A001835 -id:A001835 - cp's OEIS frontend

A084917 Positive numbers of the form 3*y^2 - x^2.

2, 3, 8, 11, 12, 18, 23, 26, 27, 32, 39, 44, 47, 48, 50, 59, 66, 71, 72, 74, 75, 83, 92, 98, 99, 104, 107, 108, 111, 122, 128, 131, 138, 143, 146, 147, 156, 162, 167, 176, 179, 183, 188, 191, 192, 194, 200, 207, 218, 219, 227, 234, 236, 239, 242, 243, 251, 263, 264, 275, 282, 284

Offset: 1

Roger Cuculière, Jul 14 2003

Positive integers k such that x^2 - 4xy + y^2 + k = 0 has integer solutions. (See the CROSSREFS section for sequences relating to solutions for particular k.)
Comments on method used, from Colin Barker, Jun 06 2014: (Start)
In general, we want to find the values of f, from 1 to 400 say, for which x^2 + bxy + y^2 + f = 0 has integer solutions for a given b.
In order to solve x^2 + bxy + y^2 + f = 0 we can solve the Pellian equation x^2 - Dy^2 = N, where D = b*b - 4 and N = 4*(b*b - 4)*f.
But since sqrt(D) < N, the classical method of solving x^2 - Dy^2 = N does not work. So I implemented the method described in the 1998 sci.math reference, which says:
"There are several methods for solving the Pellian equation when |N| > sqrt(d). One is to use a brute-force search. If N < 0 then search on y = sqrt(abs(n/d)) to sqrt((abs(n)(x1 + 1))/(2d)) and if N > 0 search on y = 0 to sqrt((n(x1 - 1))/(2d)) where (x1, y1) is the minimum positive solution (x, y) to x^2 - dy^2 = 1. If N < 0, for each positive (x, y) found by the search, also take (-x, y). If N > 0, also take (x, -y). In either case, all positive solutions are generated from these using (x1, y1) in the standard way."
Incidentally all my Pell code is written in B-Prolog, and is somewhat voluminous. (End)
Also, positive integers of the form -x^2 + 2xy + 2y^2 of discriminant 12. - N. J. A. Sloane, May 31 2014 [Corrected by Klaus Purath, May 07 2023]
The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.
The equivalent sequence for x^2 - 5xy + y^2 + k = 0 is A237351.
A positive k does not appear in this sequence if and only if there is no integer solution of x^2 - 3*y^2 = -k with (i) 0 < y^2 <= k/2 and (ii) 0 <= x^2 <= k/2. See the Nagell reference Theorems 108 a and 109, pp. 206-7, with D = 3, N = k and (x_1,y_1) = (2,1). - Wolfdieter Lang, Jan 09 2015
From Klaus Purath, May 07 2023: (Start)
There are no squares in this sequence. Products of an odd number of terms as well as products of an odd number of terms and any terms of A014209 belong to the sequence.
Products of an even number of terms are terms of A014209. The union of this sequence and A014209 is closed under multiplication.
A positive number belongs to this sequence if and only if it contains an odd number of prime factors congruent to {2, 3, 11} modulo 12. If it contains prime factors congruent to {5, 7} modulo 12, these occur only with even exponents. (End)
From Klaus Purath, Jul 09 2023: (Start)
Any term of the sequence raised to an odd power also belongs to the sequence. Proof: t^(2n+1) = t*t^2n = (3*x^2 - y^2)*t^2n = 3*(x*t^n)^2 - (y*t^n)^2. It seems that t^(2n+1) occurs only if t also is in the sequence.
Joerg Arndt has proved that there are no squares in this sequence: Assume s^2 = 3*y^2 - x^2, then s^2 + x^2 = 3 * y^2, but the sum of two squares cannot be 3 * y^2, qed. (End)
That products of any 3 terms belong to the sequence can be proved by the following identity:  (na^2 - b^2) (nc^2 - d^2) (ne^2 - f^2) = n[a(nce + df) + b(cf + de)]^2 - [na(cf + de) + b(nce + df)]^2. This can be verified by expanding both sides of the equation. - Klaus Purath, Jul 14 2023

