A033890 -id:A033890 - cp's OEIS frontend

A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.

1, 4, 25, 169, 1156, 7921, 54289, 372100, 2550409, 17480761, 119814916, 821223649, 5628750625, 38580030724, 264431464441, 1812440220361, 12422650078084, 85146110326225, 583600122205489, 4000054745112196, 27416783093579881

Offset: 0

R. K. Guy, Mar 04 2003

Indices of 12-gonal numbers which are also squares (A342709). - Bernard Schott, Mar 19 2021
Values of y in solutions of x^2 = 5*y^2 - 4*y in positive integers. See A360467 for how this relates to a problem regarding the subdivision of a square into four triangles of integer area. - Alexander M. Domashenko, Feb 26 2023
And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - Bernard Schott, Feb 26 2023

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 19.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (vii)).
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
Derek Jennings, On Sums of Reciprocals of Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 32, No. 1 (1994), pp. 18-21.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081067.
Cf. A001622, A051324, A239798, A342709.
Cf. A033890, A360467.

I:=[1, 4, 25]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

[(Lucas(4*n+2) + 2)/5: n in [0..30]]; // G. C. Greubel, Dec 17 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,(luc(4*n+2)+2)/5) od: # James Sellers, Mar 05 2003

CoefficientList[Series[-(1-4*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,4,25},50] (* Vincenzo Librandi, Jun 26 2012 *)
Table[(LucasL[4*n+2] + 2)/5, {n,0,30}] (* G. C. Greubel, Dec 17 2017 *)

main(size)={ return(concat([1],vector(size,n,fibonacci(2*n+1)^2))) } /* Anders Hellström, Jul 11 2015 */

for(n=0,30, print1(fibonacci(2*n+1)^2, ", ")) \\ G. C. Greubel, Dec 17 2017

a(n) = A001519(n+1)^2 = A122367(n)^2 = A058038(n) + 1.
a(n) = A103433(n+1) - A103433(n).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - Gary Detlefs, Apr 01 2012
G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - Peter Bala, Nov 30 2013
Sum_{n>=0} 1/a(n) = sqrt(5) * Sum_{n>=1} (-1)^(n+1)*n/Fibonacci(2*n) (Jennings, 1994). - Amiram Eldar, Oct 30 2020
Product_{n>=1} (1 + 1/a(n)) = phi^2/2 (A239798), where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

More terms from James Sellers, Mar 05 2003

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A231855 T(n,k)=Number of nXk 0..2 arrays with no element having a strict majority of its horizontal and antidiagonal neighbors equal to itself plus one mod 3, with upper left element zero (rock paper and scissors drawn positions).

Original entry on oeis.org

1, 1, 3, 3, 8, 9, 8, 34, 55, 27, 21, 144, 656, 377, 81, 55, 612, 7339, 12404, 2584, 243, 144, 2613, 85288, 360966, 234336, 17711, 729, 377, 11159, 991167, 11149456, 17726611, 4426924, 121393, 2187, 987, 47675, 11529929, 342945563, 1454768048, 870478586

Offset: 1

R. H. Hardin, Nov 14 2013

Table starts
....1......1..........3.............8...............21..................55
....3......8.........34...........144..............612................2613
....9.....55........656..........7339............85288..............991167
...27....377......12404........360966.........11149456...........342945563
...81...2584.....234336......17726611.......1454768048........118292347982
..243..17711....4426924.....870478586.....189801034186......40798265169064
..729.121393...83630516...42745416641...24762957054535...14071005227913420
.2187.832040.1579892344.2099041399895.3230773305296573.4852980371902817445

			Some solutions for n=3 k=4
..0..0..0..1....0..0..2..1....0..0..0..0....0..0..0..1....0..0..1..0
..0..2..2..2....1..2..1..1....1..1..1..1....0..0..2..2....0..2..0..1
..2..0..0..0....1..1..2..2....1..2..0..0....1..0..0..0....2..2..2..2

R. H. Hardin, Table of n, a(n) for n = 1..161

Column 1 is A000244(n-1)
Column 2 is A033890(n-1)
Row 1 is A001906(n-1)

Empirical for column k:
k=1: a(n) = 3*a(n-1)
k=2: a(n) = 7*a(n-1) -a(n-2)
k=3: a(n) = 21*a(n-1) -41*a(n-2) +22*a(n-3) for n>4
k=4: [order 9] for n>10
k=5: [order 21] for n>22
k=6: [order 52] for n>54
Empirical for row n:
n=1: a(n) = 3*a(n-1) -a(n-2) for n>3
n=2: [order 8] for n>9
n=3: [order 35] for n>39

A057085 a(n) = 9a(n-1) - 9a(n-2) for n>1, with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 9, 72, 567, 4455, 34992, 274833, 2158569, 16953624, 133155495, 1045816839, 8213952096, 64513217313, 506693386953, 3979621526760, 31256353258263, 245490585583527, 1928108090927376, 15143557548094641, 118939045114505385, 934159388097696696

Offset: 0

Wolfdieter Lang, Aug 11 2000

Scaled Chebyshev U-polynomials evaluated at 3/2.

