A003501 -id:A003501 - cp's OEIS frontend

A005386 Area of n-th triple of squares around a triangle.

1, 3, 16, 75, 361, 1728, 8281, 39675, 190096, 910803, 4363921, 20908800, 100180081, 479991603, 2299777936, 11018898075, 52794712441, 252954664128, 1211978608201, 5806938376875, 27822713276176, 133306628004003, 638710426743841, 3060245505715200

Offset: 1

Jean Meeus

a(n)*(-1)^(n+1) is the r=-3 member of the r-family of sequences S_r(n), n>=1, defined in A092184 where more information can be found.

The sequence is the case P1 = 3, P2 = -10, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 1..1000
J. Meeus, Letter to N. J. A. Sloane with attachment, Mar 1975
J. C. G. Nottrot, Vierkantenkransen rond een driehoek, Pythagoras (Netherlands), 14 (1975-1976) 77-81.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,4,-1).

Essentially the same as A003769.
First differences of A099025.
Cf. A100047.

I:=[1, 3, 16]; [n le 3 select I[n] else 4*Self(n-1) +4*Self(n-2) -Self(n-3): n in [1..41]]; // G. C. Greubel, Nov 16 2022

A005386:=-(-1+z)/(z+1)/(z**2-5*z+1); [Conjectured by Simon Plouffe in his 1992 dissertation.]
a:= n-> (Matrix([[0,1,3]]). Matrix(3, (i,j)-> if (i=j-1) then 1 elif j=1 then [4,4,-1][i] else 0 fi)^(n))[1,1]: seq(a(n), n=1..25); # Alois P. Heinz, Aug 05 2008

a[n_]:= Module[{n1=1, n2=0}, Do[{n1, n2}={Sqrt[3]*n1+n2, n1}, {n-1}];n1^2];
Table[a[n], {n,30}]
a[n_]:= Round[((5+Sqrt[21])/2)^n/7]; Table[a[n], {n, 30}]
Rest@(CoefficientList[Series[x/(1-x*(Sqrt[3]+x)), {x, 0, 30}], x])^2
Abs[ChebyshevU[Range[1,40]-1, I*Sqrt[3]/2]]^2 (* G. C. Greubel, Nov 16 2022 *)

def A005386(n): return abs(chebyshev_U(n-1, i*sqrt(3)/2))^2
[A005386(n) for n in range(1,40)] # G. C. Greubel, Nov 16 2022

G.f.: x*(1-x)/((1+x)*(1-5*x+x^2)).
a(n) = 4*a(n-1) + 4*a(n-2) - a(n-3), a(1)=1, a(2)=3, a(3)=16.
a(n) = (2/7)*(T(n, 5/2) - (-1)^n) with twice Chebyshev's polynomials of the first kind evaluated at x=5/2: 2*T(n, 5/2) = A003501(n) = ((5+sqrt(21))^n + (5-sqrt(21))^n)/2^n. - Wolfdieter Lang, Oct 18 2004
a(2*n) = A003690(n). a(2*n+1) = A004253(n)^2. - Alexander Evnin, Mar 11 2012
From Peter Bala, Apr 03 2014: (Start)
a(n) = |U(n-1, sqrt(3)*i/2)|^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 5/2; 1, 3/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)

Edited by Peter J. C. Moses, Apr 23 2004

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 09 2004

A180142 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - x^2)/(1 - 3x - 3x^2).

Original entry on oeis.org

1, 4, 14, 54, 204, 774, 2934, 11124, 42174, 159894, 606204, 2298294, 8713494, 33035364, 125246574, 474845814, 1800277164, 6825368934, 25876938294, 98106921684, 371951579934, 1410175504854, 5346381254364, 20269670277654, 76848154596054, 291353474621124

Offset: 0

Johannes W. Meijer, Aug 13 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a rook on the eight side and corner squares but on the central square the rook goes berserk and turns into a berserker, see A180140.
The sequence above corresponds to 16 A[5] vectors with decimal values between 3 and 384. These vectors lead for the corner squares to A123620 and for the central square to A155116.
This sequence appears among the members of a family of sequences with g.f. (1 + x - k*x^2)/(1 - 3*x + (k-4)*x^2). Berserker sequences that are members of this family are 4*A007482 (k=2; with leading 1 added), A180142 (k=1; this sequence), A000302 (k=0), A180140 (k=-1) and 4*A154964 (k=-2; n>=1 and a(0)=1). Some other members of this family are 2*A180148 (k=3; with leading 1 added), 4*A025192 (k=4; with leading 1 added), 2*A005248 (k=5; with leading 1 added) and A123932 (k=6).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3).

