A004146 -id:A004146 - cp's OEIS frontend

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4,0,1).

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

More terms from Michel Marcus, Feb 21 2025

A249450 a(n) = Fibonacci(2*n) - 2.

Original entry on oeis.org

-2, -1, 1, 6, 19, 53, 142, 375, 985, 2582, 6763, 17709, 46366, 121391, 317809, 832038, 2178307, 5702885, 14930350, 39088167, 102334153, 267914294, 701408731, 1836311901, 4807526974, 12586269023, 32951280097, 86267571270, 225851433715, 591286729877

Offset: 0

Vincenzo Librandi, Dec 15 2014

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A004146.

Magma
```
[Fibonacci(2*n)-2: n in [0..40]];
```

Table[Fibonacci[2 n] - 2, {n, 0, 40}] (* or *) CoefficientList[Series[(-2 + 7 x - 3 x^2) / (1 - 4 x + 4 x^2 - x^3), {x, 0, 40}], x]
LinearRecurrence[{4, -4, 1}, {-2, -1, 1}, 30] (* Robert G. Wilson v, Dec 19 2014 *)

Vec((-2+7*x-3*x^2)/(1-4*x+4*x^2-x^3) + O(x^30)) \\ Colin Barker, Nov 03 2016

G.f.: (-2+7*x-3*x^2)/(1-4*x+4*x^2-x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3) for n>2.
a(n) = 3*a(n-1) - a(n-2) + 2.
a(n) = (-2-((3-sqrt(5))/2)^n/sqrt(5)+((3+sqrt(5))/2)^n/sqrt(5)). - Colin Barker, Nov 03 2016
E.g.f.: 2*exp(3*x/2)*sinh(sqrt(5)*x/2)/sqr(5) - 2*exp(x). - Stefano Spezia, Jun 02 2024

A158753 Triangle T(n, k) = A000032(2*(n-k) + 1), read by rows.

Original entry on oeis.org

1, 4, 1, 11, 4, 1, 29, 11, 4, 1, 76, 29, 11, 4, 1, 199, 76, 29, 11, 4, 1, 521, 199, 76, 29, 11, 4, 1, 1364, 521, 199, 76, 29, 11, 4, 1, 3571, 1364, 521, 199, 76, 29, 11, 4, 1, 9349, 3571, 1364, 521, 199, 76, 29, 11, 4, 1, 24476, 9349, 3571, 1364, 521, 199, 76, 29, 11, 4, 1

Offset: 2

Roger L. Bagula and Gary W. Adamson, Mar 25 2009

			Triangle begins as:
     1;
     4,   1;
    11,   4,   1;
    29,  11,   4,  1;
    76,  29,  11,  4,  1;
   199,  76,  29, 11,  4,  1;
   521, 199,  76, 29, 11,  4,  1;
  1364, 521, 199, 76, 29, 11,  4,  1;

H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp. 159-162.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000032, A002878, A004146, A158786.

[Lucas(2*n-2*k+1): k in [2..n], n in [2..16]]; // G. C. Greubel, Dec 06 2021

Table[LucasL[2*(n-k) + 1], {n, 2, 16}, {k, 2, n}]//Flatten (* G. C. Greubel, Dec 06 2021 *)

flatten([[lucas_number2(2*(n-k)+1, 1, -1) for k in (2..n)] for n in (2..16)]) # G. C. Greubel, Dec 06 2021

T(n, k) = 5*e(n, k), where e(n,k) = (e(n-1, k)*e(n, k-1) + 1)/e(n-1, k-1), and e(n, 0) = GoldenRatio^(n) + GoldenRatio^(-n).
Sum_{k=0..n} T(n, k) = A004146(n-1).
T(n, k) = A000032(2*(n-k) + 1). - G. C. Greubel, Dec 06 2021

Edited by G. C. Greubel, Dec 06 2021

A158786 Irregular triangle T(n, k) = A000032(n-2k+1) if (n-2k) mod 2 = 0, otherwise 25A000032(n-2k), read by rows.

Original entry on oeis.org

1, 25, 4, 1, 100, 25, 11, 4, 1, 275, 100, 25, 29, 11, 4, 1, 725, 275, 100, 25, 76, 29, 11, 4, 1, 1900, 725, 275, 100, 25, 199, 76, 29, 11, 4, 1, 4975, 1900, 725, 275, 100, 25, 521, 199, 76, 29, 11, 4, 1, 13025, 4975, 1900, 725, 275, 100, 25, 1364, 521, 199, 76, 29, 11, 4, 1, 34100, 13025, 4975, 1900, 725, 275, 100, 25

Offset: 2

Roger L. Bagula and Gary W. Adamson, Mar 26 2009

			Irregular triangle begins as:
     1;
    25;
     4,    1;
   100,   25;
    11,    4,   1;
   275,  100,  25;
    29,   11,   4,   1;
   725,  275, 100,  25;
    76,   29,  11,   4,  1;
  1900,  725, 275, 100, 25;
   199,   76,  29,  11,  4,   1;
  4975, 1900, 725, 275, 100, 25;

H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp. 159-162.