Colin Barker, Table of n, a(n) for n = 0..1000
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=9, q=-9.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs.(38) and (45),lhs, m=9.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-9).

Cf. A000045, A000244, A001906, A004187, A030191, A033890, A049310, A109466.

[3^(n-1)*Fibonacci(2*n): n in [0..30]]; // G. C. Greubel, May 02 2022

f[n_]:= Fibonacci[2n]*3^(n-1); Table[f@n, {n, 0, 20}] (* or *)
a[0]=0; a[1]=1; a[n_]:= a[n]= 9(a[n-1] -a[n-2]); Table[a[n], {n, 0, 20}] (* or *)
CoefficientList[Series[x/(1-9x +9x^2), {x, 0, 20}], x] (* Robert G. Wilson v Sep 21 2006 *)

a(n)=(1/3)*sum(k=0,n,binomial(n,k)*fibonacci(4*k)) \\ Benoit Cloitre

concat(0, Vec(x/(1-9*x+9*x^2) + O(x^30))) \\ Colin Barker, Jun 14 2015

[lucas_number1(n,9,9) for n in range(0, 21)] # Zerinvary Lajos, Apr 23 2009

a(n) = A001906(n)*3^(n-1).
a(n) = S(n, 3)*3^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
a(n) = A001906(n)*A000244(n)/3. - Robert G. Wilson v, Sep 21 2006
a(2k) = A004187(k)*9^k/3, a(2k-1) = A033890(k)*9^k.
G.f.: x/(1-9*x+9*x^2).
a(n) = (1/3)*Sum_{k=0..n} binomial(n, k)*Fibonacci(4*k). - Benoit Cloitre, Jun 21 2003
a(n+1) = Sum_{k=0..n} A109466(n,k)*9^k. - Philippe Deléham, Oct 28 2008
E.g.f.: 2*exp(9*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Stefano Spezia, Sep 01 2025

Edited by N. J. A. Sloane, Sep 16 2005

A103134 a(n) = Fibonacci(6n+4).

Original entry on oeis.org

3, 55, 987, 17711, 317811, 5702887, 102334155, 1836311903, 32951280099, 591286729879, 10610209857723, 190392490709135, 3416454622906707, 61305790721611591, 1100087778366101931, 19740274219868223167, 354224848179261915075, 6356306993006846248183

Offset: 0

Creighton Dement, Jan 24 2005

Gives those numbers which are Fibonacci numbers in A103135.
Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013
a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Subsequence of A033887.
Cf. A000032, A000045, A001906, A001519, A015448, A014445, A033888, A033889, A033890, A033891, A049310, A049660, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.
Cf. A103135.

[Fibonacci(6*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+4], {n, 0, 30}]
LinearRecurrence[{18,-1},{3,55},20] (* Harvey P. Dale, Mar 29 2023 *)
Table[ChebyshevU[3*n+1, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+4) \\ Charles R Greathouse IV, Feb 05 2013

G.f.: (x+3)/(x^2-18*x+1).
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=3, a(1)=55. - Philippe Deléham, Nov 17 2008
a(n) = A007805(n) + A075796(n), as follows from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((15-7*sqrt(5)+(9+4*sqrt(5))^(2*n)*(15+7*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n+1, 3) = 3*S(n,18) + S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Edited by N. J. A. Sloane, Aug 10 2010

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Philippe Deléham, Jun 09 2007

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020

Vincenzo Librandi, Rows n = 1..101, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and Aoife Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, example 15.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Cf. A039599, A085478, A123970, A321620.

# The function RiordanSquare is defined in A321620.
RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019

max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)

@CachedFunction
def A129818(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
    return A129818(n-1,k-1) - A129818(n-2,k) - h
for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A134504 a(n) = Fibonacci(7n + 6).

Original entry on oeis.org

8, 233, 6765, 196418, 5702887, 165580141, 4807526976, 139583862445, 4052739537881, 117669030460994, 3416454622906707, 99194853094755497, 2880067194370816120, 83621143489848422977, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.

[Fibonacci(7*n +6): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[7n+6], {n, 0, 30}]
LinearRecurrence[{29,1},{8,233},20] (* Harvey P. Dale, Jul 21 2021 *)

a(n)=fibonacci(7*n+6) \\ Charles R Greathouse IV, Jun 11 2015

G.f.: (-8-x) / (-1 + 29*x + x^2). - R. J. Mathar, Jul 04 2011
a(n) = A000045(A017053(n)). - Michel Marcus, Nov 08 2013
a(n) = 29*a(n-1) + a(n-2). - Wesley Ivan Hurt, Mar 15 2023

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A105531 Decimal expansion of arctan(1/3).

