A049664 -id:A049664 - cp's OEIS frontend

A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

1, 23, 506, 11110, 243915, 5355021, 117566548, 2581109036, 56666832245, 1244089200355, 27313295575566, 599648413462098, 13164951800590591, 289029291199530905, 6345479454589089320, 139311518709760434136, 3058507932160140461673

Offset: 0

Bruno Berselli, Jun 08 2012

Partial sums of A077421.

Bruno Berselli, Table of n, a(n) for n = 0..200
Robert K. Moniot, A Family of Integer Sequences, 2021.
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3): A334673 (k=24), A212336 (k=23), A212335 (k=22), A097833 (k=21), A097832 (k=20), A049664 (k=19), A097831-A097829 (k=18,17,16), A076139 (k=15), A097828-A097826 (k=14,13,12), A097784 (k=11), A092420 (k=10), A076765 (k=9), A092521 (k=8), A053142 (k=7), A089817(k=6), A061278 (k=5), A027941 (k=4), A000217 (k=3), A021823 (k=2), A133872 (k=1), A079978 (k=0).

m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-23*x+23*x^2-x^3)));

I:=[1,23,506]; [n le 3 select I[n] else 23*Self(n-1)-23*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Aug 18 2013

a:= n-> (<<0|1|0>, <0|0|1>, <1|-23|23>>^n. <<1, 23, 506>>)[1, 1]:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 15 2012

CoefficientList[Series[1/(1 - 23 x + 23 x^2 - x^3), {x, 0, 16}], x]
LinearRecurrence[{23, -23, 1}, {1, 23, 506}, 20] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(coeff(taylor(1/(1-23*x+23*x^2-x^3), x, 0, n), x, n), n, 0, 16);

PARI
```
Vec(1/(1-23*x+23*x^2-x^3)+O(x^17))
```

[(1/20)*(-1 +21*chebyshev_U(n, 11) -chebyshev_U(n-1, 11)) for n in (0..30)] # G. C. Greubel, Feb 07 2022

G.f.: 1/((1-x)*(1 - 22*x + x^2)).
a(n) = (((6+sqrt(30))^(2*n+3) + (6-sqrt(30))^(2*n+3))/6^(n+1) - 12)/240.
a(n) = a(-n-3) = 23*a(n-1) - 23*a(n-2) + a(n-3).
a(n)*a(n+2) = a(n+1)*(a(n+1)-1).
a(n+1) - 11*a(n) = A133285(n+2).
11*a(n+1) - a(n) = (1/5)*A157096(n+2).
a(n) = (1/20)*(-1 + 21*ChebyshevU(n, 11) - ChebyshevU(n-1, 11)). - G. C. Greubel, Feb 07 2022