			11 is in the sequence because 3 * 3^2 - 4^2 = 27 - 16 = 11.
12 is in the sequence because 3 * 4^2 - 6^2 = 48 - 36 = 12.
13 is not in the sequence because there is no solution in integers to 3y^2 - x^2 = 13.
From _Wolfdieter Lang_, Jan 09 2015: (Start)
Referring to the Jan 09 2015 comment above.
k = 1 is out because there is no integer solution of (i) 0 < y^2 <= 1/2.
For k = 4, 5, 6, and 7 one has y = 1, x = 0, 1 (and the negative of this). But x^2 - 3 is not -k for these k and x values. Therefore, these k values are missing.
For k = 8 .. 16 one has y = 1, 2 and x = 0, 1, 2. Only y = 2 has a chance and only for k = 8, 11 and 12 the x value 2, 1 and 0, respectively, solves x^2 - 12 = -k. Therefore 9, 10, 13, 14, 15, 16 are missing.
... (End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Sci.math, General Pell equation: x^2 - N*y^2 = D, 1998
Sci.math, General Pell equation: x^2 - N*y^2 = D, 1998 (Edited and cached copy)
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

With respect to solutions of the equation in the early comment, see comments etc. in: A001835 (k = 2), A001075 (k = 3), A237250 (k = 11), A003500 (k = 12), A082841 (k = 18), A077238 (k = 39).
Cf. A031363, A237351.
A141123 gives the primes.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

r[n_] := Reduce[n == 3*y^2 - x^2 && x > 0 && y > 0, {x, y}, Integers]; Reap[For[n = 1, n <= 1000, n++, rn = r[n]; If[rn =!= False, Print["n = ", n, ", ", rn /. C[1] -> 1 // Simplify]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jan 21 2016 *)
Select[Range[300],Length[FindInstance[3y^2-x^2==#,{x,y},Integers]]>0&] (* Harvey P. Dale, Apr 23 2023 *)

Terms 26 and beyond from Colin Barker, Feb 06 2014

A082841 a(n) = 4*a(n-1) - a(n-2) for n>1, a(0)=3, a(1)=9.

Original entry on oeis.org

3, 9, 33, 123, 459, 1713, 6393, 23859, 89043, 332313, 1240209, 4628523, 17273883, 64467009, 240594153, 897909603, 3351044259, 12506267433, 46674025473, 174189834459, 650085312363, 2426151414993, 9054520347609, 33791929975443

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Apr 14 2003

y-values in the solutions to 3*x^2+6 = y^2. - Sture Sjöstedt, Nov 25 2011
Positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 18 = 0. - Colin Barker, Feb 04 2014
Positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 288 = 0. - Colin Barker, Feb 16 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 (4,-1).

First differences of A005320.

Cf. A001834.

a:=[3,9];; for n in [3..30] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Feb 25 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (3-6*x+3*x^2)/((1-x)*(1-4*x+x^2)) )); // G. C. Greubel, Feb 25 2019

a:=proc(n) option remember; if n=0 then 3 elif n=1 then 9 else 4*a(n-1)-a(n-2); fi; end: seq(a(n), n=0..40); # Wesley Ivan Hurt, Jan 21 2017

CoefficientList[Series[(3-6x+3x^2)/((1-x)(1-4x+x^2)), {x, 0, 25}], x]
LinearRecurrence[{4,-1},{3,9},30] (* Harvey P. Dale, Aug 28 2019 *)

my(x='x+O('x^30)); Vec((3-6*x+3*x^2)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

((3-6*x+3*x^2)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: (3 -6*x +3*x^2)/((1-x)*(1-4*x+x^2)).
a(n) = sqrt(3/2)*(a^(n+1/2) + b^(n+1/2)), with a=2+sqrt(3) and b=2-sqrt(3).
a(n) = sqrt(3*(11 +12*A082840(n) +4*A082840(n)^2)).
a(n) = sqrt((3/2)*(A003500(2n+1) +2)).
a(n) - a(n-1) = 6*A001353(n).
a(n) == 3 (mod 6).
a(n) = 3 * A001835(n+1).
a(n) = 3*x(n) + 3*y(n) for x(n)= A001075(n) and y(n) = A001353(n) the solutions to x^2 - 3*y^2 = 1. - Greg Dresden and his Math 222 Linear Algebra class, Oct 05 2022

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

Original entry on oeis.org

1, 1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 28 2005

Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is an integer. For this sequence K=4. - Andrey Vyshnevyy, Sep 18 2015
The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - N. J. A. Sloane, Sep 08 2018

Seiichi Manyama, Table of n, a(n) for n = 0..1750
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A061278, A001353, A001835, A005246, A276122, A276271.