Cf. A180141 (corner squares), A180140 (side squares), A180147 (central square).

with(LinearAlgebra): nmax:=23; m:=2; A[5]:=[0,0,0,0,0,0,0,1,1]: A:= Matrix([[0,1,1,1,0,0,1,0,0], [1,0,1,0,1,0,0,1,0], [1,1,0,0,0,1,0,0,1], [1,0,0,0,1,1,1,0,0], A[5], [0,0,1,1,1,0,0,0,1], [1,0,0,1,0,0,0,1,1], [0,1,0,0,1,0,1,0,1], [0,0,1,0,0,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);
# second Maple program:
a:= n-> ceil((<<0|1>, <3|3>>^n. <<2/3, 4>>)[1,1]):
seq(a(n), n=0..25);  # Alois P. Heinz, Jul 14 2021

LinearRecurrence[{3, 3}, {1, 4, 14}, 26] (* Jean-François Alcover, Jan 18 2025 *)

G.f.: (1 + x - x^2)/(1 - 3*x - 3*x^2).
a(n) = 3*a(n-1) + 3*a(n-2) for n >= 2 with a(0)=1, a(1)=4 and a(2)=14.
a(n) = (6-2*A)*A^(-n-1)/21 + (6-2*B)*B^(-n-1)/21 with A=(-3+sqrt(21))/6 and B=(-3-sqrt(21))/6.
Lim_{k->infinity} a(2*n+k)/a(k) = 2*A000244(n)/(A003501(n) - A004254(n)*sqrt(21)) for n >= 1.
Lim_{k->infinity} a(2*n-1+k)/a(k) = 2*A000244(n)/(A004253(n)*sqrt(21) - 3*A030221(n-1)) for n >= 1.

A099025 Expansion of 1 / ((1+x) * (1-5*x+x^2)).

Original entry on oeis.org

1, 4, 20, 95, 456, 2184, 10465, 50140, 240236, 1151039, 5514960, 26423760, 126603841, 606595444, 2906373380, 13925271455, 66719983896, 319674648024, 1531653256225, 7338591633100, 35161304909276, 168467932913279, 807178359657120, 3867423865372320

Offset: 0

Ralf Stephan, Sep 26 2004

			1 + 4*x + 20*x^2 + 95*x^3 + 456*x^4 + 2184*x^5 + 10465*x^6 + ...

Colin Barker, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
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 (4,4,-1).

First differences of A089927. First differences are in A003769 and A005386. Pairwise sums are in A004254.

I:=[1, 4, 20]; [n le 3 select I[n] else 4*Self(n-1) + 4*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 31 2017

CoefficientList[Series[1/((1+x)*(1-5*x+x^2)), {x,0,50}], x] (* or *) LinearRecurrence[{4,4,-1}, {1,4,20}, 30] (* G. C. Greubel, Dec 31 2017 *)

Vec(1/(1+x)/(1-5*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

{a(n) = (3 * (-1)^n + 38 * subst( poltchebi(n), x, 5/2) - 8 * subst( poltchebi(n-1), x, 5/2)) / 21} /* Michael Somos, Jan 25 2013 */

a(n) = (1/7)*[A030221(n+2) - A003501(n+2) + (-1)^n].
a(n) = 5*a(n-1) -a(n-2) +(-1)^n, a(0)=1, a(1)=4. - Vincenzo Librandi, Mar 22 2011
G.f.: 1 / ((1 + x) * (1 - 5*x + x^2)).
a(-3-n) = -a(n). - Michael Somos, Jan 25 2013
a(n) = (2^(-n)*(3*(-2)^n+(9-2*sqrt(21))*(5-sqrt(21))^n+(5+sqrt(21))^n*(9+2*sqrt(21))))/21. - Colin Barker, Nov 02 2016

A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

Original entry on oeis.org

5, 527, 77132286527, 35395236908668169265765137996816180039862527

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003487.

The next term -- a(4) -- has 175 digits. - Harvey P. Dale, Jun 09 2017

Cf. A001999, A003487, A003501, A219162, A219163, A219165.

