A070997 -id:A070997 - cp's OEIS frontend

A133607 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = -1, with 0 <= k <= n.

1, 0, 1, 0, 1, -1, 0, 1, -1, -1, 0, 1, -1, -2, 1, 0, 1, -1, -3, 2, 1, 0, 1, -1, -4, 3, 3, -1, 0, 1, -1, -5, 4, 6, -3, -1, 0, 1, -1, -6, 5, 10, -6, -4, 1, 0, 1, -1, -7, 6, 15, -10, -10, 4, 1, 0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1

Offset: 0

Philippe Deléham, Dec 27 2007

Previous name: Triangle T(n,k), 0<=k<=n, read by rows given by [0, 1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, -2, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

			Triangle begins:
  1;
  0, 1;
  0, 1, -1;
  0, 1, -1, -1;
  0, 1, -1, -2, 1;
  0, 1, -1, -3, 2, 1;
  0, 1, -1, -4, 3, 3, -1;
  0, 1, -1, -5, 4, 6, -3, -1;
  0, 1, -1, -6, 5, 10, -6, -4, 1;
  0, 1, -1, -7, 6, 15, -10, -10, 4, 1;
  0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  0, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...
Triangle A103631 begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 1, 2, 1;
  0, 1, 1, 3, 2, 1;
  0, 1, 1, 4, 3, 3, 1;
  0, 1, 1, 5, 4, 6, 3, 1;
  0, 1, 1, 6, 5, 10, 6, 4, 1;
  0, 1, 1, 7, 6, 15, 10, 10, 4, 1;
  0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1;
  0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
  0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1;
  ...
Triangle A108299 begins:
  1;
  1, -1;
  1, -1, -1;
  1, -1, -2, 1;
  1, -1, -3, 2, 1;
  1, -1, -4, 3, 3, -1;
  1, -1, -5, 4, 6, -3, -1;
  1, -1, -6, 5, 10, -6, -4, 1;
  1, -1, -7, 6, 15, -10, -10, 4, 1;
  1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...

Another version is A108299.

Unsigned version is A103631 (T(n,k) = A103631(n,k)*A057077(k)).

m = 13
(* DELTA is defined in A084938 *)
DELTA[Join[{0, 1}, Table[0, {m}]], Join[{1, -2, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k-1) qStirling2[n-1, k-1, q] + Sum[q^j, {j, 0, k-1}] qStirling2[n-1, k, q];
qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
qStirling2[0, k_, _] := KroneckerDelta[0, k];
qStirling2[, , _] = 0;
Table[qStirling2[n, k, -1], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *)

from sage.combinat.q_analogues import q_stirling_number2
for n in (0..9):
    print([q_stirling_number2(n,k).substitute(q=-1) for k in [0..n]])
# Peter Luschny, Mar 09 2020

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)= A057077(n), A010892(n), A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n-1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 respectively .
Sum_{k, 0<=k<=n}T(n,k)*x^k = A000007(n), A010892(n), A133631(n), A133665(n), A133666(n), A133667(n), A133668(n), A133669(n), A133671(n), A133672(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively .
G.f.: (1-x+y*x)/(1-x+y^2*x^2). - Philippe Deléham, Mar 14 2012
T(n,k) = T(n-1,k) - T(n-2,k-2), T(0,0) = T(1,1) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(2,2) = -1 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 14 2012

New name from Peter Luschny, Mar 09 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)

A071049 Number of 1's in n-th generation of 1-D CA using Rule 110, started with a single 1.

Original entry on oeis.org

1, 2, 3, 3, 5, 3, 5, 6, 8, 5, 6, 8, 8, 8, 11, 11, 13, 9, 11, 11, 13, 14, 16, 14, 14, 13, 13, 17, 22, 20, 16, 17, 24, 19, 14, 19, 25, 18, 20, 25, 24, 19, 24, 31, 27, 26, 24, 22, 32, 31, 28, 24, 29, 34, 30, 31, 37, 34, 34, 36, 35, 34, 35, 36, 43, 40, 36, 38, 37, 39, 40

Offset: 0

Hans Havermann, May 26 2002

nonn

Number of 1's in n-th row of triangle in A070887.