I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* Vincenzo Librandi, Sep 18 2015 *)
CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 11 2016 *)
a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* Michael Somos, Jul 09 2017 *)

M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))

{a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* Michael Somos, Jul 09 2017 */

a(n) = A101265(n), n>0. - R. J. Mathar, Aug 30 2008
a(n) = A079935(n+1) - A001571(n). - Gerry Martens, Jun 05 2015
a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - Seiichi Manyama, Aug 11 2016
From Ilya Gutkovskiy, Aug 11 2016: (Start)
G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
a(n) = 4*a(n-1) - a(n-2) - 1. - Seiichi Manyama, Aug 26 2016
From Seiichi Manyama, Sep 03 2016: (Start)
a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
a(n) = A005246(n)*A005246(n+1). (End)
From Michael Somos, Jul 09 2017: (Start)
0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)

a(26)-a(27) from Vincenzo Librandi, Sep 18 2015

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

Original entry on oeis.org

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

A046184 Indices of octagonal numbers which are also squares.

Original entry on oeis.org

1, 9, 121, 1681, 23409, 326041, 4541161, 63250209, 880961761, 12270214441, 170902040409, 2380358351281, 33154114877521, 461777249934009, 6431727384198601, 89582406128846401, 1247721958419651009, 17378525011746267721, 242051628206028097081

Offset: 1

Eric W. Weisstein

The equation a(t)*(3*a(t)-2) = m*m is equivalent to the Pell equation (3*a(t)-1)*(3*a(t)-1) - 3*m*m = 1. - Paul Weisenhorn, May 12 2009
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 16 2011
Also numbers n such that the octagonal number N(n) is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Also nonnegative integers y in the solutions to 2*x^2 - 6*y^2 + 4*x + 4*y + 2 + 2 = 0, the corresponding values of x being A251963. - Colin Barker, Dec 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Octagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A028230, A036428, A251963.

Cf. A006253.

I:=[1, 9, 121]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Nov 17 2011

LinearRecurrence[ {15, -15, 1}, {1, 9, 121}, 17 ] (* Ant King, Nov 16 2011 *)
CoefficientList[Series[x (1-6x+x^2)/((1-x)(1-14x+x^2)),{x,0,30}],x] (* Harvey P. Dale, Sep 01 2021 *)

Vec(x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)) + O(x^100)) \\ Colin Barker, Dec 11 2014

{n: A000567(n) in A000290}.
Nearest integer to (1/6) * (2+sqrt(3))^(2n-1). - Ralf Stephan, Feb 24 2004
a(n) = A045899(n-1) + 1 = A051047(n+1) + 1 = A003697(2n-2). - N. J. A. Sloane, Jun 12 2004
a(n) = A001835(n)^2. - Lekraj Beedassy, Jul 21 2006
From Paul Weisenhorn, May 12 2009: (Start)
With A=(2+sqrt(3))^2=7+4*sqrt(3) the equation x*x-3*m*m=1 has solutions
x(t) + sqrt(3)*m(t) = (2+sqrt(3))*A^t and the recurrences
x(t+2) = 14*x(t+1) - x(t) with  = 2, 26, 362, 5042
m(t+2) = 14*m(t+1) - m(t) with  = 1, 15, 209, 2911
a(t+2) = 14*a(t+1) - a(t) - 4 with  = 1, 9, 121, as above. (End)
From Ant King, Nov 15 2011: (Start)
a(n) = 14*a(n-1) - a(n-2) - 4.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (1/6)*( (2+sqrt(3))^(2n-1) + (2-sqrt(3))^(2n-1) + 2 ).
a(n) = ceiling( (1/6)*(2 + sqrt(3))^(2n-1) ).
a(n) = (1/6)*( (tan(5*Pi/12))^(2n-1) + (tan(Pi/12))^(2n-1) + 2 ).
a(n) = ceiling ( (1/6)*(tan(5*Pi/12))^(2n-1) ).
G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)
a(n) = A006253(2n-2). - Andrey Goder, Oct 17 2021

A180147 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + 3x)/(1 - 4x - 3x^2 + 6x^3).

Original entry on oeis.org

1, 7, 31, 139, 607, 2659, 11623, 50827, 222223, 971635, 4248247, 18574555, 81213151, 355086787, 1552539271, 6788138539, 29679651247, 129767784979, 567381262423, 2480750497147, 10846539065983, 47424120180835

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 the central square (m = 5) 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.
On a 3 X 3 chessboard there are 2^9 = 512 ways to go berserk on the central square (we assume here that a berserker might behave like a rook). The berserker is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the central squares the 512 berserkers lead to 42 berserker sequences, see the cross-references for some examples.
The sequence above corresponds to six A[5] vectors with decimal values between 191 and 506. These vectors lead for the corner squares to A180145 and for the side squares to A180146.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, 3, -6).

