A049685 -id:A049685 - cp's OEIS frontend

A111216 a(n) = 31*a(n-1)-a(n-2).

1, 30, 929, 28769, 890910, 27589441, 854381761, 26458245150, 819351217889, 25373429509409, 785756963573790, 24333092441278081, 753540108716046721, 23335410277756170270, 722644178501725231649, 22378634123275726010849, 693015013643045781104670

Offset: 0

N. J. A. Sloane, following a suggestion from R. K. Guy, Oct 26 2005

Take 31 numbers consisting of 29 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 31 times their product. (Guy)

Positive values of x (or y) satisfying x^2 - 31xy + y^2 + 29 = 0. - Colin Barker, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (31,-1).

Cf. A049685.

Cf. similar sequences listed in A238379.

I:=[1,30]; [n le 2 select I[n] else 31*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 26 2014

CoefficientList[Series[(1 - x)/(1 - 31 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 26 2014 *)

Vec((1-x)/(1-31*x+x^2) + O(x^100)) \\ Colin Barker, Feb 24 2014

G.f.: (1-x)/(1-31*x+x^2). [Philippe Deléham, Nov 18 2008]

a(n) = A200442(n) - A200442(n-1). - R. J. Mathar, Feb 13 2016

A227214 Generalized Markoff numbers: largest of 7-tuple of positive numbers a, b, c, d, e, f, g satisfying the Markoff(7) equation a^2+b^2+c^2+d^2+e^2+f^2+g^2 = 7abcdefg.

Original entry on oeis.org

1, 6, 41, 281, 1721, 1926, 13201, 72241, 80646, 90481, 493921, 620166, 2963561, 3032401, 3788441, 4250681, 23145121, 29134601, 127288601, 141753606, 158630641, 177975881, 199691526, 870287321

Offset: 1

Shanzhen Gao, Sep 19 2013

For n = 1, 2, 3, 4, 6, 7, 10, 12, 16, 18, 23, a(n) is in A049685.

			a(1)=1 and a(2)=6 since (1, 1, 1, 1, 1, 1, 1) and (6, 1, 1, 1, 1, 1, 1) satisfying a^2+b^2+c^2+d^2+e^2+f^2+g^2=7abcdefg.

Cf. A049685, A000032.

Cf. A227210, A227211, A227212.

A123971 Triangle T(n,k), read by rows, defined by T(n,k)=3*T(n-1,k)-T(n-1,k-1)-T(n-2,k), T(0,0)=1, T(1,0)=2, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n.

Original entry on oeis.org

1, 2, -1, 5, -5, 1, 13, -19, 8, -1, 34, -65, 42, -11, 1, 89, -210, 183, -74, 14, -1, 233, -654, 717, -394, 115, -17, 1, 610, -1985, 2622, -1825, 725, -165, 20, -1, 1597, -5911, 9134, -7703, 3885, -1203, 224, -23, 1, 4181, -17345, 30691, -30418, 18633, -7329

Offset: 0

Gary W. Adamson and Roger L. Bagula, Oct 30 2006

This entry is the result of merging two sequences, this one and a later submission by Philippe Deléham, Nov 29 2013 (with edits from Ralf Stephan, Dec 12 2013). Most of the present version is the work of Philippe Deléham, the only things remaining from the original entry are the sequence data and the Mathematica program. - N. J. A. Sloane, May 31 2014
Subtriangle of the triangle given by (0, 2, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
Apart from signs, equals A126124.
Row sums = 1.
Sum_{k=0..n} T(n,k)*(-x)^k = A001519(n+1), A079935(n+1), A004253(n+1), A001653(n+1), A049685(n), A070997(n), A070998(n), A072256(n+1), A078922(n+1), A077417(n), A085260(n+1), A001570(n+1) for x=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.

			Triangle begins:
  1
  2, -1
  5, -5, 1
  13, -19, 8, -1
  34, -65, 42, -11, 1
  89, -210, 183, -74, 14, -1
  233, -654, 717, -394, 115, -17, 1
Triangle (0, 2, 1/2, 1/2, 0, 0, ...) DELTA (1, -2, 0, 0, ...) begins:
  1
  0, 1
  0, 2, -1
  0, 5, -5, 1
  0, 13, -19, 8, -1
  0, 34, -65, 42, -11, 1
  0, 89, -210, 183, -74, 14, -1
  0, 233, -654, 717, -394, 115, -17, 1

Cf. A094954, A098495, A123971, A126124, A152063, A001519, A079935, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570, A001870, A126124.

