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 11-17 of 17 results.

A129720 Number of 0's in odd position in all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

0, 1, 1, 4, 5, 14, 19, 46, 65, 145, 210, 444, 654, 1331, 1985, 3926, 5911, 11434, 17345, 32960, 50305, 94211, 144516, 267384, 411900, 754309, 1166209, 2116936, 3283145, 5914310, 9197455, 16458034, 25655489, 45638101, 71293590, 126159156
Offset: 0

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Author

Emeric Deutsch, May 13 2007

Keywords

Examples

			a(4)=5 because in 1110, 1111, 110'1, 1010, 1011, 0'110, 0'111 and 0'10'1 one has altogether five 0's in odd position (marked by ').
		

Crossrefs

Programs

  • Maple
    g:=z*(1-z^2)/(1-z-z^2)^2/(1+z-z^2): gser:=series(g,z=0,43): seq(coeff(gser,z,n),n=0..40);

Formula

G.f.: z(1-z^2)/((1-z-z^2)^2*(1+z-z^2)).
a(2n) = a(2n-1) + a(2n-2) (n >= 1).
a(2n-1) = A030267(n).
a(2n) = A129722(2n) = A001870(n-1).
a(n) = Sum_{k=0..ceiling(n/2)} k*A129719(n,k).

A131321 Triangle read by rows: A168561^2.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 5, 0, 6, 0, 1, 0, 14, 0, 8, 0, 1, 13, 0, 27, 0, 10, 0, 1, 0, 46, 0, 44, 0, 12, 0, 1, 34, 0, 107, 0, 65, 0, 14, 0, 1, 0, 145, 0, 204, 0, 90, 0, 16, 0, 1, 89, 0, 393, 0, 345, 0, 119, 0, 18, 0, 1, 0, 444, 0, 854, 0, 538, 0, 152, 0, 20, 0, 1
Offset: 0

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Author

Gary W. Adamson, Jun 28 2007

Keywords

Comments

Left border, nonzero terms = odd indexed Fibonacci numbers: (1, 2, 5, 13, ...). Next column, nonzero terms = A030267: (1, 4, 14, 46, 145, ...). Row sums = A131322: (1, 1, 3, 5, 12, 23, 51, ...).
Riordan array (f(x),x*f(x)) where f(x) = (1-x^2)/(1-3*x^2+x^4). Aerated version of triangle in A188137. - Philippe Deléham, Jan 26 2012

Examples

			First few rows of the triangle are:
   1;
   0,  1;
   2,  0,  1;
   0,  4,  0,  1;
   5,  0,  6,  0,  1;
   0, 14,  0,  8,  0,  1;
  13,  0, 27,  0, 10,  0,  1;
  ...
		

Crossrefs

Programs

  • Maple
    F:= (n, k)-> coeff(combinat[fibonacci](n+1, x), x, k):
    T:= (n, k)-> add(F(n, j)*F(j, k), j=0..n):
    seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Dec 12 2019

Formula

A168561 squared, as an infinite lower triangular matrix.

A279282 Self-composition of the cubes; g.f.: A(x) = G(G(x)), where G(x) = g.f. of A000578.

Original entry on oeis.org

0, 1, 16, 182, 1720, 14149, 106944, 760463, 5160488, 33756514, 214369376, 1328496947, 8065970016, 48125315989, 282851349184, 1640791635086, 9409099218712, 53408767286521, 300417148670400, 1676056809217283, 9282172245277448, 51062759750186170, 279196558362482192, 1518068927980989575
Offset: 0

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Author

Ilya Gutkovskiy, Dec 09 2016

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x (1 - x)^4 (1 + 4 x + x^2) (1 - 4 x + 29 x^2 - 84 x^3 + 152 x^4 - 84 x^5 + 29 x^6 - 4 x^7 + x^8)/((1 + x^2)^4 (1 - 5 x + x^2)^4), {x, 0, 23}], x]
    LinearRecurrence[{20,-158,640,-1553,2920,-4806,5700,-6820,5700,-4806,2920,-1553,640,-158,20,-1},{0,1,16,182,1720,14149,106944,760463,5160488,33756514,214369376,1328496947,8065970016,48125315989,282851349184,1640791635086},30] (* Harvey P. Dale, Sep 27 2024 *)

Formula

G.f.: x*(1 - x)^4*(1 + 4*x + x^2)*(1 - 4*x + 29*x^2 - 84*x^3 + 152*x^4 - 84*x^5 + 29*x^6 - 4*x^7 + x^8)/((1 + x^2)^4*(1 - 5*x + x^2)^4).

A307415 Total number of parts in all symmetric m-color cyclic compositions of n (that is, the total number of parts in all achiral cyclic compositions of n where a part with size m can be colored with one of m colors).