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

Cf. Berserker sequences central square [numerical values A[5]]: A000007 [0], A000012 [16], 2*A001835 [17, n>=1 and a(0)=1], A155116 [3], A077829 [7], A000302 [15], 6*A179606 [111, with leading 1 added], 2*A033887 [95, n>=1 and a(0)=1], A180147 [191, this sequence], 2*A180141 [495, n>=1 and a(0)=1], 4*A107979 [383, with leading 1 added].

with(LinearAlgebra): nmax:=22; m:=5; A[5]:=[0,1,0,1,1,1,1,1,1]: 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);

CoefficientList[Series[(1+3x)/(1-4x-3x^2+6x^3),{x,0,40}],x] (* or *) LinearRecurrence[{4,3,-6},{1,7,31},40] (* Harvey P. Dale, Oct 10 2011 *)

G.f.: (1+3*x)/(1 - 4*x - 3*x^2 + 6*x^3).
a(n) = 4*a(n-1) + 3*a(n-2) - 6*a(n-3) with a(0)=1, a(1)=7 and a(2)=31.
a(n) = -1/2 + (7+6*A)*A^(-n-1)/22 + (7+6*B)*B^(-n-1)/22 with A=(-3+sqrt(33))/12 and B=(-3-sqrt(33))/12.
a(n) = A180146(n) + 3*A180146(n-1) with A180146(-1) = 0.

A350920 a(0) = 5, a(1) = 5, and a(n) = 4*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

5, 5, 11, 35, 125, 461, 1715, 6395, 23861, 89045, 332315, 1240211, 4628525, 17273885, 64467011, 240594155, 897909605, 3351044261, 12506267435, 46674025475, 174189834461, 650085312365, 2426151414995, 9054520347611, 33791929975445, 126113199554165, 470660868241211, 1756530273410675, 6555460225401485, 24465310628195261, 91305782287379555

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001835, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350921, A350922, A350923, A350924, A350925, A350926.

a(n) = 3*A001835(n) + 2. - Hugo Pfoertner, Jan 22 2022

G.f.: (5 - 20*x + 11*x^2)/((1 - x)*(1 - 4*x + x^2)). - Stefano Spezia, Jan 22 2022

A085376 Ratio-dependent insertion sequence I(0.36704) (see the link below).

Original entry on oeis.org

1, 3, 11, 30, 109, 297, 1079, 2940, 10681, 29103, 105731, 288090, 1046629, 2851797, 10360559, 28229880, 102558961, 279447003, 1015229051, 2766240150, 10049731549, 27382954497, 99482086439, 271063304820, 984771132841

Offset: 1

John W. Layman, Jun 26 2003

nonn

This sequence is the ratio-determined insertion sequence (RDIS) "twin" of I(0.37802)=A080874 and "child" of I(0.33344)=A001835 and I(0.38208)=A001906 in the RDIS recurrence tree (see the link for an explanation of terms). See A082630, A082981, A085348 and A085349 for recent examples of RDIS sequences.

Conjecture: partial sums of A129445. - Sean A. Irvine, Jul 14 2022

John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001835, A001906, A080874, A082630, A082981, A085348, A085349.

It is conjectured that a(n) = 10*a(n-2) - a(n-4).

Apparently a(n)*a(n+3) = -3 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004

A121314 Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 5, 6, 1, 0, 1, 7, 15, 10, 1, 0, 1, 9, 28, 35, 15, 1, 0, 1, 11, 45, 84, 70, 21, 1, 0, 1, 13, 66, 165, 210, 126, 28, 1, 0, 1, 15, 91, 286, 495, 462, 210, 36, 1

Offset: 0

Philippe Deléham, Aug 25 2006

A054142 with first diagonal 1, 0, 0, 0, 0, 0, 0, 0, ...

Mirror image of triangle in A165253.

			Triangle begins
  1;
  0,  1;
  0,  1,  1;
  0,  1,  3,  1;
  0,  1,  5,  6,  1;
  0,  1,  7, 15, 10,  1;
  0,  1,  9, 28, 35, 15,  1;
  0,  1, 11, 45, 84, 70, 21,  1;

Alois P. Heinz, Rows n = 0..140, flattened
F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Cf. A054142, A165253.