NestList[#^4-4#^2+2&,5,5] (* Harvey P. Dale, Jun 09 2017 *)

Let alpha = 1/2*(5 + sqrt(21)). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003487(2*n) = A003501(4^n).
Product_{n>=0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(7/3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 5. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

A060964 Table by antidiagonals where T(n,k) = n*T(n,k-1) - T(n,k-2) with T(n,0) = 2 and T(n,1) = n.

Original entry on oeis.org

2, 0, 2, -2, 1, 2, 0, -1, 2, 2, 2, -2, 2, 3, 2, 0, -1, 2, 7, 4, 2, -2, 1, 2, 18, 14, 5, 2, 0, 2, 2, 47, 52, 23, 6, 2, 2, 1, 2, 123, 194, 110, 34, 7, 2, 0, -1, 2, 322, 724, 527, 198, 47, 8, 2, -2, -2, 2, 843, 2702, 2525, 1154, 322, 62, 9, 2, 0, -1, 2, 2207, 10084, 12098, 6726, 2207, 488, 79, 10, 2

Offset: 0

Henry Bottomley, May 09 2001

			Square array begins as:
  2, 0, -2,   0,   2,    0,    -2, ...
  2, 1, -1,  -2,  -1,    1,     2, ...
  2, 2,  2,   2,   2,    2,     2, ...
  2, 3,  7,  18,  47,  123,   322, ...
  2, 4, 14,  52, 194,  724,  2702, ...
  2, 5, 23, 110, 527, 2525, 12098, ...

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows include A003499, A003500, A003501, A005248, A007395, A056854, A056918, A057079.

Columns include A001477, A007395, A008865, A058794.

T:= function(n,k)
    if k=0 then return 2;
    elif k=1 then return n;
    else return n*T(n,k-1) - T(n,k-2);
    fi; end;
Flat(List([0..12], n-> List([0..n], k-> T(k,n-k) ))); # G. C. Greubel, Jan 15 2020

function T(n,k)
  if k eq 0 then return 2;
  elif k eq 1 then return n;
  else return n*T(n, k-1) - T(n, k-2);
  end if; return T; end function;
[T(k,n-k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020

seq(seq( simplify(k*ChebyshevU(n-k-1, k/2) -2*ChebyshevU(n-k-2, k/2)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020

Table[k*ChebyshevU[n-k-1, k/2] -2*ChebyshevU[n-k-2, k/2], {n,0,12}, {k,0,n} ]//Flatten

T(n,k) = n*polchebyshev(k-1,2,n/2) -2*polchebyshev(k-2,2,n/2);
for(n=0,12, for(k=0,n, print1(T(k,n-k), ", "))) \\ G. C. Greubel, Jan 15 2020

[[k*chebyshev_U(n-k-1, k/2) -2*chebyshev_U(n-k-2, k/2) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020

For all m, T(n, k) = T(n, |m|)*T(n, |k - m|) - T(n, |k - 2m|).
T(n, 2k) = T(n, k)^2 - 2.
T(n, 2k + 1) = T(n, k)*T(n, k + 1) - n.
T(n, 3k) = T(n, k)^3 - 3*T(n, k).
T(n, 4k) = T(n, k)^4 - 4*T(n, k)^2 + 2.
T(n, 5k) = T(n, k)^5 - 5*T(n, k)^3 + 5*T(n, k) etc.
T(n, -k) = T(n, k).
T(-n, k) = T(-n, -k) = (-1)^k * T(n, k).
T(n, k) = ( n*( ((n + sqrt(n^2 -4))/2)^k - ((n - sqrt(n^2 -4))/2)^k ) - 2*( ((n + sqrt(n^2 -4))/2)^(k-1) - ((n - sqrt(n^2 -4))/2)^(k-1) ) )/sqrt(n^2 -4).
T(n, k) = n*ChebyshevU(k-1, n/2) - 2*ChebyshevU(k-2, n/2). - G. C. Greubel, Jan 15 2020

A099867 a(n) = 5*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=9.