G. C. Greubel, Rows n = 0..50 of the irregular triangle, flattened

Cf. A000032, A002878, A004146, A158753.

(* First program *)
e[n_, 0]:= Sqrt[5]*(GoldenRatio^(n) + GoldenRatio^(-n));
e[n_, k_]:= If[k>n-1, 0, (e[n-1, k]*e[n, k-1] +1)/e[n-1, k-1]];
T[n_,k_]:= 5*Rationalize[N[e[n, k]]];
Table[T[n, k], {n, 2, 16}, {k, Mod[n, 2] +1, n-1,2}]//Flatten
(* Second program *)
f[n_]:= f[n]= If[EvenQ[n], LucasL[n-1], 25*LucasL[n-2]];
T[n_, k_]:= f[n-2*k];
Table[T[n, k], {n, 2, 16}, {k, 0, (n-2)/2}]//Flatten (* G. C. Greubel, Dec 06 2021 *)

def A158786(n,k): return lucas_number2(n-2*k-1,1,-1) if ((n-2*k)%2==0) else 25*lucas_number2(n-2*k-2,1,-1)
flatten([[A158786(n,k) for k in (0..((n-2)//2))] for n in (2..16)]) # G. C. Greubel, Dec 06 2021

T(n, k) = 5*e(n, k), where e(n,k) = (e(n-1, k)*e(n, k-1) + 1)/e(n-1, k-1), and e(n, 0) = sqrt(5)*(GoldenRatio^(n) + GoldenRatio^(-n)).
From G. C. Greubel, Dec 06 2021: (Start)
T(n, k) = A000032(n-2*k+1) if (n-2*k) mod 2 = 0, otherwise 25*A000032(n-2*k).
Sum_{k=0..floor(n/2)} T(n, k) = A000032(n) - 2 if (n mod 2 = 0), otherwise 25*(A000032(n-1) - 2). (End)

Edited by G. C. Greubel, Dec 06 2021

A256233 a(n) = L(2*n+1) - 2, where L is A000032.

Original entry on oeis.org

-1, 2, 9, 27, 74, 197, 519, 1362, 3569, 9347, 24474, 64077, 167759, 439202, 1149849, 3010347, 7881194, 20633237, 54018519, 141422322, 370248449, 969323027, 2537720634, 6643838877, 17393795999, 45537549122, 119218851369, 312119004987, 817138163594

Offset: 0

Vincenzo Librandi, Mar 20 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A004146.

Magma
```
[Lucas(n)-2: n in [1..70 by 2]];
```

Table[LucasL[n] - 2, {n, 1, 70, 2}] (* or *) LinearRecurrence[{4, -4, 1}, {-1, 2, 9}, 40]

Vec((-1+6*x-3*x^2)/((1-x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Nov 03 2016

L(n) = round(((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n)
vector(30, n, n--; L(2*n+1)-2) \\ Colin Barker, Nov 03 2016

G.f.: (-1+6*x-3*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
a(n) = (-2+(2^(-1-n)*((3-sqrt(5))^n*(-5+sqrt(5))+(3+sqrt(5))^n*(5+sqrt(5))))/sqrt(5)). - Colin Barker, Nov 03 2016

Incorrect comment about A004146 removed by Georg Fischer, Sep 04 2020

A328055 Expansion of e.g.f. -log(1 - x / (1 - x)^2).

Original entry on oeis.org

0, 1, 5, 32, 270, 2904, 38400, 605520, 11113200, 232888320, 5488560000, 143704108800, 4138573824000, 130020673305600, 4425201196416000, 162194862064435200, 6369479157000960000, 266808274486161408000, 11874724379464826880000, 559591797303082672128000

Offset: 0

Ilya Gutkovskiy, Oct 03 2019

a(n) is the number of ways to choose one element from each branch of labeled octopuses with n nodes (cf. A029767 and example below). - Enrique Navarrete, Oct 29 2023

			For n=2, the 3 labeled octopuses are the following, and there are 2+2+1 ways to choose one element from each branch:
O-1-2;
O-2-1;
1-O-2. - _Enrique Navarrete_, Oct 29 2023

Cf. A001906, A004146, A005248, A005443, A029767, A052567 (exponential transform), A100404, A226968, A328054.

