A086347 -id:A086347 - cp's OEIS frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720

Offset: 0

N. J. A. Sloane

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003
This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]
n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003
For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003
a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004
Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005
Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005
Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006
Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006
If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007
If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006
If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007
a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007
For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011
Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009
Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010
If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
  for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
  for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1,  64 = 35*2 - 6, 373 = 204*2 - 35;
  for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1;  99 = 35*3 - 6; 577 = 204*3 - 35;
  for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
  for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
  See also A002315, A054488, A038761, A054489, A054490.
  - Charlie Marion, Dec 08 2010
See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012
a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012
a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012
From Richard R. Forberg, Aug 30 2013: (Start)
The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:
a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and  2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

			G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - _Michael B. Porter_, Jul 02 2016

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Index entries for sequences related to Chebyshev polynomials.
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Cf. A000217, A000290, A001108, A001542, A001653, A001850, A002315, A002965, A278310.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

a:=[0,1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018

a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
   (map (* 6) $ tail a001109_list) a001109_list
-- Reinhard Zumkeller, Dec 17 2011

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015

a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n],n=0..26); # Emeric Deutsch
with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
a[ n_]:= ChebyshevU[n-1, 3]; (* Michael Somos, Sep 02 2012 *)
Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)

{a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */

is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016

[lucas_number1(n,6,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).
a(n) = sqrt(A001110(n)).
a(n) = A001542(n)/2.
a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
a(-n) = -a(n). - Michael Somos, Apr 07 2003
For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - Charlie Marion, Jul 01 2003
For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004
For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, Sep 20 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006
The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007
For k = 0..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+1-k) - a(k) = A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007
[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - Charlie Marion, Dec 14 2010
a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011
16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011
A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011
In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012
PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012
a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012
a(n) + a(n+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*x/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Additional comments from Wolfdieter Lang, Feb 10 2000

Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

Original entry on oeis.org

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747

Offset: 0

R. H. Hardin, Dec 27 2006

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
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 king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): 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); # Johannes W. Meijer, Aug 01 2010
# second Maple program:
a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021

LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)

a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Edited by Johannes W. Meijer, Aug 10 2010

A057087 Scaled Chebyshev U-polynomials evaluated at i. Generalized Fibonacci sequence.

Original entry on oeis.org

1, 4, 20, 96, 464, 2240, 10816, 52224, 252160, 1217536, 5878784, 28385280, 137056256, 661766144, 3195289600, 15428222976, 74494050304, 359689093120, 1736732573696, 8385686667264, 40489676963840, 195501454524416

Offset: 0

Wolfdieter Lang, Aug 11 2000

a(n) gives the length of the word obtained after n steps with the substitution rule 0->1111, 1->11110, starting from 0. The number of 1's and 0's of this word is 4*a(n-1) and 4*a(n-2), respectively.
Inverse binomial transform of odd Pell bisection A001653. With a leading zero, inverse binomial transform of even Pell bisection A001542, divided by 2. - Paul Barry, May 16 2003
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 4's along the main diagonal, and 2's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths: 1, 1, 8, 1, 3, 8, 6, 1, 24, 3, 120, 8, 21, 6, 24, 1, 16, 24, 360, 3, ... . - R. J. Mathar, Aug 10 2012
Exponential convolution of Pell numbers (A000129) and companion Pell numbers (A002203), divided by 2 and leading zero dropped. - Vladimir Reshetnikov, Oct 07 2016

Indranil Ghosh, Table of n, a(n) for n = 0..1459
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=4, q=4.
Tanya Khovanova, Recursive Sequences.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419; Eqs.(39) and (45),rhs, m=4.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,4)

Pairwise sums are in A086347.

Appears in A086346, A086347 and A086348. - Johannes W. Meijer, Aug 01 2010

I:=[1,4]; [n le 2 select I[n] else 4*Self(n-1) + 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018

A057087 := n -> `if`(n=0, 1, 4^n*hypergeom([1/2-n/2, -n/2], [-n], -1)):
seq(simplify(A057087(n)), n=0..21); # Peter Luschny, Dec 17 2015

Table[Fibonacci[n + 1, 2] 2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 08 2016 *)
LinearRecurrence[{4,4},{1,4},30] (* Harvey P. Dale, Aug 17 2017 *)

a(n)=if(n<0, 0, (2*I)^n*subst(I*poltchebi(n+1)+poltchebi(n),'x,-I)/2) /* Michael Somos, Sep 16 2005 */