Original entry on oeis.org

1, 9, 44, 211, 1011, 4844, 23209, 111201, 532796, 2552779, 12231099, 58602716, 280782481, 1345309689, 6445765964, 30883520131, 147971834691, 708975653324, 3396906431929, 16275556506321, 77980876099676, 373628823992059, 1790163243860619, 8577187395311036

Offset: 0

Creighton Dement, Oct 28 2004

From Klaus Purath, Mar 07 2023: (Start)
For any two terms (a(n), a(n+1)) = (x, y), x^2 - 5*x*y + y^2 = 37 = A082111(4). This is valid in general for all recursive sequences (t) with constant coefficients (5,-1) and t(0) =  1: x^2 - 5*x*y + y^2 = A082111(t(1)-5). This includes and interprets the Feb 04 2014 comment in A004253 by Colin Barker.
By analogy to all this, for three consecutive terms (x, y, z) of any sequence (t) of the form (5,-1) with t(0) = 1: y^2 - x*z = A082111(t(1)-5). (End)

Colin Barker, Table of n, a(n) for n = 0..1000
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A099868, A003501, A004254.

I:=[1,9]; [n le 2 select I[n] else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 30 2015

a[0] = 1; a[1] = 9; a[n_] := a[n] = 5 a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 21}] (* Robert G. Wilson v, Dec 14 2004 *)
LinearRecurrence[{5, -1}, {1, 9}, 30] (* or *) CoefficientList[Series[(1 + 4 x)/(1 - 5 x + x^2), {x, 0, 30}], x] (* Harvey P. Dale, Jun 26 2011 *)

Vec((1+4*x) / (1-5*x+x^2) + O(x^30)) \\ Colin Barker, Mar 31 2017

|2*a(n) + A099868(n) - A003501(n+1)| = 20*A004254(n).
From R. J. Mathar, Sep 11 2008: (Start)
G.f.: (1+4*x) / (1-5*x+x^2).
a(n) = A004254(n+1) + 4*A004254(n).
(End)
a(n) = 2^(-1-n)*((5-sqrt(21))^n*(-13+sqrt(21)) + (5+sqrt(21))^n*(13+sqrt(21))) / sqrt(21). - Colin Barker, Mar 31 2017

A099868 a(n) = 5*a(n-1) - a(n-2), a(0) = 3, a(1) = 25.

Original entry on oeis.org

3, 25, 122, 585, 2803, 13430, 64347, 308305, 1477178, 7077585, 33910747, 162476150, 778470003, 3729873865, 17870899322, 85624622745, 410252214403, 1965636449270, 9417930031947, 45124013710465, 216202138520378, 1035886678891425, 4963231255936747

Offset: 0

Creighton Dement, Oct 28 2004

Colin Barker, Table of n, a(n) for n = 0..1000
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A003501, A004254.

a:=[3,25];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Nov 20 2018

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (3 +10*x)/(1-5*x+x^2))); // G. C. Greubel, Nov 20 2018

a[0]:=3: a[1]:=25: for n from 2 to 30 do a[n]:=5*a[n-1]-a[n-2] od: seq(a[n],n=0..25);

LinearRecurrence[{5,-1}, {3,25}, 30] (* G. C. Greubel, Nov 20 2018 *)

Vec((3+10*x) / (1-5*x+x^2) + O(x^30)) \\ Colin Barker, Mar 28 2017

s=((3+10*x)/(1-5*x+x^2)).series(x,30); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 20 2018

|2*A099867(n) + a(n) - A003501(n+1)| = 20*A004254(n)
G.f.: (3 + 10*x) / (1 - 5*x + x^2). - Emeric Deutsch, Dec 03 2004
a(n) = (2^(-1-n)*((5-sqrt(21))^n*(-35+3*sqrt(21)) + (5+sqrt(21))^n*(35+3*sqrt(21)))) / sqrt(21). - Colin Barker, Mar 28 2017

A217787 a(n) = (a(n-1)*a(n-3) + 1) / a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 4, 9, 14, 19, 43, 67, 91, 206, 321, 436, 987, 1538, 2089, 4729, 7369, 10009, 22658, 35307, 47956, 108561, 169166, 229771, 520147, 810523, 1100899, 2492174, 3883449, 5274724, 11940723, 18606722, 25272721, 57211441, 89150161, 121088881

Offset: 0

Michael Somos, Mar 25 2013

This sequence is similar to A005246 whose recursion is a(n) = (a(n-1)*a(n-2) + 1) / a(n-3). - Michael Somos, Feb 10 2017

			G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 9*x^7 + 14*x^8 + 19*x^9 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..4411
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016. See (better) also. See Example 6.2.18.
Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-1).

