A002203 -id:A002203 - cp's OEIS frontend

A124696 Number of base-3 circular n-digit numbers with adjacent digits differing by 1 or less.

1, 3, 7, 15, 35, 83, 199, 479, 1155, 2787, 6727, 16239, 39203, 94643, 228487, 551615, 1331715, 3215043, 7761799, 18738639, 45239075, 109216787, 263672647, 636562079, 1536796803, 3710155683, 8957108167, 21624372015, 52205852195

Offset: 0

R. H. Hardin, Dec 28 2006

These are the number of smooth cyclic words of length n over the alphabet {1,2,3}. See theorem 3.3 in Knopfmacher and others. - Peter Luschny, Aug 13 2012

This is the main entry for 234 similar sequences. Cf. the link to the OEIS Wiki for a list, the programs and a derivation of the linear recurrences. - Georg Fischer, Apr 09 2021

R. H. Hardin, Table of n, a(n) for n = 0..210 [uploaded by Georg Fischer, Apr 05 2021]
Arnold Knopfmacher, Toufik Mansour, Augustine Munagi, and Helmut Prodinger, Smooth words and Chebyshev polynomials, arXiv:0809.0551v1 [math.CO], 2008.
Stephan Mertens, Domination Polynomials of the Grid, the Cylinder, the Torus, and the King Graph, arXiv:2408.08053 [math.CO], 2024. See p. 5.
OEIS Wiki, Number of base k circular n-digit numbers with adjacent digits differing by d or less
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A002426, Row 3 of A276562.

T := (n, k) -> `if`(n=0, 1, add((1 + 2*cos(j*Pi/(k + 1)))^n, j=1..k)):
a := n -> simplify(T(n, 3)): seq(a(n), n=0..28); # Peter Luschny, Mar 28 2021

[Empirical] a(base,n) = a(base-1,n) + A002426(n+1) for base = 1..floor(n/2)+1.
a(n) = T(n,3) for n > 0, where T(n,k) = Sum_{j=1..k} (1 + 2*cos(j*Pi/(k + 1)))^n. - Peter Luschny, Aug 13 2012
From Colin Barker, Nov 26 2012: (Start)
a(n) = 1 + (1 - sqrt(2))^n + (1 + sqrt(2))^n for n > 0.
a(n) = 3*a(n-1) - a(n-2) - a(n-3) for n > 3.
G.f.: -(2*x^3 + x^2 - 1)/((x - 1)*(x^2 + 2*x - 1)). (End)
a(n) = A002203(n)+1, n>0. - R. J. Mathar, May 09 2023

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

Original entry on oeis.org

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A079291 Squares of Pell numbers.

Original entry on oeis.org

0, 1, 4, 25, 144, 841, 4900, 28561, 166464, 970225, 5654884, 32959081, 192099600, 1119638521, 6525731524, 38034750625, 221682772224, 1292061882721, 7530688524100, 43892069261881, 255821727047184, 1491038293021225

Offset: 0

Ralf Stephan, Feb 08 2003

(-1)^(n+1)*a(n) is the r=-4 member of the r-" of sequences S_r(n), n>=1, defined in A092184 where more information can be found.
Binomial transform of A086346. - Johannes W. Meijer, Aug 01 2010
In general, squaring the terms of a Horadam sequence with signature (c,d) will result in a third-order recurrence with signature (c^2+d, c^2*d+d^2, -d^3). - Gary Detlefs, Nov 11 2021
(Conjectured) For any primitive Pythagorean triple of the form (X, Y, Z=Y+1), it appears that Y or Z will always be (and only be) a square Pell number if X = A001333(n), for n > 1. If n is even, Y is always a square Pell number, and if n is odd, then Z is always a square Pell number. For example: (3, 4, 5), (7, 24, 25), (17, 144, 145), (41, 840, 841), (99, 4900, 4901). - Jules Beauchamp, Feb 02 2022
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, black half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), and white half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using black (1/4,1/4)-fences, white (1/4,1/4)-fences, and (1/4,3/4)-fences. - Michael A. Allen, Dec 29 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Joerg Arndt, Matters Computational (The Fxtbook), sect. 32.1.5, pp. 626-627
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Toufik Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
Toufik Mansour, Squaring the terms of an ell-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.
Pantelimon Stanica, Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001254, A001109, A001541, A001653, A174968.

Cf. A002203, A007598, A084158 (partial sums), A086346, A092184.