Vec(1/(1-4*x-4*x^2) + O(x^100)) \\ Altug Alkan, Dec 17 2015

[lucas_number1(n,4,-4) for n in range(1, 23)] # Zerinvary Lajos, Apr 23 2009

a(n) = 4*(a(n-1) + a(n-2)), a(-1)=0, a(0)=1.
G.f.: 1/(1 - 4*x - 4*x^2).
a(n) = S(n, 2*i)*(-2*i)^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
a(n) = Sum_{k=0..n} 3^k*A063967(n,k). - Philippe Deléham, Nov 03 2006
a(n) = A000129(n+1)*A000079(n). - R. J. Mathar, Jul 08 2009
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = A084128(n) + 2*A057087(n-1)*sqrt(2);
Limit_{n->oo} A084128(n)/A057087(n-1) = sqrt(2). (End)
a(n) = 4^n*hypergeom([1/2-n/2, -n/2], [-n], -1) for n>=1. - Peter Luschny, Dec 17 2015

A125145 a(n) = 3a(n-1) + 3a(n-2). a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 15, 57, 216, 819, 3105, 11772, 44631, 169209, 641520, 2432187, 9221121, 34959924, 132543135, 502509177, 1905156936, 7222998339, 27384465825, 103822392492, 393620574951, 1492328902329, 5657848431840, 21450532002507

Offset: 0

Tanya Khovanova, Jan 11 2007

Number of aa-avoiding words of length n on the alphabet {a,b,c,d}.
Equals row 3 of the array shown in A180165, the INVERT transform of A028859 and the INVERTi transform of A086347. - Gary W. Adamson, Aug 14 2010
From Tom Copeland, Nov 08 2014: (Start)
This array is one of a family related by compositions of C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for A000108; its inverse Cinv(x) = x(1-x); and the special Mobius transformation P(x,t) = x / (1+t*x) with inverse P(x,-t) in x. Cf. A091867.
O.g.f.: G(x) = P[P[P[-Cinv(-x),-1],-1],-1] = P[-Cinv(-x),-3] = x*(1+x)/[1-3x(1-x)]= x*A125145(x).
Ginv(x) = -C[-P(x,3)] = [-1 + sqrt(1+4x/(1+3x))]/2 = x*A104455(-x).
G(-x) = -x(1-x) * [ 1 - 3*[x*(1+x)] + 3^2*[x*(1+x)]^2 - ...] , and so this array is related to finite differences in the row sums of A030528 * Diag((-3)^1,3^2,(-3)^3,..). (Cf. A146559.)
The inverse of -G(-x) is C[-P(-x,3)]= [1 - sqrt(1-4x/(1-3x))]/2 = x*A104455(x). (End)
Number of 3-compositions of n+1 restricted to parts 1 and 2 (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2025. See p. 14.
Joerg Arndt, Matters Computational (The Fxtbook)
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 7.
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,3).

Cf. A028859 = a(n+2) = 2 a(n+1) + 2 a(n); A086347 = On a 3 X 3 board, number of n-move routes of chess king ending at a given side cell. a(n) = 4a(n-1) + 4a(n-2).
Cf. A128235.
Cf. A180165, A028859, A086347. - Gary W. Adamson, Aug 14 2010
Cf. A002605, A026150, A030195, A080040, A083337, A106435, A108898.
Cf. A000108, A091867, A125145, A104455, A030528, A146559.

a125145 n = a125145_list !! n
a125145_list =
   1 : 4 : map (* 3) (zipWith (+) a125145_list (tail a125145_list))
-- Reinhard Zumkeller, Oct 15 2011

I:=[1,4]; [n le 2 select I[n] else 3*Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 10 2014

a[0]:=1: a[1]:=4: for n from 2 to 27 do a[n]:=3*a[n-1]+3*a[n-2] od: seq(a[n],n=0..27); # Emeric Deutsch, Feb 27 2007
A125145 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]) ;
    else
        3*(procname(n-1)+procname(n-2)) ;
    end if;
end proc: # R. J. Mathar, Feb 13 2022

nn=23;CoefficientList[Series[(1+x)/(1-3x-3x^2),{x,0,nn}],x] (* Geoffrey Critzer, Feb 09 2014 *)
LinearRecurrence[{3,3},{1,4},30] (* Harvey P. Dale, May 01 2022 *)