Original entry on oeis.org

1, 4, 10, 26, 53, 116, 215, 434, 766, 1480, 2539, 4776, 8045, 14864, 24722, 45094, 74305, 134236, 219619, 393790, 640646, 1141844, 1849175, 3279696, 5291353, 9346396, 15031450, 26458994, 42438221, 74479940, 119182319, 208629386, 333170830, 581904544, 927617347, 1616924664
Offset: 1

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Author

Petros Hadjicostas, Jun 24 2019

Keywords

Comments

Cyclic compositions of a positive integer n are equivalence classes of ordered partitions of n such that two such partitions are equivalent if one can be obtained from the other by rotation. These were first studied by Sommerville (1909).
Symmetric cyclic compositions or circular palindromes or achiral cyclic compositions are those cyclic compositions that have at least one axis of symmetry. They were also studied by Sommerville (1909, pp. 301-304).
Let c = (c(m): m >= 1) be the input sequence and let b_k = (b_k(n): n >= 1) be the output sequence under Bower's CPAL[k] (circular palindrome with k boxes) transform of c; that is, b_k(n) = (CPAC[k] c)n for n >= 1. Hence, b_k(n) is the number of symmetric cyclic compositions of n with k parts, where a part of size m can be colored with one of c(m) colors. If C(x) = Sum{m >= 1} c(m)*x^m is the g.f. of the input sequence c, then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((CPAL[k] c): k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n,k >= 1} b_k(n)*x^n*y^k = (1 + y*C(x))^2/(2 * (1 - y^2*C(x^2))) - (1/2).
Here, Sum_{k=1..n} k*b_k(n) is the total number of parts in all symmetric colored cyclic compositions of n (where the coloring of parts is done according to the input sequence c). To find the g.f. of the sequence (Sum_{k=1..n} k*b_k(n): n >= 1), we differentiate the above bivariate g.f. (i.e., Sum_{n,k >= 1} b_k(n)*x^n*y^k) w.r.t. y and set y = 1. We get (1 + C(x)) * (C(x) + C(x^2))/(1 - C(x^2))^2.
For the current sequence, the input sequence is c(m) = m for m >= 1, and we are dealing with the so-called "m-color" compositions. m-color linear compositions were studied by Agarwal (2000), whereas m-color cyclic compositions were studied by Gibson (2017) and Gibson et al. (2018).
Thus, for the current sequence, a(n) is the total number of parts in all symmetric (achiral) m-color cyclic compositions of n (where a part of size m may be colored with one of m colors). Since C(x) = x/(1 - x)^2, we get that (1 + C(x)) * (C(x) + C(x^2))/(1 - C(x^2))^2 = (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2).
The roots of the denominator (x^2 + x - 1)^2 * (x^2 - x - 1)^2 are phi, -phi, 1/phi, -1/phi, where phi = (1 + sqrt(5))/2 = A001622 is the golden ratio. Each of these roots is double, so a(n) = (c_1 + d_1*n) * phi^n + (c_2 + d_2*n) * (-phi)^n + (c_3 + d_3*n) * (1/phi)^n + (c_4 + d_4*n) * (-1/phi)^n, and the constants c_i, d_i (i = 1, 2, 3, 4) can be determined from the first eight terms of (a(n): n >= 1). Since, however, the Fibonacci numbers (A000045(n): n >= 0) can be expressed in terms of phi, after some tedious algebra, we may derive the formula given below.

Examples

			We have a(1) = 1 because we only have one symmetric cyclic composition of n = 1, namely 1_1, which has 1 part.
We have a(2) = 4 because we have the following colored achiral cyclic compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 10 because we have the following colored achiral cyclic compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 3 = 10.
We have a(4) = 26 because we have the following colored achiral cyclic compositions of n = 4: 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1; hence, a(4) = 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 4 = 26.
We have a(5) = 53 because we have the following colored achiral cyclic compositions of n = 5:
(i) With one part: 5_1, 5_2, 5_3, 5_4, 5_5; i.e., a total of 5 parts.
(ii) With two parts: 1_1 + 4_1, 1_1 + 4_2, 1_1 + 4_3, 1_1 + 4_4, 2_1 + 3_1, 2_1 + 3_2, 2_1 + 3_3, 2_2 + 3_1, 2_2 + 3_2, 2_2 + 3_3; i.e., a total of 20 parts.
(iii) With three parts: 1_1 + 3_1 + 1_1, 1_1 + 3_2 + 1_1, 1_1 + 3_3 + 1_1, 2_1 + 1_1 + 2_1, 2_2 + 1_1 + 2_2; i.e., a total of 15 parts.
(iv) With four parts: 1_1 + 1_1 + 2_1 + 1_1, 1_1 + 1_1 + 2_2 + 1_1 (here, the axis of symmetry passes through one of the 1's and through 2); i.e., a total of 8 parts.
(v) With five parts: 1_1 + 1_1 + 1_1 + 1_1 + 1_1; i.e., a total of 5 parts.
Thus, a(5) = 5 + 20 + 15 + 8 + 5 = 53.
		