[0] cat [Factorial(n - 1)*(Lucas(2*n)-2):n in [1..20]]; // Marius A. Burtea, Oct 03 2019

nmax = 19; CoefficientList[Series[-Log[1 - x/(1 - x)^2], {x, 0, nmax}], x] Range[0, nmax]!
Join[{0}, Table[(n - 1)! (LucasL[2 n] - 2), {n, 1, 19}]]

my(x='x+O('x^20)); concat(0, Vec(serlaplace(-log(1 - x / (1 - x)^2)))) \\ Michel Marcus, Oct 03 2019

E.g.f.: log(1 + Sum_{k>=1} Fibonacci(2*k) * x^k).

a(n) = (n - 1)! * (Lucas(2*n) - 2) for n > 0.

A328987 The sequence C(n) defined in the comments (A and B smallest missing numbers, offset 0).

Original entry on oeis.org

3, 10, 15, 20, 27, 32, 39, 44, 51, 56, 61, 68, 73, 80, 85, 90, 97, 102, 109, 114, 119, 126, 131, 138, 143, 150, 155, 160, 167, 172, 179, 184, 189, 196, 201, 208, 213, 220, 225, 230, 237, 242, 249, 254, 259, 266, 271, 278, 283, 290, 295, 300, 307, 312, 319

Offset: 0

N. J. A. Sloane, Nov 07 2019

nonn

Define a triple of sequences A,B,C by A[0]=1, B[0]=2, C[0]=3; for n>=1, A[n] = smallest missing number from the terms of A,B,C defined so far; B[n] = = smallest missing number from the terms of A,B,C defined so far; C[n] = n+A[n]+B[n].
Then A = A086377, B = A080652, C = the present sequence.
Inspired by the triples [A003144, A003145, A004146] and [A298468, A298469, A047218].

			The initial terms are:
n: 0, 1, 2, 3, 4,  5,  6,  7,  8.  9. 10. ...
a: 1, 4, 6, 8, 11, 13, 16, 18, 21, 23,25, ...
b: 2, 5, 7, 9, 12, 14, 17, 19, 22, 24,26, ...
c: 3, 10, 15, 20, 27, 32, 39, 44, 51, 56, 61, 68, 73, ...

Cf. A086377, A080652; A003144, A003145, A003146; A298468, A298469, A047218.

Conjectures from Colin Barker, Nov 08 2019: (Start)
G.f.: (3 + 7*x + 5*x^2 + 5*x^3 + 7*x^4 + 5*x^5 + 7*x^6 + 5*x^7 + 7*x^8 + 5*x^9 + 5*x^10 + 7*x^11 + 2*x^12 - 2*x^20 + 2*x^21) / (1 - x - x^12 + x^13).
a(n) = a(n-1) + a(n-12) - a(n-13) for n>21.
(End)
Conjecture: a(n) ~ 35*n/6. - Stefano Spezia, Nov 02 2021

A340561 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals, where T(n,k) = sqrt( Product_{a=1..n-1} Product_{b=1..k-1} (4sin(aPi/n)^2 + 4cos(bPi/k)^2) ).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 12, 16, 4, 1, 1, 29, 75, 45, 5, 1, 1, 70, 361, 384, 121, 6, 1, 1, 169, 1728, 3509, 1805, 320, 7, 1, 1, 408, 8281, 31500, 30976, 8100, 841, 8, 1, 1, 985, 39675, 284089, 508805, 261725, 35287, 2205, 9, 1

Offset: 1

Seiichi Manyama, Jan 11 2021

			Square array begins:
  1, 1,   1,    1,      1,       1, ...
  1, 2,   5,   12,     29,      70, ...
  1, 3,  16,   75,    361,    1728, ...
  1, 4,  45,  384,   3509,   31500, ...
  1, 5, 121, 1805,  30976,  508805, ...
  1, 6, 320, 8100, 261725, 7741440, ...

Columns 1..4 give A000012, A000027, A004146, A006235.
Rows 1..3 give A000012, A000129, A005386.
Main diagonal gives A340563.
T(n, 2*n) gives A252767.
Cf. A173958, A340476, A340560.

default(realprecision, 120);
{T(n, k) = round(sqrt(prod(a=1, n-1, prod(b=1, k-1, 4*sin(a*Pi/n)^2+4*cos(b*Pi/k)^2))))}

A355645 The number of regions in the G-Shi arrangement when G is the cycle graph C_n.