Mathematica ( general k th center) Clear[M, T, d, a, x, k] k = 3 T[n_, m_, d_] := If[ n == m && n < d && m < d, k, If[n == m - 1 || n == m + 1, -1, If[n == m == d, k - 1, 0]]] M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}] Table[M[d], {d, 1, 10}] Table[Det[M[d]], {d, 1, 10}] Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}] a = Join[{M[1]}, Table[CoefficientList[ Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]] Flatten[a] MatrixForm[a] Table[NSolve[Det[M[d] - x*IdentityMatrix[d]] == 0, x], {d, 1, 10}] Table[x /. NSolve[Det[M[d] - x*IdentityMatrix[d]] == 0, x][[d]], {d, 1, 10}]

T(n,k)=polcoeff(polcoeff(Ser((1-x)/(1+(y-3)*x+x^2)),n,x),n-k,y) \\ Ralf Stephan, Dec 12 2013

@CachedFunction
def A123971(n,k): # With T(0,0) = 1!
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = 2*A123971(n-1,k) if n==1 else 3*A123971(n-1,k)
    return A123971(n-1,k-1) - A123971(n-2,k) - h
for n in (0..9): [A123971(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^n*A126124(n+1,k+1).
T(n,k) = (-1)^k*Sum_{m=k..n} binomial(m,k)*binomial(m+n,2*m). - Wadim Zudilin, Jan 11 2012
G.f.: (1-x)/(1+(y-3)*x+x^2).
T(n,0) = A001519(n+1) = A000045(2*n+1).
T(n+1,1) = -A001870(n).

Edited by N. J. A. Sloane, May 31 2014

A161582 The list of the k values in the common solutions to the 2 equations 5k+1=A^2, 9k+1=B^2.

Original entry on oeis.org

0, 7, 336, 15792, 741895, 34853280, 1637362272, 76921173511, 3613657792752, 169764995085840, 7975341111241735, 374671267233275712, 17601574218852716736, 826899317018844410887, 38846666325666834594960, 1824966417989322381552240, 85734574979172485098360327

Offset: 1

Paul Weisenhorn, Jun 14 2009

The 2 equations are equivalent to the Pell equation x^2-45*y^2=1, with x=(45*k+7)/2 and y= A*B/2, case C=5 in A160682.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Cf. A160682, A049685 (sequence of A), A033890 (sequence of B).

t:=0: for n from 0 to 1000000 do a:=sqrt(5*n+1); b:=sqrt(9*n+1);
if (trunc(a)=a) and (trunc(b)=b) then t:=t+1; print(t,n,a,b): end if: end do:

LinearRecurrence[{48,-48,1},{0,7,336},30] (* or *) Rest[CoefficientList[ Series[ -7x^2/((x-1)(x^2-47x+1)),{x,0,30}],x]] (* Harvey P. Dale, Mar 21 2013 *)

k(t+3) = 48*(k(t+2)-k(t+1))+k(t).
With w = sqrt(5),
k(t) = ((7+3*w)*((47+21*w)/2)^(t-1)+(7-3*w)*((47-21*w)/2)^(t-1))/90.
k(t) = floor((7+3*w)*((47+21*w)/2)^(t-1)/90) = 7*|A156093(t-1)|.
G.f.: -7*x^2/((x-1)*(x^2-47*x+1)).
a(1)=0, a(2)=7, a(3)=336, a(n) = 48*a(n-1)-48*a(n-2)+a(n-3). - Harvey P. Dale, Mar 21 2013

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

A172511 a(n) = a(n-1) * (11a(n-1) - a(n-2)) / (a(n-1) + 4a(n-2)), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 7, 35, 210, 1365, 9165, 62322, 425867, 2915551, 19974626, 136884937, 938162617, 6430103330, 44072167855, 302074043195, 2070443441970, 14191023001437, 97266699113157, 666675822475026, 4569463931720051

Offset: 0

Michael Somos, Feb 05 2010

			G.f. = 1 + x + 2*x^2 + 7*x^3 + 35*x^4 + 210*x^5 + 1365*x^6 + 9165*x^7 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-33,33,-11,1).

Cf. A005247, A001519, A049685.

RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1] (11a[n-1]-a[n-2])/(a[n-1]+ 4a[n-2])},a,{n,30}] (* or *) LinearRecurrence[{11,-33,33,-11,1},{1,1,2,7,35},30] (* Harvey P. Dale, Nov 18 2013 *)

{a(n) = (4 + fibonacci(4*n - 1)/3 + fibonacci(4*n - 3)/3 + 5 * fibonacci(2*n - 1)) / 10};