Crossrefs

Formula

a(n) = (n/5) * (11*F(n) + 7*F(n-1)) + (-1)^n * (n/5) * (-4*F(n) + 7*F(n-1)) - (2/5) * (5*F(n+1) + 3*F(n)) - (-1)^n * (2/5) * (3*F(n) - 5*F(n-1)) for n >= 1, where F(n) = A000045(n) is the n-th Fibonacci number.
G.f.: (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2).

A377857 Number of subwords of the form UUUD in nondecreasing Dyck paths of length 2n.

Original entry on oeis.org

0, 0, 0, 1, 5, 18, 60, 191, 589, 1775, 5257, 15360, 44394, 127171, 361595, 1021693, 2871245, 8031246, 22372344, 62096135, 171797257, 473928875, 1304007889, 3579517116, 9804791910, 26804181643, 73145473655, 199276078201, 542076556949, 1472491141770, 3994615719732
Offset: 0

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Author

Rigoberto Florez, Nov 09 2024

Keywords

Comments

A Dyck path is nondecreasing if the y-coordinates of its valleys form a nondecreasing sequence.

Crossrefs

Programs

  • Mathematica
    Table[If[n<3,0,n Fibonacci[2n-5]-LucasL[2n-6]], {n,0,30}]

Formula

a(n) = n*F(2*n-5) - L(2*n-6) for n>=3, where F(n) = A000045(n) and L(n) = A000032(n).
G.f.: x^3*(1 - x)^2*(1 + x)/(1 - 3*x + x^2)^2.
a(n) = A317408(n-2)-A317408(n-3) = A030267(n-2)+A030267(n-3). - R. J. Mathar, Dec 16 2024

A276644 Self-composition of the repunits; g.f.: A(x) = G(G(x)), where G(x) = g.f. of A002275.

Original entry on oeis.org

0, 1, 22, 464, 9658, 199666, 4112922, 84558014, 1736623658, 35646098566, 731452470122, 15006822709814, 307859627711658, 6315326642698966, 129547066718721322, 2657377349777550614, 54509922224486463658, 1118139793621467673366, 22935894163202834676522, 470473020119757115115414
Offset: 0

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Author

Ilya Gutkovskiy, Sep 08 2016

Keywords

Crossrefs

Cf. A030267 (self-composition of the natural numbers), A030279 (self-composition of the squares), A030280 (self-composition of the triangular numbers).

Programs

  • Magma
    I:=[0,1,22,464]; [n le 4 select I[n] else 33*Self(n-1)-272*Self(n-2)+330*Self(n-3)-100*Self(n-4): n in [1..20]]; // Vincenzo Librandi, Sep 09 2016
  • Mathematica
    LinearRecurrence[{33, -272, 330, -100}, {0, 1, 22, 464}, 20]
  • PARI
    concat(0, Vec(x*(1-x)*(1-10*x)/((1-21*x+10*x^2)*(1-12*x+10*x^2)) + O(x^99))) \\ Altug Alkan, Sep 08 2016
    

Formula

O.g.f.: x*(1 - x)*(1 - 10*x)/((1 - 21*x + 10*x^2)*(1 - 12*x + 10*x^2)).
a(n) = 33*a(n-1) - 272*a(n-2) + 330*a(n-3) - 100*a(n-4) for n > 3.
a(n) = ((6 - sqrt(26))^n - (6 + sqrt(26))^n)/(18*sqrt(26)) + 10*(((21 + sqrt(401))/2)^n - ((21 - sqrt(401))/2)^n)/(9*sqrt(401)).
A000035(a(n)) = A063524(n).

A308817 Total number of parts in all m-color dihedral compositions of n (that is, the total number of parts in all dihedral compositions of n where a part of size m may be colored with one of m colors).

Original entry on oeis.org

1, 4, 10, 26, 56, 138, 299, 726, 1686, 4158, 10130, 25678, 64725, 166538, 428456, 1112226, 2888604, 7533750, 19653903, 51367462, 134277878, 351284164, 919080550, 2405427698, 6295780309, 16480373968, 43141303978, 112939105716, 295664584064, 774042041090, 2026429360115, 5305210333758
Offset: 1