G.f.: (1+z)/(1-3z-3z^2). - Emeric Deutsch, Feb 27 2007
a(n) = (5*sqrt(21)/42 + 1/2)*(3/2 + sqrt(21)/2)^n + (-5*sqrt(21)/42 + 1/2)*(3/2 - sqrt(21)/2)^n. - Antonio Alberto Olivares, Mar 20 2008
a(n) = A030195(n)+A030195(n+1). - R. J. Mathar, Feb 13 2022
E.g.f.: exp(3*x/2)*(21*cosh(sqrt(21)*x/2) + 5*sqrt(21)*sinh(sqrt(21)*x/2))/21. - Stefano Spezia, Aug 04 2022
a(n) = (((3 + sqrt(21)) / 2)^(n+2) - ((3 - sqrt(21)) / 2)^(n+2)) / (3 * sqrt(21)). - Werner Schulte, Dec 17 2024

A214992 Power ceiling-floor sequence of (golden ratio)^4.

Original entry on oeis.org

7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711

Offset: 0

Clark Kimberling, Nov 08 2012, Jan 24 2013

Let f = floor and c = ceiling.  For x > 1, define four sequences as functions of x, as follows:
p1(0) = f(x), p1(n) = f(x*p1(n-1));
p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;
p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;
p4(0) = c(x), p4(n) = c(x*p4(n-1)).
The present sequence is given by a(n) = p3(n).
Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences.  In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms.  Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).
x ......p1..... p2..... p3..... p4.......Limit
r.......A000045 A000045 A000045 A000045..r
r^2.....A001519 A001654 A061646 A001906..-1+S(5)
r^3.....A024551 A001076 A015448 A049652..-1+S(5)
r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)
r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2
r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)
1+S(2)..A024537 A000129 A001333 A048739..S(2)
2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2
1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2
2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)
S(5)....A214999 A215091 A218982 A218983..1.26879683...
2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)
2+S(6)..A218984 A090017 A123347 A218985..S(3/2)
2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3
2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2
3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)
3+S(10).A218991 A005668 A015451 A218992..-2+S(10)
...
Properties of p1, p2, p3, p4:
(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.
(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.
(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer.  If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences.  Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?
(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent?  If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?
(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

			a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.

Clark Kimberling, Table of n, a(n) for n = 0..250
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
Index entries for linear recurrences with constant coefficients, signature (6,6,-1).
Index entries for sequences related to the Josephus Problem.

Cf. A001622, A214986, A049685, A157335, A004187.

(* Program 1.  A214992 and related sequences *)
x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}]  (* A049685 *)
Table[p2[n], {n, 0, z}]  (* A157335 *)
Table[p3[n], {n, 0, z}]  (* A214992 *)
Table[p4[n], {n, 0, z}]  (* A004187 *)
Table[p4[n] - p1[n], {n, 0, z}]  (* A004187 *)
Table[p3[n] - p2[n], {n, 0, z}]  (* A098305 *)
(* Program 2.  Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
(* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]

a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012
a(n) = 7*A157335(n) + 5*A157335(n-1) - A157335(n-2). - R. J. Mathar, Feb 05 2020
E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024

A084128 a(n) = 4a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 12, 56, 272, 1312, 6336, 30592, 147712, 713216, 3443712, 16627712, 80285696, 387653632, 1871757312, 9037643776, 43637604352, 210700992512, 1017354387456, 4912221519872, 23718303629312, 114522100596736, 552961616904192, 2669934870003712

Offset: 0

Paul Barry, May 16 2003

Original name was: Generalized Fibonacci sequence.

Binomial transform of A084058.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4,4).

Cf. A001333 A001541, A002203, A057087, A084058, A084128, A201730.

Appears in A086346, A086347 and A086348. - Johannes W. Meijer, Aug 01 2010

[2^(n-1)*Evaluate(DicksonFirst(n,-1), 2): n in [0..40]]; // G. C. Greubel, Oct 13 2022

a:=proc(n) option remember; if n=0 then 1 elif n=1 then 2 else
4*a(n-1)+4*a(n-2); fi; end: seq(a(n), n=0..40); # Wesley Ivan Hurt, Jan 31 2017
a := n -> (2*I)^n*ChebyshevT(n, -I):
seq(simplify(a(n)), n = 0..23); # Peter Luschny, Dec 03 2023

CoefficientList[Series[(2 z - 1)/(4 z^2 + 4 z - 1), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Table[2^(n-1) LucasL[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 07 2016 *)
LinearRecurrence[{4,4},{1,2},30] (* Harvey P. Dale, Mar 01 2018 *)

a(n)=if(n<0,0,polsym(4+4*x-x^2,n)[n+1]/2)