I:=[0,1,4]; [n le 3 select I[n] else 5*Self(n-1)+ 5*Self(n-2) - Self(n-3): n in [1..31]]; // Vincenzo Librandi, May 17 2013

with(combinat):seq(fibonacci(i,2)^2, i=0..31); # Zerinvary Lajos, Mar 20 2008

CoefficientList[Series[x(1-x)/((1+x)*(1-6x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, May 17 2013 *)
LinearRecurrence[{5,5,-1},{0,1,4},40] (* Harvey P. Dale, Dec 20 2015 *)
Fibonacci[Range[0, 30], 2]^2 (* G. C. Greubel, Sep 17 2021 *)

[lucas_number1(n, 2, -1)^2 for n in (0..30)] # G. C. Greubel, Sep 17 2021

G.f.: x*(1-x)/((1+x)*(1-6*x+x^2)).
a(n) = (r^n + (1/r)^n - 2*(-1)^n)/8, with r = 3 + sqrt(8).
a(n+3) = 5*a(n+2) + 5*a(n+1) - a(n).
L.g.f.: (1/8)*log((1+2*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} (a(n)/n)*x^n, see p. 627 of the Fxtbook link; special case of the following: let v(0)=0, v(1)=1, and v(n) = u*v(n-1) + v(n-2), then (1/A)*log((1+2*x+x^2)/(1-(2-A)*x+x^2)) = Sum_{n>=0} v(n)^2/n*x^n where A = u^2 + 4. - Joerg Arndt, Apr 08 2011
a(n+1) = Sum_{k=0..n} ( (-1)^(n-k)*A001653(k) ); e.g., 144 = -1 + 5 - 29 + 169; 25 = 1 - 5 + 29. - Charlie Marion, Jul 16 2003
a(n) = A000129(n)^2.
a(n) = (T(n, 3) - (-1)^n)/4 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3) = A001541(n) = ((3 + 2*sqrt(2))^n + (3 - 2*sqrt(2))^n )/2. - Wolfdieter Lang, Oct 18 2004
a(n) is the rightmost term of M^n * [1 0 0] where M is the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. a(n+1) = leftmost term. E.g., a(6) = 4900, a(5) = 841 since M^5 * [1 0 0] = [4900 2030 841]. - Gary W. Adamson, Oct 31 2004
a(n) = ( (-1)^(n+1) + A001109(n+1) - 3*A001109(n) )/4. - R. J. Mathar, Nov 16 2007
a(n) = ( (((1 - sqrt(2))^n + (1 + sqrt(2))^n) /2 )^2 + (-1)^(n+1) )/2. - Antonio Pane (apane1(AT)spc.edu), Dec 15 2007
Lim_{k -> infinity} ( a(n+k)/a(k) ) = A001541(n) + 2*A001109(n)*sqrt(2). - Johannes W. Meijer, Aug 01 2010
For n>0, a(2*n) = 6*a(2*n-1) - a(2*n-2) - 2, a(2*n+1) = 6*a(2*n) - a(2*n-1) + 2. - Charlie Marion, Sep 24 2011
a(n) = (1/8)*(A002203(2*n) - 2*(-1)^n). - G. C. Greubel, Sep 17 2021
Conjectured formula for (X, Y, Z) for primitive Pythagorean triple of the form (X, Y, Z=Y+1) is (A001333(n)^2, A079291(n)^2, A079291(n)^2-1) or (A001333(n)^2, A079291(n)^2-1, A079291(n)^2). As a closed formula (X, Y, Z) = (((1-sqrt(2))^n + (1+sqrt(2))^n)/2, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2 - 4)/8, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2 + 4)/8). - Jules Beauchamp, Feb 02 2022
From Michael A. Allen, Dec 29 2022: (Start)
a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = (1 + (-1)^n)/2 + 4*Sum_{k=1..n} ( k*a(n+1-k) ). (End)
Product_{n>=2} (1 + (-1)^n/a(n)) = (1 + sqrt(2))/2 (A174968) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A204270 a(n) = tau(n)*Pell(n), where tau(n) = A000005(n), the number of divisors of n.