A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

Original entry on oeis.org

0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432

Offset: 0

N. J. A. Sloane, James Sellers, Jan 20 2000

Define a(1)=0, a(2)=8 with 5*(a(1)^2) + 5*a(1) + 1 = j(1)^2 = 1^2 and 5*(a(2)^2) + 5*a(2) + 1 = j(2)^2 = 19^2. Then a(n) = a(n-2) + 8*sqrt(5*(a(n-1)^2) + 5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2) + 5*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey, Apr 03 2005
sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey, Apr 19 2005
a(n) is such that 5*a(n)^2 + 5*a(n) + 1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI, Apr 21 2005
From Russell Jay Hendel, Apr 25 2015: (Start)
We prove the two McGarvey-CAMI conjectures mentioned at the beginning of the Comments section. Let, as usual, F(n) = A000045(n), the Fibonacci numbers. In the sequel we indicate equations with upper case letters ((A), (B), (C), (D)) for ease of reference.
Then we must prove (A), 5*((F(6*n+3)-2)/4)^2 + 5*((F(6*n+3)-2)/4) + 1 = ((F(6*n+5)-F(6*n+1))/4)^2. Let m = 3*n+1 so that 6*n+1, 6*n+3, and 6*n+5 are 2*m-1, 2*m+1, and 2*m+3 respectively. Define G(m) = F(6*n+3) = F(2*m+1) = A001519(m+1), the bisected Fibonacci numbers. We can now simplify equation (A) by i) multiplying the LHS and RHS by 16, ii) expanding squares, and iii) gathering like terms. This shows proof of (A) equivalent to proving (B), 5*G(m)^2-4 = (G(m+1)-G(m-1))^2.
By Jarden's theorem (D. Jarden, Recurring sequences, 2nd ed. Jerusalem, Riveon Lematematika, (1966)), if {H(n)}{n >=1} is any recursive sequence satisfying (C), H(n)=3H(n-1)-H(n-2), then {H(n)}^2{n >=1} is also a recursive sequence satisfying (D), H(n)^2=8H(n-1)^2-8H(n-2)^2+H(n-3)^2. As noted in the Formula section of A001519, {G(m)}_{m >= 1} satisfies (C).
Proof of (B) is now straightforward. Since {G(m)}{m >=1} satisfies (C), it follows that {G(m)^2}{m >=1} satisfies (D), and therefore, {5G(m)^2-4}_{m >=1} also satisfies (D).
Similarly, since {G(m)}{m >=1} satisfies (C), it follows that both {G(m+1)}{m >=1}, {G(m-1)}{m >=1} and their difference {G(m+1)-G(m-1)}{m >=1} satisfy (C), and therefore {G(m+1)-G(m-1)}^2_{m >=1} satisfies (D).
But then the LHS and RHS of (B) are equal for m=1,2,3 and satisfy the same recursion, (D). Hence the LHS and RHS of (B) are equal for all m. This completes the proof. (End)

G. C. Greubel, Table of n, a(n) for n = 0..790
F. Ellermann, Illustration of binomial transforms
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A049629.

Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.

List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # G. C. Greubel, May 16 2019

[(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // Vincenzo Librandi, Apr 20 2011

A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
seq(A053606(n),n=0..30) ;

Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

a(n)=fibonacci(6*n+3)\4 \\ Charles R Greathouse IV, Jul 02 2013

[(fibonacci(6*n+3)-2)/4 for n in (0..30)] # G. C. Greubel, May 16 2019

a(n) = 8*A049664(n).
a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - Richard Choulet, Aug 30 2007
G.f.: 8*x/((1-x)*(1-18*x+x^2)). - Richard Choulet, Oct 09 2007
a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - Gary Detlefs, Dec 07 2010
a(n) = Sum_{k=0..n} A134492(k). - Gary Detlefs, Dec 07 2010
a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - Gary Detlefs, Dec 08 2010

A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 7, 12, 5, 1, 12, 33, 22, 8, 1, 20, 88, 94, 56, 12, 1, 33, 232, 399, 385, 134, 19, 1, 54, 609, 1691, 2640, 1487, 342, 30, 1, 88, 1596, 7164, 18096, 16492, 6138, 872, 48, 1, 143, 4180, 30348, 124033, 182900, 110143, 25319, 2256, 77, 1

Offset: 1

Clark Kimberling, Oct 28 2012

col 1: A001612 (except for initial term)
row 1: A000071
row 2: A027941
row 3: A049652
row 4: A092521
row 6: A049664
row 8: A156093 without minus signs

			Northwest corner:
1...2....4.....7......12......20
1...4....12....33.....88......232
1...5....22....94.....399.....1691
1...8....56....385....2640....18096
1...12...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A214978, A214985, A214986.

F[n_] := Fibonacci[n]; L[n_] := LucasL[n];
t[m_, n_] := (1/F[m])*Sum[F[m*k], {k, 1, n}]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
Flatten[Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}]]

For odd-numbered rows (m odd):
T(m,n) = (F(m*n+m) + F(m*n) - F(m))/(F(m)*L(m)).
For even-numbered rows (m even):
T(m,n) = (F(m*n+m) - F(m*n) - F(m))/(F(m)*(L(m)-2)).