{a(n) = my(A); if( n<1, n = 1-n); if( n<3, n, A = vector(n, k, k); for(k=3, n, A[k] = A[k-1] * (11*A[k-1] - A[k-2]) / (A[k-1] + 4*A[k-2])); A[n])}; /* Michael Somos, Sep 22 2014 */

a(n) = (4 + A049685(n-1) + 5 * A001519(n)) / 10 = a(1 - n).
G.f.: (4 / (1 - x) + (1 - 6*x) / (1 - 7*x + x^2) + (5 - 10*x) / (1 - 3*x + x^2)) / 10.
a(0)=1, a(1)=1, a(2)=2, a(3)=7, a(4)=35, a(n)=11*a(n-1)-33*a(n-2)+ 33*a(n-3)- 11*a(n-4)+a(n-5). - Harvey P. Dale, Nov 18 2013
a(n) = a(1-n) for all n in Z. - Michael Somos, Sep 22 2014
0 = a(n)*(+a(n+1) + 4*a(n+2)) + a(n+1)*(-11*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 22 2014
a(n) = b(n+1) * b(n) * b(n-1) * b(n-2) / 6 for all n in Z where b = A005247. - Michael Somos, Sep 22 2014

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.

A276529 a(n) = (a(n-1) * a(n-5) + 1) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 13, 20, 27, 34, 41, 89, 137, 185, 233, 281, 610, 939, 1268, 1597, 1926, 4181, 6436, 8691, 10946, 13201, 28657, 44113, 59569, 75025, 90481, 196418, 302355, 408292, 514229, 620166, 1346269, 2072372, 2798475, 3524578, 4250681, 9227465

Offset: 0

Seiichi Manyama, Nov 16 2016

Thanks to the linear recurrence signature, we see that this is actually five separate linear recurrence sequences, each with signature (7,-1), interwoven together. - Greg Dresden, Oct 16 2021

Seiichi Manyama, Table of n, a(n) for n = 0..5985
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,7,0,0,0,0,-1).

Cf. A275173, A102276, A276530.

5th-sections: A049685, A033891, A033889.

LinearRecurrence[{0,0,0,0,7,0,0,0,0,-1}, {1,1,1,1,1,1,2,3,4,5}, 50] (* G. C. Greubel, Nov 18 2016 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==a[4]==a[5]==1,a[n]==(a[n-1]a[n-5]+ 1)/a[n-6]},a,{n,50}] (* Harvey P. Dale, Oct 08 2020 *)
Flatten[Table[{LucasL[4 n - 2]/3, Fibonacci[4 n - 1], LucasL[4 n + 2]/3 - Fibonacci[4 n], LucasL[4 n - 2]/3 + Fibonacci[4 n], Fibonacci[4 n + 1]}, {n, 0, 10}]] (* Greg Dresden, Oct 16 2021 *)

Vec((1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9)/(1 -7*x^5 +x^10) + O(x^50)) \\ Colin Barker, Nov 16 2016

def A(k, m, n)
  a = Array.new(2 * k, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[-1] * a[1] + a[k] ** m
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276529(n)
  A(3, 0, n)
end

a(n) + a(n+10) = 7*a(n+5).
a(5-n) = a(n).
G.f.: (1 +x +x^2 +x^3 +x^4 -6*x^5 -5*x^6 -4*x^7 -3*x^8 -2*x^9) / (1 -7*x^5 +x^10). - Colin Barker, Nov 16 2016
From Greg Dresden, Oct 16 2021: (Start)
a(5*n) = L(4*n-2)/3 = A049685(n-1),
a(5*n+1) = F(4*n-1) = A033891(n-1),
a(5*n+2) = L(4*n+2)/3 - F(4*n),
a(5*n+3) = L(4*n-2)/3 + F(4*n),
a(5*n+4) = F(4*n+1) = A033889(n). (End)

A328697 Rectangular array R read by descending antidiagonals: divide the multiples of 3 in the Wythoff array (A035513) by 3, and delete all others.

Original entry on oeis.org

1, 7, 6, 48, 41, 2, 329, 281, 14, 3, 2255, 1926, 96, 5, 4, 15456, 13201, 658, 8, 28, 20, 105937, 90481, 4510, 13, 192, 137, 15, 726103, 620166, 30912, 21, 1316, 939, 103, 27, 4976784, 4250681, 211874, 34, 9020, 6436, 706, 185, 12, 34111385, 29134601, 1452206

Offset: 1

Clark Kimberling, Oct 29 2019

Every positive integer occurs in R exactly once, and every row of R is a linear recurrence sequence.
Row 1 of R is essentially A004187.
Row 2 of R is essentially A049685.
Row 4 of R is essentially A000045.

			Row 1 of the Wythoff array is (1,2,3,5,8,13,21,34,55,89,144,...), so that row 1 of R is (1,7,48,329,2255,...).
=====================
Northwest corner of R:
   1,   7,  48,  329,  2255,  15456,  105937,   726103
   6,  41, 281, 1926, 13201,  90481,  620166,  4250681
   2,  14,  96,  658,  4510,  30912,  211874,  1452206
   3,   5,   8,   13,    21,     34,      55,       89
   4,  28, 192, 1316,  9020,  61824,  423748,  2904412
  20, 137, 939, 6436, 44113, 302355, 2072372, 14204249
  15, 103, 706, 4839, 33167, 227330, 1558143, 10679671

Cf. A035513, A004187, A049685, A000045, A328695, A328696.