[lucas_number2(n,4,-4)/2 for n in range(0, 23)] # Zerinvary Lajos, May 14 2009

a(n) = 2^n * A001333(n).
G.f.: (1-2*x)/(1-4*x-4*x^2).
a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.
a(n) = (2 + 2*sqrt(2))^n/2 + (2 - 2*sqrt(2))^n/2.
E.g.f.: exp(2*x)*cosh(2*x*sqrt(2)).
From Johannes W. Meijer, Aug 01 2010: (Start)
Lim_{k->infinity} a(n+k)/a(k) = A084128(n) + 2*A057087(n-1)*sqrt(2).
Lim_{n->infinity} A084128(n)/A057087(n-1) = sqrt(2). (End)
a(n) = Sum_{k=0..n} A201730(n,k)*7^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(4*k-2)/(x*(4*k+2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = 2^(n-1)*A002203(n). - Vladimir Reshetnikov, Oct 07 2016

A086346 On a 3 X 3 board, the number of n-move paths for a chess king ending in a given corner square.

Original entry on oeis.org

1, 3, 18, 80, 400, 1904, 9248, 44544, 215296, 1039104, 5018112, 24227840, 116985856, 564850688, 2727354368, 13168803840, 63584665600, 307013812224, 1482394042368, 7157631156224, 34560101318656, 166870928850944, 805724122775552, 3890380202311680, 18784417308737536, 90699190027419648

Offset: 0

Zak Seidov, Jul 17 2003

From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move paths of a chess king on a 3 X 3 board that end or start in a given corner square m (m = 1, 3, 7, 9). To determine the a(n) we can either sum the components of the column vector A^n[k,m], with A the adjacency matrix of the king's graph, or we can sum the components of the row vector A^n[m,k], see the Maple program.
Inverse binomial transform of A079291 (without the leading 0).
(End)
From R. J. Mathar, Oct 12 2010: (Start)
The row n=3 of an array counting king walks on an n X n board with k steps, starting from a corner:
  1, 3,  9,  27,  81,  243,   729,   2187,    6561,    19683,    59049, ...;
  1, 3, 18,  80, 400, 1904,  9248,  44544,  215296,  1039104,  5018112, ...;
  1, 3, 18, 105, 615, 3600, 21075, 123375,  722250,  4228125, 24751875, ...;
  1, 3, 18, 105, 684, 4359, 28278, 182349, 1179792,  7622667, 49283802, ...;
  1, 3, 18, 105, 684, 4550, 30807, 209867, 1434279,  9815190, 67209723, ...;
  1, 3, 18, 105, 684, 4550, 31340, 218056, 1533712, 10829360, 76720288, ...;
  1, 3, 18, 105, 684, 4550, 31340, 219555, 1559835, 11177190, 80573373, ...;
  1, 3, 18, 105, 684, 4550, 31340, 219555, 1564080, 11259785, 81765550, ...;
  1, 3, 18, 105, 684, 4550, 31340, 219555, 1564080, 11271876, 82025163, ...;
  1, 3, 18, 105, 684, 4550, 31340, 219555, 1564080, 11271876, 82059768, ...;
  1, 3, 18, 105, 684, 4550, 31340, 219555, 1564080, 11271876, 82059768, ...;
The partial sums along the rows are documented in A123109 (king walks with between 1 and k steps). (End)

Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984. [From Johannes W. Meijer, Aug 01 2010]

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mike Oakes, KingMovesForPrimes.
Zak Seidov et al., New puzzle? King moves for primes, digest of 28 messages in primenumbers group, Jul 13 - Jul 23, 2003. [Cached copy]
Zak Seidov, KingMovesForPrimes.
Sleephound, KingMovesForPrimes.
Index entries for linear recurrences with constant coefficients, signature (2,12,8).

Cf. A086347, A086348, A086349.
Cf. A179596. - Johannes W. Meijer, Aug 01 2010
Cf. A002203, A057087, A079291, A084128, A110048, A123109.