A221076 Continued fraction expansion of product_{n>=0} (1-sqrt(5)[sqrt(5)-2]^{4n+3})/(1-sqrt(5)[sqrt(5)-2]^{4n+1}).

Original entry on oeis.org

2, 16, 1, 32, 1, 320, 1, 608, 1, 5776, 1, 10944, 1, 103680, 1, 196416, 1, 1860496, 1, 3524576, 1, 33385280, 1, 63245984, 1, 599074576, 1, 1134903168, 1, 10749957120, 1, 20365011072, 1, 192900153616, 1, 365435296160, 1

Offset: 0

Peter Bala, Jan 06 2013

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 5. For other cases see A221073 (m = 2), A221074 (m = 3) and A221075 (m = 4).

If we denote the present sequence by [2; 16, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+3)]/[1 - sqrt(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+1)]. An example is given below.

			Product {n >= 0} {1 - sqrt(5)*(sqrt(5) - 2)^(4*n+3)}/{1 - sqrt(5)*(sqrt(5) - 2)^(4*n+1)} = 2.05892 54859 32105 82744 ...
= 2 + 1/(16 + 1/(1 + 1/(32 + 1/(1 + 1/(320 + 1/(1 + 1/(608 + ...))))))).
Since (sqrt(5) - 2)^3 = 17*sqrt(5) - 38 we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+3)}/{1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+1)} = 1.03030 31892 29728 52318 ... = 1 + 1/(32 + 1/(1 + 1/(5776 + 1/(1 + 1/(196416 + 1/(1 + 1/(33385280 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,18,0,-18,0,-1,0,1).

Cf. A049664, A049863, A053606, A132584, A174500, A221073 (m = 2), A221074 (m = 3), A221075 (m = 4).

LinearRecurrence[{0,1,0,18,0,-18,0,-1,0,1},{2,16,1,32,1,320,1,608,1,5776,1},40] (* or *) Join[{2},Riffle[LinearRecurrence[{1,18,-18,-1,1},{16,32,320,608,5776},20],1]] (* Harvey P. Dale, Jun 05 2023 *)

a(2*n) = 1 for n >= 1. For n >= 1 we have:
a(4*n - 3) = (sqrt(5) + 2)^(2*n) + (sqrt(5) - 2)^(2*n) - 2;
a(4*n - 1) = 1/sqrt(5)*{(sqrt(5) + 2)^(2*n + 1) + (sqrt(5) - 2)^(2*n + 1)} - 2.
a(4*n - 3) = 16*A049863(n) = 4*A132584(n);
a(4*n - 1) = 32*A049664(n) = 4*A053606(n).
O.g.f.: 2 + x^2/(1 - x^2) + 16*x*(1 + x^2)^2/(1 - 19*x^4 + 19*x^8 - x^12) = 2 + 16*x + x^2 + 32*x^3 + x^4 + 320*x^5 + ....
O.g.f.: (x^10-2*x^8-18*x^6+36*x^4-16*x^3+x^2-16*x-2) / ((x-1)*(x+1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)). - Colin Barker, Jan 10 2014

A077832 Expansion of 1/(1-2x-3x^2-3*x^3).

Original entry on oeis.org

1, 2, 7, 23, 73, 236, 760, 2447, 7882, 25385, 81757, 263315, 848056, 2731328, 8796769, 28331690, 91247671, 293880719, 946499521, 3048384212, 9817909144, 31620469487, 101839819042, 327994773977, 1056370413541, 3402244606139, 10957584774832, 35291014608704

Offset: 0

N. J. A. Sloane, Nov 17 2002

Index entries for linear recurrences with constant coefficients, signature (2,3,3).

Partial sums of S(n, x), x=1..17 see A077831 with links. Partial sums of S(n, 18) see A049664.

Vec(1/(1-2*x-3*x^2-3*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

A214985 Array: T(m,n) = (F(n) + F(2n) + ... + F(nm))/F(n), by antidiagonals; transpose of A214984.