Original entry on oeis.org

1, 4, 10, 36, 58, 280, 338, 1632, 2955, 9512, 11482, 83160, 66922, 323128, 780100, 2354160, 2273378, 16465260, 13250218, 95966568, 154455860, 372889432, 450117362, 4346717760, 3935214363, 12667263848, 30581480180, 110745336312, 89120964298

Offset: 1

Paul D. Hanna, Jan 14 2012

nonn

Compare g.f. to the Lambert series identity: Sum_{n>=1} x^n/(1-x^n) = Sum_{n>=1} tau(n)*x^n.
Related identities:
(1) Sum_{n>=1} n^k*Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} sigma_{k}(n)*Pell(n)*x^n for k>=0.
(2) Sum_{n>=1} phi(n)*Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} n*Pell(n)*x^n.
(3) Sum_{n>=1} moebius(n)*Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) = x.
(4) Sum_{n>=1} lambda(n)*Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} Pell(n^2)*x^(n^2).

			G.f.: A(x) = 1 + 4*x + 10*x^2 + 36*x^3 + 58*x^4 + 280*x^5 + 338*x^6 +...
where A(x) = x/(1-2*x-x^2) + 2*x^2/(1-6*x^2+x^4) + 5*x^3/(1-14*x^3-x^6) + 12*x^4/(1-34*x^4+x^8) + 29*x^5/(1-82*x^5-x^10) + 70*x^6/(1-198*x^6+x^12) +...+ Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) +...

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A203847, A204271, A204272, A204273, A204274, A204275, A000005 (tau), A002203, A000045.

Table[DivisorSigma[0, n] Fibonacci[n, 2], {n, 1, 50}] (* G. C. Greubel, Jan 05 2018 *)

/* Subroutines used in PARI programs below: */
{Pell(n)=polcoeff(x/(1-2*x-x^2+x*O(x^n)), n)}
{A002203(n)=polcoeff(2*(1-x)/(1-2*x-x^2+x*O(x^n)), n)}

PARI
```
{a(n)=sigma(n,0)*Pell(n)}
```

{a(n)=polcoeff(sum(m=1,n,Pell(m)*x^m/(1-A002203(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n)}

G.f.: Sum_{n>=1} Pell(n)*x^n/(1 - A002203(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} tau(n)*Pell(n)*x^n, where Pell(n) = A000129(n) and A002203 is the companion Pell numbers.

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes, and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

A059020 Number of 2 X n checkerboards (with at least one red square) in which the set of red squares is edge connected.

Original entry on oeis.org

0, 3, 13, 40, 108, 275, 681, 1664, 4040, 9779, 23637, 57096, 137876, 332899, 803729, 1940416, 4684624, 11309731, 27304157, 65918120, 159140476, 384199155, 927538873, 2239276992, 5406092952, 13051462995, 31509019045, 76069501192, 183648021540, 443365544387

Offset: 0

John W. Layman, Dec 14 2000

In other words, the number of connected (non-null) induced subgraphs in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, May 02 2017

Also, the number of cycles in the grid graph P_3 X P_{n+1}. - Andrew Howroyd, Jun 12 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Connected Graph.
Eric Weisstein's World of Mathematics, Induced Subgraph.
Eric Weisstein's World of Mathematics, Ladder Graph.
Index entries for linear recurrences with constant coefficients, signature (4,-4,0,1).

Row 2 of A287151 and row 2 of A231829.
See also A059021, A059524.
Cf. A000129. - Jaume Oliver Lafont, Sep 28 2009
Other sequences counting connected induced subgraphs: A020873, A059525, A286139, A286182, A286183, A286184, A286185, A286186, A286187, A286188, A286189, A286191, A285765, A285934, A286304.

I:=[0, 3, 13, 40];[n le 4 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-4):n in [1..30]]; // Marius A. Burtea, Aug 25 2019

Join[{0},LinearRecurrence[{4, -4, 0, 1}, {3, 13, 40, 108}, 20]] (* Eric W. Weisstein, May 02 2017 *) (* adapted by Vincenzo Librandi, May 09 2017 *)
Table[(LucasL[n + 3, 2] - 8 n - 14)/4, {n, 0, 20}] (* Eric W. Weisstein, May 02 2017 *)

a(n) = 2*a(n-1) + a(n-2) + 4*n - 1.
From Jaume Oliver Lafont, Nov 23 2008: (Start)
a(n) = 3*a(n-1) - a(n-2) - a(n-3) + 4;
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4). (End)
G.f.: x*(3+x)/((1-2*x-x^2)*(1-x)^2). - Jaume Oliver Lafont, Sep 28 2009
Empirical observations (from Superseeker):
(1) if b(n) = a(n) + n then {b(n)} is A048777;
(2) if b(n) = a(n+3) - 3*a(n+2) - 3*a(n+1) + a(n) then {b(n)} is A052542;
(3) if b(n) = a(n+2) - 2*a(n+1) + a(n) then {b(n)} is A001333.
4*a(n) = A002203(n+3) - 8*n - 14. - Eric W. Weisstein, May 02 2017
a(n) = 3*A048776(n-1) + A048776(n-2). - R. J. Mathar, May 12 2019
E.g.f.: (1/2)*exp(x)*(-7-4*x+7*cosh(sqrt(2)*x)+5*sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, Aug 25 2019

A006266 A continued cotangent.