A342710 Solutions x to the Pell-Fermat equation x^2 - 5*y^2 = 4.

Original entry on oeis.org

3, 18, 123, 843, 5778, 39603, 271443, 1860498, 12752043, 87403803, 599074578, 4106118243, 28143753123, 192900153618, 1322157322203, 9062201101803, 62113250390418, 425730551631123, 2918000611027443, 20000273725560978, 137083915467899403, 939587134549734843

Offset: 0

Bernard Schott, Mar 19 2021

This Pell equation is used to find the 12-gonal square numbers (see A342709).
The corresponding solutions y are in A033890.
Essentially the same as A246453. - R. J. Mathar, Mar 24 2021

			a(1)^2 - 5 * A033890(1)^2 = 18^2 - 5 * 8^2 = 4.

Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A033890, A342709.

a(n) = 3*A049685(n). - Hugo Pfoertner, Mar 19 2021

LinearRecurrence[{7, -1}, {3, 18}, 20] (* Amiram Eldar, Mar 19 2021 *)
Table[2 ChebyshevT[2 n + 1, 3/2], {n, 0, 20}] (* Eric W. Weisstein, Sep 02 2025 *)
Table[2 Cos[(2 n + 1) ArcCos[3/2]], {n, 0, 20}] // FunctionExpand (* Eric W. Weisstein, Sep 02 2025 *)

a(n) = 7*a(n-1) - a(n-2).
a(n) = 2*T(2*n+1, 3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 02 2022
From Stefano Spezia, Apr 14 2025: (Start)
G.f.: 3*(1 - x)/(1 - 7*x + x^2).
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2)). (End)
a(n) = 2*cos((2*n+1)*arccos(3/2)). - Eric W. Weisstein, Sep 02 2025

A152119 a(n) = Product_{k=1..(n-1)/2} (5 + 4cos(kPi/n)^2).

Original entry on oeis.org

1, 1, 1, 6, 7, 41, 48, 281, 329, 1926, 2255, 13201, 15456, 90481, 105937, 620166, 726103, 4250681, 4976784, 29134601, 34111385, 199691526, 233802911, 1368706081, 1602508992, 9381251041, 10983760033, 64300051206, 75283811239

Offset: 0

Roger L. Bagula and Gary W. Adamson, Nov 24 2008

Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Index entries for linear recurrences with constant coefficients, signature (0,7,0,-1).

Cf. A004187 (bisection), A049685 (bisection).

a = Table[Product[5 + 4*Cos[k*Pi/n]^2, {k, 1, (n - 1)/2}], {n, 0, 10}]; FullSimplify[ExpandAll[a]]
Denominator[NestList[(5/(5+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
LinearRecurrence[{0,7,0,-1},{1,1,1,6,7},30] (* Harvey P. Dale, May 17 2025 *)

a(n) = Product_{k=1..(n-1)/2} (5 + 4*cos(k*Pi/n)^2).
From Joerg Arndt, Jan 24 2013: (Start)
a(n) = 7*a(n-2) - a(n-4).
G.f.: (x^4 - x^3 - 6*x^2 + x + 1)/((x^2 - 3*x + 1)*(x^2 + 3*x + 1)). (End)

cp's OEIS Frontend

A111216 a(n) = 31*a(n-1)-a(n-2).

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A227214 Generalized Markoff numbers: largest of 7-tuple of positive numbers a, b, c, d, e, f, g satisfying the Markoff(7) equation a^2+b^2+c^2+d^2+e^2+f^2+g^2 = 7abcdefg.

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A123971 Triangle T(n,k), read by rows, defined by T(n,k)=3*T(n-1,k)-T(n-1,k-1)-T(n-2,k), T(0,0)=1, T(1,0)=2, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n.

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A161582 The list of the k values in the common solutions to the 2 equations 5*k+1=A^2, 9*k+1=B^2.

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A172511 a(n) = a(n-1) * (11*a(n-1) - a(n-2)) / (a(n-1) + 4*a(n-2)), with a(0) = a(1) = 1.

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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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A276529 a(n) = (a(n-1) * a(n-5) + 1) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

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A161582 The list of the k values in the common solutions to the 2 equations 5k+1=A^2, 9k+1=B^2.

A172511 a(n) = a(n-1) * (11a(n-1) - a(n-2)) / (a(n-1) + 4a(n-2)), with a(0) = a(1) = 1.

A152119 a(n) = Product_{k=1..(n-1)/2} (5 + 4cos(kPi/n)^2).