Search: id:A001333 Results 1-1 of 1 results found. %I A001333 M2665 N1064 %S A001333 1,1,3,7,17,41,99,239,577,1393,3363,8119,19601,47321,114243,275807, %T A001333 665857,1607521,3880899,9369319,22619537,54608393,131836323,318281039, %U A001333 768398401,1855077841,4478554083,10812186007,26102926097,63018038201 %N A001333 Numerators of continued fraction convergents to sqrt(2). %C A001333 Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley]. %C A001333 Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180 degree rotational symmetry - Erich Friedman (erich.friedman(AT)stetson.edu) %C A001333 a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other. %C A001333 Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row. - Ron Hardin (rhhardin(AT)att.net), Mar 16 2002 %C A001333 a(2*n+1) with b(2*n+1) := A000129(2*n+1), n>=0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1. %C A001333 a(2*n) with b(2*n) := A000129(2*n), n>=1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference). %C A001333 Bisection: a(2*n)= T(n,3)=A001541(n), n>=0 and a(2*n+1)=S(2*n,2*sqrt(2))= A002315(n), n>=0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first,resp. second kind. See A053120, resp. A049310. %C A001333 Binomial transform of A077957. - Paul Barry (pbarry(AT)wit.ie), Feb 25 2003 %C A001333 For n>0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,....,n, s(0) = 2, s(n) = 2. - Herbert Kociemba (kociemba(AT)t-online.de), Jun 02 2004 %C A001333 x satisfying x^2 - 2*y^2 = -+1. Corresponding y is given by A000129(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 24 2004 %C A001333 For n>1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 06 2004 %C A001333 Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 18 2004 %C A001333 Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006 %C A001333 Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2,7/5,17/12,41/ 29,... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006] %C A001333 a(n) mod 10 = A131707. See A131711. - Paul Curtz (bpcrtz(AT)free.fr), Apr 08 2008 %C A001333 Starting (1, 3, 7, 17,...) = row sums of triangle A140750 - Gary W. Adamson (qntmpkt(AT)yahoo.com), May 26 2008 %C A001333 Prime numerators with an odd index are prime RMS numbers(A140480) and also NSW primes(A088165). [From Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), Aug 13 2008] %C A001333 A001333(2*n+1)=A002315(n); A001333(2*n)=A001541(n). [From Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), Aug 13 2008] %C A001333 The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators=A001333. - Clark Kimberling (ck6(AT)evansville.edu), Aug 26 2008 %C A001333 Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 06 2008: (Start) %C A001333 Equals right border of triangle A143966. Starting (1, 3, 7,...) equals %C A001333 INVERT transform of (1, 2, 2, 2,...) and row sums of triangle A143966. (End) %C A001333 Inverse binomial transform of A006012 ; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...] . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 04 2008] %C A001333 Contribution from Charlie Marion (charliemath(AT)optonline.net), Jan 07 2009: (Start) %C A001333 In general, denominators, a(k,n) and numerators, b(k,n), of continued %C A001333 fraction convergents to sqrt((k+1)/k) may be found as follows: %C A001333 let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1)+a(k,2n-2) %C A001333 and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1); %C A001333 let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1)+b(k,2n-2) %C A001333 and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1). %C A001333 For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, %C A001333 17/12, 41/29. %C A001333 In general, if a(k,n) and b(k,n) are the denominators and numerators, %C A001333 respectively, of continued fraction convergents to sqrt((k+1)/k) %C A001333 as defined above, then %C A001333 k*a(k,2n)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and %C A001333 b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n); %C A001333 for example, if k=1 and n=3, then b(1,n)=a(n+1) and %C A001333 1*a(1,6)^2-a(1,5)*a(1,7)=1*169^2-70*408=1; %C A001333 1*a(1,4)*a(1,6)-a(1,5)^2=1*29*169-70^2=1; %C A001333 b(1,5)*b(1,7)-1*b(1,6)^2=99*577-1*239^2=2; %C A001333 