathematica
    RealDigits[Root[ -1 - #1^3 + #1^4 &, 2], 10, 110][[1]]
  • PARI
    polrootsreal( x^4-x^3-1)[2] \\ Charles R Greathouse IV, Apr 14 2014
    
  • PARI
    default(realprecision, 20080); x=solve(x=1, 2, x^4 - x^3 - 1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b086106.txt", n, " ", d));  \\ Iain Fox, Oct 23 2017

Formula

Equals (1 + (A^2 + sqrt(A^4 - 16*u*A^2 + 2*A))/A)/4 with A = sqrt(8*u + 3/2), u = (-(Bp/2)^(1/3) + (Bm/2)^(1/3)*(1 - sqrt(3)*i)/2 - 3/8)/6, with Bp = 27 + 3*sqrt(3*283), Bm = 27 - 3*sqrt(3*283), and i = sqrt(-1). (Standard computation of a quartic.) The other (negative) real root -A230151 is obtained by using in the first formula the negative square root. The other two complex roots are obtained by replacing A by -A in these two formulas. - Wolfdieter Lang, Aug 19 2022

A376033 Number A(n,k) of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 4, 5, 16, 1, 2, 3, 6, 8, 32, 1, 2, 4, 4, 9, 13, 64, 1, 2, 3, 8, 6, 15, 21, 128, 1, 2, 4, 5, 12, 9, 25, 34, 256, 1, 2, 3, 6, 7, 18, 13, 40, 55, 512, 1, 2, 4, 4, 8, 11, 27, 19, 64, 89, 1024, 1, 2, 3, 8, 5, 11, 16, 45, 28, 104, 144, 2048
Offset: 0

Views

Author

Alois P. Heinz, Sep 09 2024

Keywords

Comments

Also the number of subsets of [n] avoiding distance (i+1) between elements if the i-th bit is set in the binary representation of k. A(6,3) = 13: {}, {1}, {2}, {3}, {4}, {5}, {6}, {1,4}, {1,5}, {1,6}, {2,5}, {2,6}, {3,6}.
Each column sequence satisfies a linear recurrence with constant coefficients.
The sequence of row n is periodic with period A011782(n) = ceiling(2^(n-1)).

Examples

			A(6,6) = 17: 000000, 000001, 000010, 000011, 000100, 000110, 001000, 001100, 010000, 010001, 011000, 100000, 100001, 100010, 100011, 110000, 110001 because 6 = 110_2 and no two "1" digits have distance 2 or 3.
A(6,7) = 10: 000000, 000001, 000010, 000100, 001000, 010000, 010001, 100000, 100001, 100010.
A(7,7) = 14: 0000000, 0000001, 0000010, 0000100, 0001000, 0010000, 0010001, 0100000, 0100001, 0100010, 1000000, 1000001, 1000010, 1000100.
Square array A(n,k) begins:
     1,  1,   1,  1,   1,  1,  1,  1,   1,  1, ...
     2,  2,   2,  2,   2,  2,  2,  2,   2,  2, ...
     4,  3,   4,  3,   4,  3,  4,  3,   4,  3, ...
     8,  5,   6,  4,   8,  5,  6,  4,   8,  5, ...
    16,  8,   9,  6,  12,  7,  8,  5,  16,  8, ...
    32, 13,  15,  9,  18, 11, 11,  7,  24, 11, ...
    64, 21,  25, 13,  27, 16, 17, 10,  36, 17, ...
   128, 34,  40, 19,  45, 25, 27, 14,  54, 25, ...
   256, 55,  64, 28,  75, 37, 41, 19,  81, 37, ...
   512, 89, 104, 41, 125, 57, 60, 26, 135, 57, ...
		

Crossrefs

Columns k=0-20 give: A000079, A000045(n+2), A006498(n+2), A000930(n+2), A006500, A130137, A079972(n+3), A003269(n+4), A031923(n+1), A263710(n+1), A224809(n+4), A317669(n+4), A351873, A351874, A121832(n+4), A003520(n+4), A208742, A374737, A375977, A375980, A375978.
Rows n=0-2 give: A000012, A007395(k+1), A010702(k+1).
Main diagonal gives A376091.
A(n,2^k-1) gives A141539.
A(2^n-1,2^n-1) gives A376697.
A(n,2^k) gives A209435.

Programs

  • Maple
    h:= proc(n) option remember; `if`(n=0, 1, 2^(1+ilog2(n))) end:
    b:= proc(n, k, t) option remember; `if`(n=0, 1, add(`if`(j=1 and
          Bits[And](t, k)>0, 0, b(n-1, k, irem(2*t+j, h(k)))), j=0..1))
        end:
    A:= (n, k)-> b(n, k, 0):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
  • PARI
    step(v,b)={vector(#v, i, my(j=(i-1)>>1); if(bittest(i-1,0), if(bitand(b,j)==0, v[1+j], 0), v[1+j] + v[1+#v/2+j]));}
    col(n,k)={my(v=vector(2^(1+logint(k,2))), r=vector(1+n)); v[1]=r[1]=1; for(i=1, n, v=step(v,k); r[1+i]=vecsum(v)); r}
    A(n,k)=if(k==0, 2^n, col(n,k)[n+1]) \\ Andrew Howroyd, Oct 03 2024

Formula

A(n,k) = A(n,k+ceiling(2^(n-1))).
A(n,ceiling(2^(n-1))-1) = n+1.
A(n,ceiling(2^(n-2))) = ceiling(3*2^(n-2)) = A098011(n+2).
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