Although the initial behavior is chaotic, it is an astonishing fact, pointed out by Wolfram [2002, p. 39], that after about three thousand terms all the irregularities disappear. - N. J. A. Sloane, May 15 2015

Matthew Cook, A Concrete View of Rule 110 Computation, in "The Complexity of Simple Programs", T. Neary, D. Woods, A. K. Seda, and N. Murphy (Eds.), 2008, pp. 31-55.
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Vincenzo Librandi and N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (First 1000 terms from Vincenzo Librandi)
Matthew Cook, A Concrete View of Rule 110 Computation, arXiv:0906.3248 [cs.CC], 2009.
Matthew Cook, Universality in Elementary Cellular Automata, Complex Systems 15 (2004), 1-40.
N. J. A. Sloane, Illustration of first 20 generations
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Rule 110
Wikipedia, Rule 110
Index entries for sequences related to cellular automata

Cf. A070950, A070952, A070997, A071029, A071044, A151931, A246027.

A071049 := proc(n)
    add( A070887(n+1,k),k=1..n+1) ;
end proc:
seq(A071049(n),n=0..20) ; # R. J. Mathar, Feb 18 2015

Total /@ CellularAutomaton[110,{{1},0},100] (* N. J. A. Sloane, Aug 10 2009 *)

For n >= 2854, a(n+469) = -a(n+453) + a(n+256) + a(n+240) + a(n+229) + a(n+213) - a(n+16) - a(n). - N. J. A. Sloane, May 15 2015

Added references and links. - N. J. A. Sloane, Aug 09 2014

Changed offset to make consistent with A070952, etc. - N. J. A. Sloane, Aug 15 2014

A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321

Offset: 0

Peter Bala, Apr 30 2012

The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1) + r) with initial conditions a(0) = 1, a(1) = 1 + r has the solution a(n) = 1/2 + (1/2)*Sum_{k = 0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613 (r = -2), A146983 (r = -3) and A084159(r = -4).
Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - Colin Barker, Jan 01 2015

Seiichi Manyama, Table of n, a(n) for n = 0..1116 (first 201 terms from Vincenzo Librandi)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A011900, A054318, A070997, A101265, A109613, A133872, A146983, A211955.

I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // Vincenzo Librandi, May 18 2012

RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* Harvey P. Dale, May 14 2012 *)

Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n) = 1/2 + (1/2)*Sum_{k = 0..n} 6^k*binomial(n+k,2*k).
a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.
a(n) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the n-th Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2) - 3 with a(0) = 1 and a(1) = 4.
O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....
Sum_{n >= 0} 1/a(n) = sqrt(5/3); 5 - 3*(Sum_{k = 0..2*n} 1/a(k))^2 = 2/A070997(n)^2.
a(0) = 1, a(1) = 4, a(2) = 28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - Harvey P. Dale, May 14 2012

A080871 a(n)a(n+3) - a(n+1)a(n+2) = 3, given a(0)=a(1)=1, a(2)=4.

Original entry on oeis.org

1, 1, 4, 7, 31, 55, 244, 433, 1921, 3409, 15124, 26839, 119071, 211303, 937444, 1663585, 7380481, 13097377, 58106404, 103115431, 457470751, 811826071, 3601659604, 6391493137, 28355806081, 50320119025, 223244789044

Offset: 0

Paul D. Hanna, Feb 22 2003

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0, 8, 0, -1).

Cf. A001519, A079496, A080872, A080873, A080874, A080875.

Bisections are A001091 and A070997.

RecurrenceTable[{a[0]==a[1]==1,a[2]==4,a[n]==(3+a[n+1]a[n+2])/a[n+3]},a,{n,30}] (* Harvey P. Dale, Jun 08 2017 *)

a(n) = (3 + a(n-1)*a(n-2))/a(n-3) for n>2.
G.f.: (-x^3 - 4*x^2 + x + 1)/(x^4 - 8*x^2 + 1)
a(n+4) = 8*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]
a(n) = (0.25 + sqrt(10)/20)*(sqrt(4 + sqrt(15)))^n + (0.25 + sqrt(10)/20)*(sqrt(4 - sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - sqrt(4 + sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(4 - sqrt(15))))^n. [Richard Choulet, Dec 06 2008]

A237262 Values of x in the solutions to x^2 - 8xy + y^2 + 11 = 0, where 0 < x < y.