T(0,0)=1; T(n,0)=0 for n > 0; T(n+1,k+1) = binomial(2*n-k,k)for n >= 0 and k >= 0.
Sum_{k=0..n} T(n,k)*x^k = A001519(n), A047849(n), A165310(n), A165311(n), A165312(n), A165314(n), A165322(n), A165323(n), A165324(n) for x = 1,2,3,4,5,6,7,8,9 respectively.
Sum_{k=0..n} 2^k*T(n,k) = (4^n+2)/3.
Sum_{k=0..n} 2^(n-k)*T(n,k) = A001835(n).
Sum_{k=0..n} 3^k*4^(n-k)*T(n,k) = A054879(n). - Philippe Deléham, Aug 26 2006
Sum_{k=0..n} T(n,k)*(-1)^k*2^(3n-2k) = A143126(n). - Philippe Deléham, Oct 31 2008
Sum_{k=0..n} T(n,k)*(-1)^k*3^(n-k) = A138340(n)/4^n. - Philippe Deléham, Nov 01 2008
G.f.: (1-(y+1)*x)/(1-(2y+1)*x+y^2*x^2). - Philippe Deléham, Nov 01 2011
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2), T(0,0) = T(1,1) = 1, T(1,0) = 0. - Philippe Deléham, Feb 19 2012

A028471 Number of perfect matchings (or domino tilings) in the graph P_9 X P_2n.

Original entry on oeis.org

1, 55, 6336, 817991, 108435745, 14479521761, 1937528668711, 259423766712000, 34741645659770711, 4652799879944138561, 623139489426439754945, 83456125990631342400791, 11177167872295392172767936, 1496943834332592837945956455, 200483802581126644843760725601

Offset: 0

Per H. Lundow

nonn

Sean A. Irvine, Table of n, a(n) for n = 0..150
J. L. Hock and R. B. McQuistan, A note on the occupational degeneracy for dimers on a saturated two-dimensional lattice space, Discrete Applied Mathematics, 1984, v.8, 101-104.
Per Hakan Lundow, Computation of matching polynomials and the number of 1-factors in polygraphs, Research report, No 12, 1996, Department of Math., Umea University, Sweden.
Per Hakan Lundow, Enumeration of matchings in polygraphs, 1998.
Index entries for linear recurrences with constant coefficients, signature (209, -11936, 274208, -3112032, 19456019, -70651107, 152325888, -196664896, 152325888, -70651107, 19456019, -3112032, 274208, -11936, 209, -1).

Cf. A000045, A001835, A005178, A003775, A028468, A028469, A028470.

Row 9 of array A099390.

T[?OddQ, ?OddQ] = 0;
T[m_, n_] := Product[2(2+Cos[2 j Pi/(m+1)]+Cos[2 k Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}];
a[n_] := T[2n, 9] // Round;
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 28 2022 *)

{a(n) = sqrtint(polresultant(polchebyshev(2*n, 2, x/2), polchebyshev(9, 2, I*x/2)))} \\ Seiichi Manyama, Apr 13 2020

a(n) = 209*a(n-1) - 11936*a(n-2) + 274208*a(n-3) - 3112032*a(n-4) + 19456019*a(n-5) - 70651107*a(n-6) + 152325888*a(n-7) - 196664896*a(n-8) + 152325888*a(n-9) - 70651107*a(n-10) + 19456019*a(n-11) - 3112032*a(n-12) + 274208*a(n-13) - 11936*a(n-14) + 209*a(n-15) - a(n-16). - Jay Anderson (horndude77(AT)gmail.com), Apr 07 2007

G.f.: (1 - 154x + 6777x^2 - 123961x^3 + 1132714x^4 - 5684515x^5 + 16401668x^6 - 27757938x^7 + 27757938*x^8 - 16401668x^9 + 5684515x^10 - 1132714x^11 + 123961x^12 -6777x^13 + 154x^14 - x^15)/(1 - 209x + 11936x^2 - 274208x^3 + 3112032x^4 - 19456019x^5 + 70651107x^6 - 152325888x^7 + 196664896x^8 - 152325888x^9 + 70651107x^10 -19456019x^11 + 3112032x^12 - 274208x^13 + 11936x^14 - 209x^15 + x^16). - Sergey Perepechko, Nov 23 2012

Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

cp's OEIS Frontend

A084917 Positive numbers of the form 3*y^2 - x^2.

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A082841 a(n) = 4*a(n-1) - a(n-2) for n>1, a(0)=3, a(1)=9.

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A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).

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

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A046184 Indices of octagonal numbers which are also squares.

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A180147 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + 3*x)/(1 - 4*x - 3*x^2 + 6*x^3).

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A350920 a(0) = 5, a(1) = 5, and a(n) = 4*a(n-1) - a(n-2) - 4 for n >= 2.

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A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

A180147 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + 3x)/(1 - 4x - 3x^2 + 6x^3).