Cf. A003501.

Cf. A005246, A048736, A006720, A072876, A072877, A111459.

[n le 3 select 1 else (Self(n)*Self(n-2)+1)/Self(n-3): n in [0..40]]; // Bruno Berselli, Mar 25 2013

a[ n_] := With[{m = If [n < 0, 3 - n, n]}, SeriesCoefficient[ (1 + x + x^2 - 4 x^3 - 3 x^4 - 2 x^5) / (1 - 5 x^3 + x^6), {x, 0, m}]]; (* Michael Somos, Jan 18 2015 *)
LinearRecurrence[{0,0,5,0,0,-1},{1,1,1,1,2,3},40] (* Harvey P. Dale, Nov 20 2016 *)

{a(n) = if( n<0, n = 3-n); polcoeff( (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6) + x * O(x^n), n)};

G.f.: (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6).
a(n) = a(3-n) for all n in Z.
a(n+3) + a(n-3) = 5*a(n) for all n in Z.
a(n+1) + a(n-1) = a(n) * (2 + [n mod 3 == 0]) for all n in Z.
a(n+3k)+a(n-3k) = A003501(k)*a(n) for n>=3k. - Bruno Berselli, Mar 25 2013

A164582 a(n) = 5*a(n - 1) - a(n - 2), with n>2, a(1)=2, a(2)=3.

Original entry on oeis.org

2, 3, 13, 62, 297, 1423, 6818, 32667, 156517, 749918, 3593073, 17215447, 82484162, 395205363, 1893542653, 9072507902, 43468996857, 208272476383, 997893385058, 4781194448907, 22908078859477, 109759199848478, 525887920382913, 2519680402066087

Offset: 1

Vincenzo Librandi, Aug 17 2009

From Klaus Purath, Aug 18 2024: (Start)
For any two consecutive terms (x,y), x^2 - 5xy + y^2 = -17 = A127147(6) always applies. In general, the following applies to all recurrences (t) with constant coefficients (5,-1) and t(0) = 2 and two consecutive terms (x,y): x^2 - 5xy + y^2 = A127147(t(1)+3) for any integer t(1). This includes and interprets the Feb 08 2014 comment on A003501 by Colin Barker.
By analogy to this, for three consecutive terms (x,y,z) of any recurrence (t) of the  form (5,-1) with t(0) = 2: y^2 - xz = A127147(t(1)+3).
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/6, where (t) is any third order recurrence with constant coefficients (6,-6,1) and initial values t(0) = x, t(1) = x + 2, t(2) = x + 5 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/4, where (t) is any third order recurrence with constant coefficients (4,4,-1) and initial values t(0) = x, t(1) = 2 - x, t(2) = x + 1 for any integer x. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-1).

[n le 2 select n+1 else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 12 2013

CoefficientList[Series[(2 - 7 x) / (1 - 5 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 12 2013 *)
LinearRecurrence[{5,-1},{2,3},30] (* Harvey P. Dale, Apr 06 2016 *)

Vec(x*(2 - 7*x) / (1 - 5*x + x^2) + O(x^30)) \\ Colin Barker, Nov 08 2017

a(n) = 5*a(n-1) - a(n-2) = 2*A004254(n) - 7*A004254(n-1).
G.f.: x*(2-7*x) / (1-5*x+x^2).
a(n) = (2^(-1-n)*((5+sqrt(21))^n*(-31+7*sqrt(21)) + (5-sqrt(21))^n*(31+7*sqrt(21)))) / sqrt(21). - Colin Barker, Nov 08 2017
a(n) = (a(n-1)^2 + 17)/a(n-2). - Klaus Purath, Aug 30 2020

Extended by R. J. Mathar, Aug 19 2009

cp's OEIS Frontend

A005386 Area of n-th triple of squares around a triangle.

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A180142 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - x^2)/(1 - 3*x - 3*x^2).

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A099025 Expansion of 1 / ((1+x) * (1-5*x+x^2)).

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A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

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A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

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A060964 Table by antidiagonals where T(n,k) = n*T(n,k-1) - T(n,k-2) with T(n,0) = 2 and T(n,1) = n.

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A099867 a(n) = 5*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=9.

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A180142 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - x^2)/(1 - 3x - 3x^2).