[2^(n-3)*(Evaluate(DicksonFirst(n+2,-1), 2) +2*(-1)^n): n in [0..30]]; // G. C. Greubel, Aug 18 2022

with(LinearAlgebra):
nmax:=19; m:=1;
A[5]:= [1, 1, 1, 1, 0, 1, 1, 1, 1]:
A:=Matrix([[0, 1, 0, 1, 1, 0, 0, 0, 0], [1, 0, 1, 1, 1, 1, 0, 0, 0], [0, 1, 0, 0, 1, 1, 0, 0, 0], [1, 1, 0, 0, 1, 0, 1, 1, 0], A[5], [0, 1, 1, 0, 1, 0, 0, 1, 1], [0, 0, 0, 1, 1, 0, 0, 1, 0], [0, 0, 0, 1, 1, 1, 1, 0, 1], [0, 0, 0, 0, 1, 1, 0, 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); # Johannes W. Meijer, Aug 01 2010

Table[(1/32)(2(-2)^(n+2)+(2+Sqrt[8])^(n+2)+(2-Sqrt[8])^(n+2)), {n, 0, 19}] // FullSimplify
LinearRecurrence[{2,12,8}, {1,3,18}, 31] (* G. C. Greubel, Aug 18 2022 *)

Vec((1+x)/((1+2*x)*(1-4*x-4*x^2))+O(x^30)) \\ Joerg Arndt, Jan 29 2024

[2^(n-3)*(lucas_number2(n+2,2,-1) +2*(-1)^n) for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = (1/32)*(2*(-2)^(n+2) + (2+sqrt(8))^(n+2) + (2-sqrt(8))^(n+2)).
From R. J. Mathar, Jul 22 2010: (Start)
a(n) = 2*a(n-1) + 12*a(n-2) + 8*a(n-3).
G.f.: (1+x) / ( (1+2*x)*(1-4*x-4*x^2) ).
a(n) = (2*A057087(n-1) + 3*A057087(n) + (-2)^n)/4. (End)
Limit_{k->oo} a(n+k)/a(k) = A084128(n) + 2*A057087(n-1)*sqrt(2). - Johannes W. Meijer, Aug 01 2010
a(n) = A110048(n) + A110048(n-1). - R. J. Mathar, Mar 08 2021
a(n) = 2^(n-3)*(A002203(n+2) + 2*(-1)^n). - G. C. Greubel, Aug 18 2022

Offset changed and edited by Johannes W. Meijer, Jul 15 2010

A154929 A Fibonacci convolution triangle.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 5, 10, 6, 1, 8, 22, 21, 8, 1, 13, 45, 59, 36, 10, 1, 21, 88, 147, 124, 55, 12, 1, 34, 167, 339, 366, 225, 78, 14, 1, 55, 310, 741, 976, 770, 370, 105, 16, 1, 89, 566, 1557, 2422, 2337, 1443, 567, 136, 18, 1, 144, 1020, 3174, 5696, 6505, 4920, 2485

Offset: 0

Paul Barry, Jan 17 2009

Row sums are A028859. Diagonal sums are A141015(n+1). Inverse is A154930. Product of A030528 and A007318.
Transforms sequence m^n with g.f. 1/(1-m*x) to the sequence with g.f. (1+x)/(1-(m+1)x-(m+1)x^2).
Subtriangle of triangle T(n,k), given by (0, 2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. This triangle is the Riordan array (1, x(1+x)/(1-x-x^2)). - Philippe Deléham, Jan 25 2012

			Triangle begins
   1;
   2,   1;
   3,   4,   1;
   5,  10,   6,   1;
   8,  22,  21,   8,   1;
  13,  45,  59,  36,  10,   1;
  21,  88, 147, 124,  55,  12,   1;
  34, 167, 339, 366, 225,  78,  14,  1;
  55, 310, 741, 976, 770, 370, 105, 16, 1;
Production array is
     2,    1;
    -1,    2,   1;
     3,   -1,   2,   1;
   -10,    3,  -1,   2,  1;
    36,  -10,   3,  -1,  2,  1;
  -137,   36, -10,   3, -1,  2, 1;
   543, -137,  36, -10,  3, -1, 2, 1;
or ((1+x+sqrt(1+6x+5x^2))/2,x) beheaded.
T(5,3) = T(4,3) + T(4,2) + T(3,3) + T(3,2) = 8 + 21 + 1 + 6 = 36. - _Philippe Deléham_, Jan 18 2009
From _Philippe Deléham_, Jan 25 2012: (Start)
Triangle (0,2,-1/2,-1/2,0,0,0,...) DELTA (1,0,0,0,0,0,...) begins:
  1;
  0,   1;
  0,   2,   1;
  0,   3,   4,   1;
  0,   5,  10,   6,   1;
  0,   8,  22,  21,   8,   1;
  0,  13,  45,  59,  36,  10,   1;
  0,  21,  88, 147, 124,  55,  12,   1; (End)

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Table[Sum[Binomial[j + 1, n - j] Binomial[j, k], {j, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

Riordan array ((1+x)/(1-x-x^2), x(1+x)/(1-x-x^2));
Triangle T(n,k) = Sum_{j=0..n} C(j+1,n-j)*C(j,k).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(n,k)=0 if k > n. - Philippe Deléham, Jan 18 2009
Sum_{k=0..n} T(n,k)*x^k = A000045(n+1), A028859(n), A125145(n), A086347(n+1) for x=0,1,2,3 respectively. - Philippe Deléham, Jan 19 2009

A155117 a(n) = 4a(n-1) + 4a(n-2), n>2, a(0)=1, a(1)=3, a(2)=15.