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 1, 5, 12, 7, 1, 8, 22, 33, 12, 1, 12, 56, 94, 88, 20, 1, 19, 134, 385, 399, 232, 33, 1, 30, 342, 1487, 2640, 1691, 609, 54, 1, 48, 872, 6138, 16492, 18096, 7164, 1596, 88, 1, 77, 2256, 25319, 110143, 182900, 124033, 30348, 4180, 143

Offset: 1

Clark Kimberling, Oct 28 2012

row 1: A001612 (except for initial term)
col 1: A000071
col 2: A027941
col 3: A049652
col 4: A092521
col 6: A049664
col 8: A156093 without minus signs

			Northwest corner:
1....1.....1......1.......1
2....4.....5......8.......12
4....12....22.....56......134
7....33....94.....385.....1487
12...88....399....2640....16492
20...232...1691...18096...182900

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A214978, A214984, A214986.

F[n_] := Fibonacci[n]; L[n_] := LucasL[n];
t[m_, n_] := (1/F[n])*Sum[F[k*n], {k, 1, m}]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
Flatten[Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}]]

For odd-numbered columns (m odd):
T(m,n) = (F(m*n+m) + F(m*n) - F(m))/(F(m)*L(m)).
For even-numbered columns (m even):
T(m,n) = (F(m*n+m) - F(m*n) - F(m))/(F(m)*(L(m)-1)).

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A049663 a(n) = (F(6*n+5) - 1)/4, where F = A000045 (the Fibonacci sequence).

Original entry on oeis.org

1, 22, 399, 7164, 128557, 2306866, 41395035, 742803768, 13329072793, 239180506510, 4291920044391, 77015380292532, 1381984925221189, 24798713273688874, 444994854001178547, 7985108658747524976, 143286961003454271025, 2571180189403429353478

Offset: 0

Clark Kimberling

Colin Barker, Table of n, a(n) for n = 0..700
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000045, A001622, A049664.

[(Fibonacci(6*n+5) - 1)/4: n in [0..30]]; // G. C. Greubel, Dec 02 2017

(Fibonacci[6*Range[0,20]+5]-1)/4 (* or *) LinearRecurrence[{19,-19,1},{1,22,399},20] (* Harvey P. Dale, Sep 22 2016 *)

Vec((1+3*x)/((1-x)*(1-18*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 04 2016

for(n=0,30, print1((fibonacci(6*n+5) - 1)/4, ", ")) \\ G. C. Greubel, Dec 02 2017

From R. J. Mathar, Oct 26 2015: (Start)
G.f.: (1+3*x)/( (1-x)*(x^2-18*x+1) ).
a(n) = A049664(n+1) + 3*A049664(n). (End)
From Colin Barker, Mar 04 2016: (Start)
a(n) = (-1/4+1/40*(9+4*sqrt(5))^(-n)*(25-11*sqrt(5)+(9+4*sqrt(5))^(2*n)*(25+11*sqrt(5)))).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3) for n>2. (End)
Product_{n>=1} (1 - 1/a(n)) = 4*phi^2/11 = 2*(sqrt(5)+3)/11, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

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A212336 Expansion of 1/(1 - 23*x + 23*x^2 - x^3).

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A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

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A214984 Array: T(m,n) = (F(m) + F(2*m) + ... + F(n*m))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

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A221076 Continued fraction expansion of product_{n>=0} (1-sqrt(5)*[sqrt(5)-2]^{4n+3})/(1-sqrt(5)*[sqrt(5)-2]^{4n+1}).

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A077832 Expansion of 1/(1-2*x-3*x^2-3*x^3).

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A214985 Array: T(m,n) = (F(n) + F(2*n) + ... + F(n*m))/F(n), by antidiagonals; transpose of A214984.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

A221076 Continued fraction expansion of product_{n>=0} (1-sqrt(5)[sqrt(5)-2]^{4n+3})/(1-sqrt(5)[sqrt(5)-2]^{4n+1}).

A077832 Expansion of 1/(1-2x-3x^2-3*x^3).

A214985 Array: T(m,n) = (F(n) + F(2n) + ... + F(nm))/F(n), by antidiagonals; transpose of A214984.