Views

Author

Petros Hadjicostas, Jun 26 2019

Keywords

Comments

Cyclic compositions of a positive integer n are equivalence classes of ordered partitions of n such that two such partitions are equivalent iff one can be obtained from the other by rotation. These were first studied by Sommerville (1909).
Symmetric cyclic compositions or circular palindromes or achiral cyclic compositions are those cyclic compositions that have at least one axis of symmetry. They were also studied by Sommerville (1909, pp. 301-304).
Dihedral compositions of a positive integer n are equivalence classes of ordered partitions of n such that two such partitions are equivalent iff one can be obtained from the other by rotation or reflection. See, for example, Knopfmacher and Robbins (2013).
The number of dihedral compositions of n (of any kind) is the average of the corresponding numbers of cyclic and symmetric cyclic compositions of n.
Let c = (c(m): m >= 1) be the input sequence and let b_k = (b_k(n): n >= 1) be the output sequence under Bower's DIK[k] ("bracelet, indistinct, unlabeled" with k boxes) transform of c; that is, b_k(n) = (DIK[k] c)n for n >= 1. Hence, b_k(n) is the number of dihedral compositions of n with k parts, where a part of size m can be colored with one of c(m) colors. If C(x) = Sum{m >= 1} c(m)*x^m is the g.f. of the input sequence c, then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((DIK[k] c): k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n,k >= 1} b_k(n)*x^n*y^k = (1 + y * C(x))^2/(4 * (1 - y^2 * C(x^2))) - (1/4) - (1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - y^d * C(x^d)).
Here, Sum_{k=1..n} k*b_k(n) is the total number of parts in all colored dihedral compositions of n (where the coloring of parts is done according to the input sequence c). To find the g.f. of the sequence (Sum_{k=1..n} k*b_k(n): n >= 1), we differentiate the above bivariate g.f. (i.e., Sum_{n,k >= 1} b_k(n)*x^n*y^k) w.r.t. y and set y = 1. We get (1 + C(x)) * (C(x) + C(x^2))/(2 * (1 - C(x^2))^2) + (1/2) * Sum_{d >= 1} phi(d) * C(x^d)/(1 - C(x^d)).
For the current sequence, the input sequence is c(m) = m for m >= 1, and we are dealing with the so-called "m-color" compositions. m-color linear compositions were studied by Agarwal (2000), whereas m-color cyclic compositions were studied by Gibson (2017) and Gibson et al. (2018).
Thus, for the current sequence, a(n) is the total number of parts in all m-color dihedral compositions of n (where a part of size m may be colored with one of m colors). Since C(x) = x/(1 - x)^2, we get that (1 + C(x)) * (C(x) + C(x^2))/(2 * (1 - C(x^2))^2) + (1/2) * Sum_{d >= 1} phi(d) * C(x^d)/(1 - C(x^d)) = (1/2) * (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2) + (1/2) * Sum_{s >= 1} phi(s) * x^s/(1 - 3*x^s + x^(2*s)).

Examples

			We have a(1) = 1 because 1_1 is the only m-color dihedral composition of n = 1 and the total number of parts is 1.
We have a(2) = 4 because 2_1, 2_2, 1_1 + 1_1 are all the m-color dihedral compositions of 2 and the total number of parts is 1 + 1 + 2 = 4.
We have a(3) = 10 because 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1 are all the m-color dihedral compositions of n = 3 and the total number of parts is 1 + 1 + 1 + 2 + 2 + 3 = 10.
We have a(4) = 26 because 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1 are all the m-color dihedral compositions of n = 4 and the total number of parts is 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 4 = 26.
Note that a(n) = A307415(n) = A308723(n) for n = 1, 2, 3, 4 because all cyclic compositions of n when 1 <= n <= 4 are symmetric as well (and thus are dihedral).
For n = 5, we have A307415(5) = 53 <> 59 = A308723(5), in which case, a(n) = (53 + 59)/2 = 56. For example, 1_1 + 2_1 + 3_1 is a dihedral composition of n = 5, but it is not symmetric, so it corresponds to two (inequivalent) cyclic compositions: 1_1 + 2_1 + 3_1 and 3_1 + 2_1 + 1_1. Similarly, 1_1 + 2_1 + 2_2 is a dihedral composition of n = 5, but it is not symmetric, so it corresponds to two (inequivalent) cyclic compositions: 1_1 + 2_1 + 2_2 and 2_2 + 2_1 + 1_1.
		

Crossrefs

Formula

a(n) = (A307415(n) + A308723(n))/2 for n >= 1.
a(n) = (n/10) * (11*F(n) + 7*F(n-1)) + (-1)^n * (n/10) * (-4*F(n) + 7*F(n-1)) - (1/5) * (5*F(n+1) + 3*F(n)) - (-1)^n * (1/5) * (3*F(n) - 5*F(n-1)) + (1/2) * Sum_{s|n} phi(n/s) * F(2*s) for n >= 1, where F(n) = A000045(n) (Fibonacci numbers).
G.f.: (1/2) * (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2) + (1/2) * Sum_{s >= 1} phi(s) * x^s/(1 - 3*x^s + x^(2*s)).
Previous Showing 11-17 of 17 results.