Original entry on oeis.org

1, 3, 15, 72, 348, 1680, 8112, 39168, 189120, 913152, 4409088, 21288960, 102792192, 496324608, 2396467200, 11571167232, 55870537728, 269766819840, 1302549430272, 6289265000448, 30367257722880, 146626090893312

Offset: 0

Philippe Deléham, Jan 20 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Sequences of the form a(n) = m*(a(n-1) + a(n-2)) with a(0)=1, a(1) = m-1, a(2) = m^2 -1: A155020 (m=2), A155116 (m=3), this sequence (m=4), A155119 (m=5), A155127 (m=6), A155130 (m=7), A155132 (m=8), A155144 (m=9), A155157 (m=10).

Cf. A000129.

m:=4; [1] cat [n le 2 select (m-1)*(m*n-(m-1)) else m*(Self(n-1) + Self(n-2)): n in [1..30]]; // G. C. Greubel, Mar 25 2021

1,seq(simplify(-3*(2*I)^(n-2)*ChebyshevU(n, -I)), n = 1..30); # G. C. Greubel, Mar 25 2021

With[{m=4}, LinearRecurrence[{m, m}, {1, m-1, m^2-1}, 30]] (* G. C. Greubel, Mar 25 2021 *)

m=4; [1]+[-(m-1)*(sqrt(m)*i)^(n-2)*chebyshev_U(n, -sqrt(m)*i/2) for n in (1..30)] # G. C. Greubel, Mar 25 2021

G.f.: (1-x-x^2)/(1-4*x-4*x^2) . a(n)=3*A086347(n), n>=1 .
From G. C. Greubel, Mar 25 2021: (Start)
a(n) = (1/4)*[n=0] - 3*(2*i)^(n-2)*ChebyshevU(n, -i).
a(n) = (1/4)*[n=0] + 3*2^(n-2)*P_{n+1}, where P_{n} = A000129(n) (Pell numbers). (End)

A329118 Array read by antidiagonals: T(m, n) is the number of simple paths from corner to diagonally opposite corner on an m X n grid with king moves allowed.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 24, 24, 1, 1, 116, 235, 116, 1, 1, 560, 2922, 2922, 560, 1, 1, 2704, 38169, 96371, 38169, 2704, 1, 1, 13056, 494596, 3764367, 3764367, 494596, 13056, 1, 1, 63040, 6375379, 150610151, 447544629, 150610151, 6375379, 63040, 1, 1, 304384, 82191766, 5898799685, 56182569218, 56182569218, 5898799685, 82191766, 304384, 1

Offset: 1

Andrew Howroyd, Nov 05 2019

			Array begins:
===================================================================
m\n | 1     2       3          4             5                6
----+--------------------------------------------------------------
  1 | 1     1       1          1             1                1 ...
  2 | 1     5      24        116           560             2704 ...
  3 | 1    24     235       2922         38169           494596 ...
  4 | 1   116    2922      96371       3764367        150610151 ...
  5 | 1   560   38169    3764367     447544629      56182569218 ...
  6 | 1  2704  494596  150610151   56182569218   22132498074021 ...
  7 | 1 13056 6375379 5898799685 6972159602221 8656506756327178 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..231

Rows 2..6 are A086347, A193168, A193170, A193171, A193172.
Main diagonal is A140518.
Cf. A158651, A272445, A288033.

cp's OEIS Frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

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GAP

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Magma

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PARI

PARI

PARI

PARI

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A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

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Maple

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PARI

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A057087 Scaled Chebyshev U-polynomials evaluated at i. Generalized Fibonacci sequence.

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A125145 a(n) = 3a(n-1) + 3a(n-2). a(0) = 1, a(1) = 4.

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Haskell

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A214992 Power ceiling-floor sequence of (golden ratio)^4.

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A084128 a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.

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A084128 a(n) = 4a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

A155117 a(n) = 4a(n-1) + 4a(n-2), n>2, a(0)=1, a(1)=3, a(2)=15.