Original entry on oeis.org

1, 2, 6, 15, 47, 118, 370, 929, 2913, 7314, 22934, 57583, 180559, 453350, 1421538, 3569217, 11191745, 28100386, 88112422, 221233871, 693707631, 1741770582, 5461548626, 13712930785, 42998681377, 107961675698, 338527902390, 849980474799, 2665224537743

Offset: 1

Colin Barker, Feb 05 2014

The corresponding values of y are given by a(n+2).

Also values of y in the solutions to the negative Pell equation x^2 - 15*y^2 = -11. - Colin Barker, Jan 25 2017

			6 is a term because (x, y) = (6, 47) is a solution to x^2 - 8xy + y^2 + 11 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-1).

Cf. A001091, A070997, A199336, A281584.

For first and second differences see A322780, A199336.

LinearRecurrence[{0,8,0,-1},{1,2,6,15},30] (* Harvey P. Dale, Sep 06 2020 *)

Vec(-x*(x-1)*(x^2+3*x+1)/(x^4-8*x^2+1) + O(x^100))

G.f.: -x*(x-1)*(x^2 + 3*x + 1) / (x^4 - 8*x^2 + 1).
a(n) = 8*a(n-2) - a(n-4) for n > 4.
a(n) = (11*a(n-1) - 4*a(n-2))/3 if n is odd; a(n) = (11*a(n-1) - 3*a(n-2))/4 if n is even. - R. J. Mathar, Jun 18 2014

A098306 Unsigned member r=-6 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 6, 49, 384, 3025, 23814, 187489, 1476096, 11621281, 91494150, 720331921, 5671161216, 44648957809, 351520501254, 2767515052225, 21788599916544, 171541284280129, 1350541674324486, 10632792110315761, 83711795208201600

Offset: 0

Wolfdieter Lang, Oct 18 2004

((-1)^(n+1))*a(n) = S_{-6}(n), n>=0, defined in A092184.

This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 6, P2 = -16, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, Linear divisibility sequences and Chebyshev polynomials
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 (7,7,-1).

Cf. A001090, A001091, A070997, A092184, A100047.

a[n_] := 1/10*((4 + Sqrt[15])^n + (4 - Sqrt[15])^n - 2*(-1)^n) // Simplify; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 28 2017 *)
LinearRecurrence[{7, 7, -1}, {0, 1, 6, 49, 384, 3025}, 50] (* G. C. Greubel, Aug 08 2017 *)

x='x+O('x^50); Vec(x*(1-x)/((1+x)*(1-8*x+x^2))) \\ G. C. Greubel, Aug 08 2017

a(n) = (T(n, 4)-(-1)^n)/5, with Chebyshev's polynomials of the first kind evaluated at x=4: T(n, 4)=A001091(n)=((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 8*a(n-1) - a(n-2) + 2*(-1)^(n+1), n>=2, a(0)=0, a(1)=1.
a(n) = 7*a(n-1) + 7*a(n-2) - a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=6.
G.f.: x*(1-x)/((1+x)*(1-8*x+x^2)) = x*(1-x)/(1-7*x-7*x^2+x^3) (from the Stephan link, see A092184).
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001090(n)^2; a(2*n+1) = A070997(n)^2.
a(n) = |u(n)|^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(-6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(-6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(-6)/2)*U(n-1,-sqrt(-6)/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = 1/10*( (4 + sqrt(15))^n + (4 - sqrt(15))^n - 2*(-1)^n ).
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 4; 1, 3] 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)

A136325 a(n) = 8*a(n-1)-a(n-2) with a(0)=0 and a(1)=3.

Original entry on oeis.org

0, 3, 24, 189, 1488, 11715, 92232, 726141, 5716896, 45009027, 354355320, 2789833533, 21964312944, 172924670019, 1361433047208, 10718539707645, 84386884613952, 664376537203971, 5230625413017816, 41180626766938557

Offset: 0

Lorenz H. Menke, Jr., Mar 26 2008

Nonnegative integers k such that 15*k^2 + 9 is a square.

From the recurrence we have a(n) = sqrt(15)*((4 + sqrt(15))^n - (4 - sqrt(15))^n)/10.

			G.f. = 3*x + 24*x^2 + 189*x^3 + 1488*x^4 + 11715*x^5 + 92232*x^6 + 726141*x^7 + ...

Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A001090.

