Search: id:A001353 Results 1-1 of 1 results found. %I A001353 M3499 N1420 %S A001353 0,1,4,15,56,209,780,2911,10864,40545,151316,564719,2107560,7865521, %T A001353 29354524,109552575,408855776,1525870529,5694626340,21252634831, %U A001353 79315912984,296011017105,1104728155436,4122901604639,15386878263120 %N A001353 a(n) = 4a(n-1)-a(n-2) with a(0) = 0, a(1) = 1. %C A001353 3*a(n)^2 + 1 is a perfect square. %C A001353 Number of spanning trees in 2 X n grid: by examining what happens at the right-hand end we see that a(n) = 3*a(n-1) + 2*a(n-2) + 2*a(n-3) + ... + 2*a(1) + 1, where the final 1 corresponds to the tree ==...=| !. Solving this we get a(n) = 4a(n-1) - a(n-2). %C A001353 Complexity of 2 X n grid. %C A001353 A016064 also describes triangles whose sides are consecutive integers and in which an inscribed circle has an integer radius. A001353 is exactly and precisely mapped to the integer radii of such inscribed circles, i.e. for each term of A016064, the corresponding term of A001353 gives the radius of the inscribed circle - Harvey P. Dale (hpd1(AT)is2.nyu.edu), Dec 28 2000 %C A001353 If M is any term of the sequence, the next one is 2M + sqrt(3M^2 + 1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Feb 18 2002 %C A001353 n such that 3*n^2=floor(sqrt(3)*n*ceil(sqrt(3)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003 %C A001353 For n>0, ratios a(n+1)/a(n) may be obtained as convergents of the continued fraction expansion of 2+sqrt(3): either as successive convergents of [4;-4] or as odd convergents of [3;1, 2]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Sep 19 2003 %C A001353 Ways of packing a 3 X (2n-1) rectangle with dominoes, after attaching an extra square to the end of one of the sides of length 3. With reference to A001835, therefore: a(n) = a(n-1) + A001835(n-1) and A001835(n) = 3*A011835(n-1) + 2*a(n-1). - Joshua Zucker and the Castilleja School Math Club (joshua_zucker(AT)castilleja.org), Oct 28 2003 %C A001353 a(n+1) is a Chebyshev transform of 4^n, where the sequence with g.f. G(x) is sent to the sequence with g.f. (1/(1+x^2))G(x/(1+x^2)). - Paul Barry (pbarry(AT)wit.ie), Oct 25 2004 %C A001353 This sequence generates many brilliant (A078972) numbers for a(p) with prime p: a(2) = 4 = 2 * 2 a(3) = 15 = 3 * 5 a(5) = 209 = 11 * 19 a(7) = 2911 = 41 * 71 a(19) = 21252634831 = 110771 * 191861 a(37) = 419245718107612602961 = 15558008491 * 26947261171. Is this a prime-free sequence? If not, what is its first prime? - Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 08 2005 %C A001353 Numbers such that there is an m with t(n+m)=3t(m), where t(n) are the triangular numbers A000217. For instance t(35)=3t(20)=630, so 35-20=15 is in the sequence. - comment by Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 13 2005 %C A001353 a(n) = number of unique matrix products in (A+B+C+D)^n where commutator [A,B]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Max Alekseyev (maxale(AT)gmail.com), Feb 01 2006 %C A001353 For n>1, middle side (or long leg) of primitive Pythagorean triangles having an angle nearing pi/3 with larger values of sides. [Complete triple (X, Y, Z), XTable of n, a(n) for n=0..200 %H A001353 Index entries for sequences related to linear recurrences with constant coefficients %H A001353 F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154. %H A001353 F. Faase, Counting Hamilton cycles in product graphs %H A001353 F. Faase, Results from the counting program %H A001353 F. Faase, Counting Hamilton cycles in product graphs %H A001353 Tanya Khovanova, Recursive Sequences %H A001353 Hojoo Lee, Problems in elementary number theory Problem I 18. %H A001353 S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992. %H A001353 