Original entry on oeis.org

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

Offset: 0

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

arctan(1/3) + A073000 = 2*arctan(1/3) + A105533 = Pi/4.

			0.3217505543966421934014046143...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

Peter Bala, New series for old functions
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
Eric Weisstein's World of Mathematics, Machin-Like Formulas
Index entries for transcendental numbers

Cf. A003881 (Pi/4), A073000 (arctan(1/2)), A105533 (arctan(1/7)).

RealDigits[ArcTan[1/3],10,120][[1]] (* Harvey P. Dale, Oct 28 2011 *)

PARI
```
atan(1/3) \\ Michel Marcus, Mar 29 2016
```

From Peter Bala, Feb 04 2015: (Start)
Equals (1/3)*Sum_{k >= 0} (-1)^k/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*9^k). Both sequences satisfy the same recurrence equation u(n) = (32*n + 20)*u(n-1) + 36*(2*n - 1)^2*u(n-2). From this observation we find the continued fraction expansion arctan(1/3) = (1/3)*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ...))))).
Equals (3/10) * Sum_{k >= 0} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ).
Define a pair of integer sequences C(n) = 10^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ). Both sequences satisfy the same recurrence equation u(n) = (44*n + 20)*u(n-1) - 80*n*(2*n - 1)*u(n-2). From this observation we obtain the continued fraction expansion arctan(1/3) = (3/10)*( 1 + 4/(60 - 480/(108 - 1200/(152 - ... - 80*n*(2*n - 1)/((44*n + 20) - ...))))). (End)
Equals Sum_{k>=1} arctan(L(4*k+2)/F(4*k+2)^2) where L=A000032 and F=A000045. See also A033890 and A246453. - Michel Marcus, Mar 29 2016 [corrected by Jason Yuen, Jan 18 2025]
From Amiram Eldar, Aug 09 2020: (Start)
Equals Sum_{k>=2} arctan(1/(2*k^2)) = Sum_{k>=2} (-1)^k arctan(2/k^2).
Equals Integral_{x=1..2} 1/(x^2 + 1) dx. (End)
Equals Sum_{n>=0} arctan(1/F(2*n+5)) = Sum_{n>=0} (-1)^n arctan(F(2*n+1)) where F=A000045. - Gleb Koloskov, Oct 01 2021

A167816 Numerator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; denominator=A167817.

Original entry on oeis.org

0, 1, 1, 2, 1, 5, 8, 13, 7, 34, 55, 89, 48, 233, 377, 610, 329, 1597, 2584, 4181, 2255, 10946, 17711, 28657, 15456, 75025, 121393, 196418, 105937, 514229, 832040, 1346269, 726103, 3524578, 5702887, 9227465, 4976784, 24157817, 39088169, 63245986, 34111385

Offset: 0

Reinhard Zumkeller, Nov 13 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 7, 0, 0, 0, -1).

Cf. A000045, A167808.

[0,1,1] cat [Numerator(Fibonacci(n)/Fibonacci(2*n-4)): n in [3..40]]; // Vincenzo Librandi, Jun 28 2016

Numerator[LinearRecurrence[{1,1},{0,1/3},40]] (* Harvey P. Dale, Dec 07 2014 *)
LinearRecurrence[{0, 0, 0, 7, 0, 0, 0, -1},{0, 1, 1, 2, 1, 5, 8, 13},39] (* Ray Chandler, Aug 03 2015 *)

a(n) = (a(n-1)*A093148(n+2) + a(n-2)*A093148(n+1))/A093148(n-1) for n>1.
a(4*n) = A004187(n) = (a(4*n-1) + a(4*n-2))/3;
a(4*n+1) = A033889(n) = 3*a(4*n-1) + a(4*n-2);
a(4*n+2) = A033890(n) = a(4*n-1) + 3*a(4*n-2);
a(4*n+3) = A033891(n) = a(4*n-1) + a(4*n-2).
Numerator of Fibonacci(n) / Fibonacci(2n-4) for n>=3. - Gary Detlefs, Dec 20 2010

Definition corrected by D. S. McNeil, May 09 2010

cp's OEIS Frontend

A081068 a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.

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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A231855 T(n,k)=Number of nXk 0..2 arrays with no element having a strict majority of its horizontal and antidiagonal neighbors equal to itself plus one mod 3, with upper left element zero (rock paper and scissors drawn positions).

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A057085 a(n) = 9*a(n-1) - 9*a(n-2) for n>1, with a(0)=0, a(1)=1.

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A103134 a(n) = Fibonacci(6n+4).

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A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

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A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A057085 a(n) = 9a(n-1) - 9a(n-2) for n>1, with a(0)=0, a(1)=1.