Original entry on oeis.org

1, 3, 16, 61, 206, 659, 2052, 6297, 19162, 58015, 175088, 527333, 1586118, 4766571, 14316124, 42981169, 129009074, 387158327, 1161737160, 3485735805, 10458256030, 31376865283, 94134790196, 282412759241, 847255054986, 2541798719439, 7625463267232

Offset: 1

Robin Truax, Jul 11 2022

The G-Shi arrangement of a graph G is the hyperplane arrangement given by hyperplanes x_i - x_j = 0 and x_i - x_j = 1 for each edge {i,j} of G with i < j.

a(n) is also the number of parking functions of length n with all cars preferring to park in the first 3 spots. - Jayden Thadani, May 20 2024

Paolo Xausa, Table of n, a(n) for n = 1..1000
Douglas M. Chen, On the Structure of Permutation Invariant Parking, arXiv:2311.15699 [math.CO], 2023. See p. 16.
Index entries for linear recurrences with constant coefficients, signature (7,-17,17,-6).

Cf. A004146 (cycle graph labels).

Cf. A000244 (path graph regions), A001906 (path graph labels).

A355645[n_] := If[n == 1, 1, 3^n - 2^n - n]; Array[A355645, 50] (* or *)
LinearRecurrence[{7, -17, 17, -6}, {1, 3, 16, 61, 206}, 50] (* Paolo Xausa, May 24 2024 *)

a(n) = 3^n - 2^n - n for n >= 2.

G.f.: x*(1 - 4*x + 12*x^2 - 17*x^3 + 6*x^4)/((1 - x)^2*(1 - 2*x)*(1 - 3*x)). - Stefano Spezia, Jul 12 2022

A365907 Smallest nonnegative integer that is not the sum of fewer than n signed Lucas numbers.

Original entry on oeis.org

0, 1, 5, 16, 63, 262, 1105, 4676, 19803, 83882, 355325, 1505176, 6376023, 27009262, 114413065, 484661516, 2053059123, 8696898002, 36840651125, 156059502496, 661078661103, 2800374146902, 11862575248705, 50250675141716, 212865275815563, 901711778403962

Offset: 0

Mike Speciner, Sep 22 2023

Signed Lucas numbers are the union of A000032 and A061084.

			a(0) = 0, the sum of 0 Lucas numbers.
a(1) = 1 = A000032(1), the sum of 1 Lucas number.
a(2) = 5 = 1+4 = A000032(1)+A000032(3), the sum of 2 Lucas numbers. (2, 3, and 4 need only one term, since they are Lucas numbers.)
a(4) = 63 = 1+4+11+47.
For comparison, 45 is the first sum requiring 4 positive Lucas numbers (45 = 1+4+11+29, see A004146), but here 45 = 47+2-4 requires only 3 signed Lucas numbers so that a(4) != 45.

Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000032, A061084, A004146 (analogous with only positive Lucas numbers).

LinearRecurrence[{5, -3, -1}, {0, 1, 5, 16, 63}, 26] (* Amiram Eldar, Sep 26 2023 *)

from sympy import lucas
a = lambda n: n if n<2 else (lucas(3*n-2)+3)//2

a(n) = n, for n<2.
a(n) = (A000032(3*n-2)+3)/2 = 1+4+Sum_{i=2..n-1} A000032(3*n-1), for n>1.
G.f.: x*(1 - 6*x^2 - x^3)/((1 - x)*(1 - 4*x - x^2)). - Stefano Spezia, Sep 25 2023

cp's OEIS Frontend

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(3 - sqrt(5)).

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A249450 a(n) = Fibonacci(2*n) - 2.

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A158753 Triangle T(n, k) = A000032(2*(n-k) + 1), read by rows.

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A158786 Irregular triangle T(n, k) = A000032(n-2*k+1) if (n-2*k) mod 2 = 0, otherwise 25*A000032(n-2*k), read by rows.

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Mathematica

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A256233 a(n) = L(2*n+1) - 2, where L is A000032.

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Magma

Mathematica

PARI

PARI

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Extensions

A328055 Expansion of e.g.f. -log(1 - x / (1 - x)^2).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A328987 The sequence C(n) defined in the comments (A and B smallest missing numbers, offset 0).

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Author

Keywords

Comments

Examples

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

A158786 Irregular triangle T(n, k) = A000032(n-2k+1) if (n-2k) mod 2 = 0, otherwise 25A000032(n-2k), read by rows.

A340561 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals, where T(n,k) = sqrt( Product_{a=1..n-1} Product_{b=1..k-1} (4sin(aPi/n)^2 + 4cos(bPi/k)^2) ).