b(1,5)^2-1*b(1,4)*b(1,6)=99^2-1*41*239=2. %C A001333 Cf. A000129, A142238-A142239, A153313-153318. %C A001333 [From Charlie Marion (charliemath(AT)optonline.net), Jan 07 2009] %C A001333 (End) %C A001333 Equals row sums of triangle A160756 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 25 2009] %D A001333 Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5. %D A001333 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964. %D A001333 F. R. K. Chung and R. L. Graham, Primitive juggling sequences, preprint, 2006. %D A001333 John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16. %D A001333 E. I. Emerson, Recurrent sequences in the equation DQ^2=R^2+N, Fib. Quart., 7 (1969), 231-242, see Ex. 1, pp. 237-238. %D A001333 Reinhardt Euler, The Fibonacci Number of a Grid Graph and a New Class of Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.6. %D A001333 David Garth and Adam Gouge, Affinely Self-Generating Sets and Morphisms, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.5. %D A001333 A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. %D A001333 Y. Kong, Ligand binding on ladder lattices, Biophysical Chemistry, Vol. 81 (1999), pp. 7-21. %D A001333 Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159 %D A001333 Munarini, Emanuele, Combinatorial properties of the antichains of a garland. Integers, 9 (2009), 353-374. %D A001333 I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10. %D A001333 H. Prodinger and R. F. Tichy, Fibonacci numbers of graphs, Fibonacci Quarterly, 20 (1982), 16-21. %D A001333 B. S. Rao, Heptagonal numbers in the associated Pell sequence ..., Fib. Quarterly, 43 (2005), 302-306. %D A001333 J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224. %D A001333 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). %D A001333 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). %D A001333 R. P. Stanley, Enumerative Combinatorics, Volume 1 (1986), p. 203, Example 4.1.2. %D A001333 A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12. %D A001333 Gy. Tasi et al., Quantum algebraic-combinatoric study of the conformational properties of n-alkanes. II, J. Math. Chemistry, 27, 2000, 191-199 (see p. 193). %D A001333 V. Thebault, Concerning two classes of remarkable perfect square pairs, Amer. Math. Monthly, 56 (1949), 443-448. %D A001333 R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339. %H A001333 T. D. Noe, Table of n, a(n) for n = 0..500 %H A001333 Joerg Arndt, Fxtbook %H A001333 C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002. %H A001333 Nick Hobson, Python program for this sequence %H A001333 INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 143 %H A001333 Tanya Khovanova, Recursive Sequences %H A001333 Kin Y. Li, Problem 159 %H A001333 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 A001333 S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992. %H A001333 Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics. %H A001333 Eric Weisstein's World of Mathematics, Pythagoras's Constant %H A001333 Eric Weisstein's World of Mathematics, Square Triangular Number %H A001333 Index entries for "core" sequences %H A001333 Index entries for sequences related to Chebyshev polynomials. %H A001333 Index entries for sequences related to linear recurrences with constant coefficients %F A001333 a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007 %F A001333 a(n) = 2a(n-1) + a(n-2); a(n) = ( (1-Sqrt[ 2 ])^n + (1+Sqrt[ 2 ])^n) /2. %F A001333 G.f.: (1-x)/(1-2*x-x^2). %F A001333 A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30, 2002 %F A001333 a(n) = ((-i)^n)*T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1. %F A001333 a(n)=a(n-1)+A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003 %F A001333 E.g.f.: exp(x)cosh(x*sqrt(2)) - Paul Barry (pbarry(AT)wit.ie), May 08 2003 %F A001333 a(n)=sum{k=0..floor(n/2), C(n, 2k)2^k } - Paul Barry (pbarry(AT)wit.ie), May 13 2003 %F A001333 For n >0, a(n )^2 - (1 + (-1)^(n ))/2 =sum_{k=0...n-1}((2k+1)*A001653(n-1-k)); e.g. 17^2-1=288=1*169+3*29+5*5+7*1; 7^2=49=1*29+3*5+5*1 - Charlie Marion (charliem(AT)bestweb.net), Jul 18 2003 %F A001333 A001333(n+2) = A078343(n+1) + A048654(n) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jan 19 2005 %F A001333 Conjecture: For prime p, A001333(p) congruent to 1 