Original entry on oeis.org

2, 14, 2786, 21624372014, 10111847525912679844192131854786, 1033930953043290626825587838528711318150300040875029341893199068078185510802565166824630504014

Offset: 0

N. J. A. Sloane

The next (6th) term is 280 digits long. - M. F. Hasler, Oct 06 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 = 0..7
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Cf. A002203, A006267, A114525, A145451, A145452, A271222, A271224.

[Evaluate(DicksonFirst(3^n, -1), 2): n in [0..7]]; // G. C. Greubel, Mar 25 2022

a := proc(n) option remember; if n = 1 then 14 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala Nov 15 2022

Table[Round[(1+Sqrt[2])^(3^n)],{n,0,10}] (* Artur Jasinski, Sep 24 2008 *)
LucasL[3^Range[0, 7], 2] (* G. C. Greubel, Mar 25 2022 *)

a(n,s=2)=for(i=2,n,s*=(s^2+3));s \\ M. F. Hasler, Oct 06 2014

[lucas_number2(3^n,2,-1) for n in (0..7)] # G. C. Greubel, Mar 25 2022

From Artur Jasinski, Sep 24 2008: (Start)
a(n+1) = a(n)^3 + 3*a(n) with a(0) = 2.
a(n) = round((1+sqrt(2))^(3^n)). [Corrected by M. F. Hasler, Oct 06 2014] (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = A002203(3^n).
a(n) = L(3^n,2), where L(n,x) denotes the n-th Lucas polynomial of A114525.
a(n) == 2 (mod 3).
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the companion Pell numbers).
The smallest positive residue of a(n) mod(3^n) = A271222(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271224. Cf. A006267. (End)

Edited by M. F. Hasler, Oct 06 2014

Offset corrected by G. C. Greubel, Mar 25 2022

A084158 a(n) = A000129(n) * A000129(n+1)/2.

Original entry on oeis.org

0, 1, 5, 30, 174, 1015, 5915, 34476, 200940, 1171165, 6826049, 39785130, 231884730, 1351523251, 7877254775, 45912005400, 267594777624, 1559656660345, 9090345184445, 52982414446326, 308804141493510, 1799842434514735, 10490250465594899, 61141660359054660, 356359711688733060

Offset: 0

Paul Barry, May 18 2003

May be called Pell triangles.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001652, A001654, A002203, A002315, A003499.

Cf. A041011, A046729, A079291, A084159, A084175, A163960, A383720.

[Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n)/16): n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

with(combinat): a:=n->fibonacci(n,2)*fibonacci(n-1,2)/2: seq(a(n), n=1..22); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{5,5,-1},{0,1,5},30] (* Harvey P. Dale, Sep 07 2011 *)

Pell(n)=([2, 1; 1, 0]^n)[2, 1];
a(n)=Pell(n)*Pell(n+1)/2 \\ Charles R Greathouse IV, Mar 21 2016

a(n)=([0,1,0; 0,0,1; -1,5,5]^n*[0;1;5])[1,1] \\ Charles R Greathouse IV, Mar 21 2016

[(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/16 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = ((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/16.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3). - Mohamed Bouhamida, Sep 02 2006; corrected by Antonio Alberto Olivares, Mar 29 2008
a(n) = (-1/8)*(-1)^n + (( sqrt(2)+1)/16)*(3+2*sqrt(2))^n + ((-sqrt(2)+1)/16)*(3-2*sqrt(2))^n. - Antonio Alberto Olivares, Mar 30 2008
sqrt(a(n) - a(n-1)) = A000129(n). - Antonio Alberto Olivares, Mar 30 2008
O.g.f.: x/((1+x)(1-6*x+x^2)). - R. J. Mathar, May 18 2008
a(n) = A041011(n)*A041011(n+1). - R. K. Guy, May 18 2008
From Mohamed Bouhamida, Aug 30 2008: (Start)
a(n) = 6*a(n-1) - a(n-2) - (-1)^n.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3) - 2*(-1)^n. (End)
In general, for n>k+1, a(n+k) = A003499(k+1)*a(n-1) - a(n-k-2) - (-1)^n A000129(k+1)^2. - Charlie Marion, Jan 04 2012
For n>0, a(2n-1)*a(2n+1) = oblong(a(2n)); a(2n)*a(2n+2) = oblong(a(2n+1)-1). - Charlie Marion, Jan 09 2012
a(n) = A046729(n)/4. - Wolfdieter Lang, Mar 07 2012
a(n) = sum of squares of first n Pell numbers A000129 (A079291). - N. J. A. Sloane, Jun 18 2012
a(n) = (A002315(n) - (-1)^n)/8. - Adam Mohamed, Sep 05 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} Pell(3*k)/Pell(k) * x^k/k ). - Seiichi Manyama, May 07 2025