Do[If[IntegerQ[Sqrt[3 (3 + 5 x^2)]], Print[{x, Sqrt[3 (3 + 5 x^2)]}]], {x, 0, 2000000}]
LinearRecurrence[{8,-1},{0,3},30] (* Harvey P. Dale, Aug 18 2014 *)
a[ n_] := 3 ChebyshevU[ n - 1, 4]; (* Michael Somos, Oct 14 2015 *)
a[ n_] := 3/2 ((4 + Sqrt[15])^n - (4 - Sqrt[15])^n) / Sqrt[15] // Simplify; (* Michael Somos, Oct 14 2015 *)

{a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 5}; /* Michael Somos, Apr 05 2008 */

{a(n) = 3 * polchebyshev(n-1, 2, 4)}; /* Michael Somos, Oct 14 2015 */

{a(n) = 3 * imag( (4 + quadgen(60))^n )}; /* Michael Somos, Oct 14 2015 */

From Colin Barker, Jan 24 2013: (Start)
a(n) = (sqrt(3/5)*(-(4-sqrt(15))^n + (4+sqrt(15))^n))/2.
G.f.: 3*x/(x^2-8*x+1). (End)
a(n) = 3 * A001090(n).
For n > 0, a(n) is the denominator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the numerators see A070997. - Greg Dresden, Sep 12 2019

Definition corrected by Bruno Berselli, Jan 24 2013
Definition, comments, formulas further corrected by Greg Dresden, Sep 13 2019
Exchanged definition and comment, in order to retain offset 0. - N. J. A. Sloane, Sep 23 2019

A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3n(j) + 1 = a(j)a(j) and 5n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 16, 1008, 62496, 3873760, 240110640, 14882985936, 922505017408, 57180428093376, 3544264036771920, 219687189851765680, 13617061506772700256, 844038126230055650208, 52316746764756677612656, 3242794261288683956334480, 201000927453133648615125120

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..559
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Cf. A057080, A070997, A157456, A245031.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(16*x^2/((1-x)*(1-62*x+x^2)))); // G. C. Greubel, Jun 02 2018

for a from 1 by 2 to 100000 do b:=sqrt((5*a*a-2)/3): if (trunc(b)=b) then
n:=(a*a-1)/3: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq((4/15)*(simplify(ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)) -1), n=1..30); # G. C. Greubel, Sep 27 2022

CoefficientList[Series[16*x/((1-x)*(1-62*x+x^2)), {x, 0, 30}], x] (* G. C. Greubel, Jun 02 2018 *)
LinearRecurrence[{63,-63,1},{0,16,1008},30] (* Harvey P. Dale, May 07 2022 *)

concat(0, Vec(16*x^2/((1-x)*(1-62*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 25 2015

[(4/15)*(-1 + chebyshev_U(n, 31) - 61*chebyshev_U(n-1, 31)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

The a(j) recurrence is a(1)=1, a(2)=7, a(t+2) = 8*a(t+1) - a(t) resulting in terms 1, 7, 55. 433, 3409, ... (A070997).
The b(j) recurrence is b(1)=1, b(2)=9, b(t+2) = 8*b(t+1) - b(t) resulting in terms 1, 9, 71, 559, 4401, ... (A057080).
The n(j) recurrence is n(0) = n(1) = 0, n(2)=16, n(t+3) = 63*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 16, 1008, 62496, ... (this sequence).
From Colin Barker, Sep 25 2015: (Start)
a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3) for n>3.
G.f.: 16*x^2 / ((1-x)*(1-62*x+x^2)). (End)
a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) -(-4+sqrt(15))*(31+8*sqrt(15))^n)/30. - Colin Barker, Mar 03 2016
a(n) = (4/15)*(-1 + ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)). - G. C. Greubel, Sep 27 2022

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

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A182432 Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1) + 3) with a(0) = 1, a(1) = 4.

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A237262 Values of x in the solutions to x^2 - 8*x*y + y^2 + 11 = 0, where 0 < x < y.

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A098306 Unsigned member r=-6 of the family of Chebyshev sequences S_r(n) defined in A092184.

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A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

A080871 a(n)a(n+3) - a(n+1)a(n+2) = 3, given a(0)=a(1)=1, a(2)=4.

A237262 Values of x in the solutions to x^2 - 8xy + y^2 + 11 = 0, where 0 < x < y.

A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3n(j) + 1 = a(j)a(j) and 5n(j) + 1 = b(j)b(j) with positive integer numbers.