S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992. %H A001353 P. Raff Analysis of the Number of Spanning Trees of K_2 x P_n. Contains sequence, recurrence, generating function, and more. [From Paul Raff (praff(AT)math.rutgers.edu), Mar 06 2009] %H A001353 D. P. Walsh, Counting n x 2 Simple Rectangular Mazes %H A001353 Index entries for sequences related to Chebyshev polynomials. %F A001353 a(n)=[(2+sqrt(3))^n-(2-sqrt(3))^n]/(2*sqrt(3)). %F A001353 Limit as n-> infinity of a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 06 2002 %F A001353 Binomial transform of A002605. E.g.f.: exp(2x)sinh(sqrt(3)x)/sqrt(3). %F A001353 G.f.: x/(1-4*x+x^2). a(n) = S(n-1, 4) = U(n-1, 2), S(-1, x) := 0, Chebyshev's polynomials of the second kind A049310. %F A001353 a(n+1)=sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*4^(n-2k)} - Paul Barry (pbarry(AT)wit.ie), Oct 25 2004 %F A001353 a(n)=sum{k=0..n-1, binomial(n+k, 2k+1)2^k} - Paul Barry (pbarry(AT)wit.ie), Nov 30 2004 %F A001353 a(n)=3*a(n-1)+3*a(n-2)-a(n-3); a(0)=0, a(1)=1, a(2)=4. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 13 2006 %F A001353 a(n) = 2*a(n-1)+sqrt[3*a(n-1)^2+1]. a(n) = -A106707(n). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2006 %F A001353 a(n) = 3*(a(n-1)+a(n-2))-a(n-3), a(n) = 5*(a(n-1)-a(n-2))+a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 2006 %F A001353 M^n * [1,0] = [A001075(n), A001353(n)], where M = the 2 X 2 matrix [2, 3; 1,2]; e.g., a(4) = 56 since M^4 * [1,0] = [97, 56] = [A001075(4), A001353(4)]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 27 2006 %F A001353 Sequence satisfies 1 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos Sep 19 2008 %F A001353 a(-n) = -a(n). - Michael Somos Sep 19 2008 %e A001353 For example, when n=3: %e A001353 **** %e A001353 .*** %e A001353 .*** %e A001353 can be packed with dominoes in 4 different ways: 3 in which the top row is tiled with two horizontal dominoes and 1 in which the top row has two vertical and one horizontal domino, as shown below, so a(2) = 4. %e A001353 ---- ---- ---- ||-- %e A001353 .||| .--| .|-- .||| %e A001353 .||| .--| .|-- .||| %p A001353 A001353 := proc(n) option remember; if n <= 1 then 1+3*n else 4*A001353(n-1)-A001353(n-2); fi; end; %p A001353 A001353:=z/(1-4*z+z**2); [S. Plouffe in his 1992 dissertation.] %t A001353 a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, 0, 23}]] (from Robert G. Wilson v Jan 13 2005) %t A001353 lst = {}; Do[AppendTo[lst, GegenbauerC[n, 1, 2]], {n, -1, 23}]; lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 14 2009] %o A001353 (PARI) M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=0,30,print1(([1,0,0]*M^i)[2], ",")) - from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005 %o A001353 sage: [lucas_number1(n,4,1) for n in range(27)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008 %o A001353 (PARI) {a(n) = real( (2 + quadgen(12))^n / quadgen(12) )} /* Michael Somos Sep 19 2008 */ %o A001353 (PARI) {a(n) = subst( polchebyshev(n-1, 2), x, 2)} /* Michael Somos Sep 19 2008 */ %o A001353 (Other) sage: [lucas_number1(n,4,1) for n in xrange(0, 25)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 22 2009] %Y A001353 a(n) = sqrt((A001075(n)^2-1)/3). %Y A001353 Cf. A003500, A001835, A001075, A001571, A001834, A002531, A005246, A016064, A082840, A079935, A078972. %Y A001353 A bisection of A002530. %Y A001353 Cf. A125077 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 21 2009] %Y A001353 Sequence in context: A060111 A077824 A010905 this_sequence A106707 A125905 A026030 %Y A001353 Adjacent sequences: A001350 A001351 A001352 this_sequence A001354 A001355 A001356 %K A001353 nonn,easy,nice %O A001353 0,3 %A A001353 N. J. A. Sloane (njas(AT)research.att.com). Search completed in 0.003 seconds