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Original entry on oeis.org

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003
Equivalently, 2*n^2 + 8 is a square.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009
The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013
a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Amiram Eldar, Table of n, a(n) for n = 1..1306
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Tanya Khovanova, Recursive Sequences
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
J. J. O'Connor and E. F. Robertson, Pell's Equation. [Broken link]
Eric Weisstein's World of Mathematics, Pell Equation.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

(A077445(n))^2 - 2*a(n) = 8.
First differences of A001541.
Pairwise sums of A001542.
Bisection of A002203 and A080039.
Cf. A001653.

[n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

LinearRecurrence[{6,-1},{2,14},30] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,20,q=(1+sqrt(2))^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - (3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).
a(n) = 2*A002315(n-1).
Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.
Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: 2*x*(1+x)/(1-6*x+x^2).
a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)
a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010
a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013
a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015
a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023
E.g.f.: 2*(1 + exp(3*x)*(sqrt(2)*sinh(2*sqrt(2)*x) - cosh(2*sqrt(2)*x))). - Stefano Spezia, Aug 27 2024

A286182 Number of connected induced (non-null) subgraphs of the prism graph with 2n nodes.

Original entry on oeis.org

3, 13, 51, 167, 503, 1441, 4007, 10923, 29355, 78037, 205659, 538127, 1399583, 3621289, 9327695, 23931603, 61186131, 155949085, 396369795, 1004904695, 2541896519, 6416348209, 16165610999, 40657256571, 102090514683, 255968753125, 640899345579, 1602640560479

Offset: 1

Giovanni Resta, May 04 2017

nonn

Cases n=1 and n=2 correspond to degenerate prism graphs, but they fit the same (conjectured) linear recurrence as the other terms.

Andrew Howroyd, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Prism Graph
Eric Weisstein's World of Mathematics, Vertex-Induced Subgraph
Index entries for linear recurrences with constant coefficients, signature (6, -11, 4, 5, -2, -1).

Cf. A020873 (wheel), A059020 (ladder), A059525 (grid), A286139 (king), A286183 (antiprism), A286184 (helm), A286185 (Möbius ladder), A286186 (friendship), A286187 (web), A286188 (gear), A286189 (rook), A285765 (queen).

a[n_] := Block[{g = Graph@ Flatten@ Table[{i <-> Mod[i,n] + 1, n+i <-> Mod[i,n] + n+1, i <-> i+n}, {i, n}]}, -1 + ParallelSum[ Boole@ ConnectedGraphQ@ Subgraph[g, s], {s, Subsets@Range[2 n]}]]; Array[a, 8]

a(n) = 6*a(n-1) - 11*a(n-2) + 4*a(n-3) + 5*a(n-4) - 2*a(n-5) - a(n-6), for n > 6 (conjectured).
a(n) = A002203(n) + 3*n*A000129(n) - 3*n + 1 (conjectured). - Eric W. Weisstein, May 08 2017
G.f.: x*(3 - 5*x + 6*x^2 - 8*x^3 - 5*x^4 - 3*x^5) / ((1 - x)^2*(1 - 2*x - x^2)^2) (conjectured). - Colin Barker, May 31 2017

Terms a(18) and beyond from Andrew Howroyd, Aug 15 2017

cp's OEIS Frontend

A124696 Number of base-3 circular n-digit numbers with adjacent digits differing by 1 or less.

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Maple

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A114525 Triangle of coefficients of the Lucas (w-)polynomials.

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A079291 Squares of Pell numbers.

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A204270 a(n) = tau(n)*Pell(n), where tau(n) = A000005(n), the number of divisors of n.

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A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

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A059020 Number of 2 X n checkerboards (with at least one red square) in which the set of red squares is edge connected.

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A006266 A continued cotangent.