mod p ( compare with similar comment for A000032 ) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Oct 11 2005 %F A001333 a(n) = A000129(n)+A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/ A000129(n)^2) = ceiling(sqrt(A001108(n))) - Henry Bottomley (se16(AT)btinternet.com), Apr 18 2000 %F A001333 Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1)+1 - Graeme McRae (g_m(AT)mcraefamily.com), Aug 03 2006 %F A001333 a(n) = Sum_{k,0<=k<=n}A122542(n,k) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 08 2006 %F A001333 For another recurrence see A000129. %F A001333 Starting (1, 3, 7, 17, 41,...), = row sums of triangle A135837. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 01 2007 %F A001333 a(n)=Sum_{k, 0<=k<=n}A098158(n,k)*2^(n-k). - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 26 2007 %F A001333 a(n) = upper left and lower right terms of [1,1; 2,1]^n - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 12 2008 %F A001333 (a)=((sqrt2*(1+sqrt2)^n)-(-sqrt2*(1-sqrt2)^n))/sqrt8. Offset 0. a(3)=7 [From Al Hakanson (hawkuu(AT)gmail.com), Apr 11 2009] %e A001333 Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/ 985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129 %e A001333 The 15 3 X 2 crossword grids, with white squares represented by an o: %e A001333 ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo %e A001333 ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo. %p A001333 A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*A001333(n-1)+A001333(n-2) fi end; # version 1 %p A001333 Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n); # version 2 %p A001333 with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)]; # version 3 %p A001333 with(numtheory):cf := cfrac (sin(Pi/4),100): seq(nthdenom (cf,i), i=0..29 ); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 07 2007 %p A001333 A001333:=-(z+1)/(-1+2*z+z**2); [S. Plouffe in his 1992 dissertation.] %p A001333 (Maple) a := proc(n) local M; M := (Matrix([[2,1],[1,0]])^n); M[2,1]+M[2, 2] end; seq (a(n), n=0..29); [From Alois P. Heinz (heinz(AT)hs-heilbronn.de), Aug 01 2008] %t A001333 Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 08 2006 %t A001333 f[n_] := ((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2; Table[Simplify@ f@n, {n, 0, 29}] (* Or *) %t A001333 a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (from Robert G. Wilson v (rgwv(at)rgwv.com), May 02 2006) %t A001333 a=0;b=1;c=0;lst={b};Do[c=a+b+c;AppendTo[lst,c];a=b;b=c,{n,5!}];lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Mar 23 2009] %o A001333 (PARI) a(n)=if(n<0,0,contfracpnqn(vector(n,i,1+(i>1)))[1,1]) %o A001333 sage: from sage.combinat.sloane_functions import recur_gen2 sage: it = recur_gen2(1,1,2,1) sage: [it.next() for i in range(30)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 24 2008 %o A001333 (Other) sage: [lucas_number2(n,2,-1)/2 for n in xrange(0, 30)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 30 2009] %o A001333 (PARI) { default(realprecision, 2000); for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jun 12 2009] %Y A001333 For denominators see A000129. %Y A001333 See A040000 for the continued fraction expansion of sqrt(2). %Y A001333 a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n) (companion Pell numbers). %Y A001333 See also A078057 which is the same sequence without the initial 1. %Y A001333 Cf. also A152113. %Y A001333 Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle). %Y A001333 Essentially the same as A078057. Equals A034182(n-1) + 2 and A084128(n)/ 2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965. %Y A001333 The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519. %Y A001333 Second row of the array in A135597. %Y A001333 Cf. A135837, A140750, A143966. %Y A001333 A160756 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 25 2009] %Y A001333 Sequence in context: A131056 A077851 A089737 this_sequence A078057 A123335 A089742 %Y A001333 Adjacent sequences: A001330 A001331 A001332 this_sequence A001334 A001335 A001336 %K A001333 nonn,cofr,easy,core,nice,frac %O A001333 0,3 %A A001333 N. J. A. Sloane (njas(AT)research.att.com), R. K. Guy %E A001333 Chebyshev comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Jan 10 2003 Search completed in 0.003 seconds