A123480 -id:A123480 - cp's OEIS frontend

A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

Offset: 0

Don N. Page

Triangular numbers that are twice other triangular numbers. - Don N. Page
Triangular numbers that are also pronic numbers. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002
In other words, triangular numbers which are products of two consecutive numbers. E.g., a(2) = 210: 210 is a triangular number which is the product of two consecutive numbers: 14 * 15. - Shyam Sunder Gupta, Oct 26 2002
Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2; 1, 4, 1], [2; 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1, 4, 1] and so forth. - Gene Ward Smith, Sep 30 2006
This sequence satisfy the same recurrence as A165518. - Ant King, Dec 13 2010
Intersection of A000217 and A002378.
This is the sequence of areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + 1,z(n)) with x(0) = 0, y(0) = 1, z(0) = 1, a(0) = 0 and x(1) = 3, y(1) = 4, z(1) = 5, a(1) = 6. - George F. Johnson, Aug 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Shyam Sunder Gupta Fascinating Triangular Numbers
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).
Cf. A000129, A001108, A002203, A009111, A245031.
Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # G. C. Greubel, Jan 13 2020

a029549 n = a029549_list !! n
a029549_list = [0,6,210] ++
   zipWith (+) a029549_list
               (map (* 35) $ tail delta)
   where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011

(makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* Bill Gosper, Feb 07 2010 */

R:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // G. C. Greubel, Jul 15 2018

A029549 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,6]) ;
    else
        34*procname(n-1)-procname(n-2)+6 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by Ray Chandler, Jul 09 2015 *)
CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* Bill Gosper, Feb 07 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* Harvey P. Dale, Jun 06 2011 *)
(LucasL[4Range[20] - 2, 2] -6)/32 (* G. C. Greubel, Jan 13 2020 *)

concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ Charles R Greathouse IV, Jun 13 2013

[(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # G. C. Greubel, Jan 13 2020

val triNums = (0 to 39999).map(n => (n * n + n)/2)
triNums.filter( % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte, Jan 12 2020

G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).
a(n) = 6*A029546(n-1) = 2*A075528(n).
a(n) = -3/16 + ((3+2*sqrt(2))/32) *(17 + 12*sqrt(2))^n + ((3-2*sqrt(2))/32) *(17 - 12*sqrt(2))^n. - Gene Ward Smith, Sep 30 2006
From Bill Gosper, Feb 07 2010: (Start)
a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.
a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)
a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009
a(n) = ceiling((3 + 2*sqrt(2))^(2n + 1) - 6)/32 = floor((1/32) (1+sqrt(2))^(4n+2)). - Ant King, Dec 13 2010
Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - Ant King, Dec 13 2010
a(n) = a(n - 1) + A001109(2n). - Charlie Marion, Feb 10 2011
a(n+2) = 34*a(n + 1) - a(n) + 6. - Charlie Marion, Feb 11 2011
From George F. Johnson, Aug 20 2012: (Start)
a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.
8*a(n) + 1 = (A002315(n))^2, 4*a(n) + 1 = (A000129(2*n + 1))^2, 32*a(n)^2 + 12*a(n) + 1 are perfect squares.
a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).
a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).
Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).
Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.
Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)
a(n) = 3 * T( b(n) ) + (2*b(n) + 1)*sqrt( T( b(n) ) ) where b(n) = A001108(n) (indices of the square triangular numbers), T(n) = A000217(n) (the n-th triangular number). - Dimitri Papadopoulos, Jul 07 2017
a(n) = (Pell(2*n + 1)^2 - 1)/4 = (Q(4*n + 2) - 6)/32, where Q(n) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Jan 13 2020
a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Christian G. Bower, Sep 19 2002; T. D. Noe, Nov 07 2006; and others

Edited by N. J. A. Sloane, Apr 18 2007, following suggestions from Andrew S. Plewe and Tanya Khovanova

A076139 Triangular numbers that are one-third of another triangular number: T(m) such that 3*T(m) = T(k) for some k.

Original entry on oeis.org

0, 1, 15, 210, 2926, 40755, 567645, 7906276, 110120220, 1533776805, 21362755051, 297544793910, 4144264359690, 57722156241751, 803965923024825, 11197800766105800, 155965244802456376, 2172315626468283465, 30256453525753512135, 421418033734080886426

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

Both triangular and generalized pentagonal numbers: intersection of A000217 and A001318. - Vladeta Jovovic, Aug 29 2004

Partial sums of Chebyshev polynomials S(n,14).

			G.f. = x + 15*x^2 + 210*x^3 + 2926*x^4 + 40755*x^5 + 567645*x^6 + ...
a(3)=210=T(20) and 3*210=630=T(35).

Colin Barker, Table of n, a(n) for n = 0..874
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

The m values are in A061278, the k values are in A001571.
Cf. A014979, A076140, A108281.
Cf. A045899, A123480, A217855.
Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

[(Evaluate(ChebyshevU(n+1), 7) - Evaluate(ChebyshevU(n), 7) - 1)/12 : n in [0..30]]; // G. C. Greubel, Feb 03 2022

a[n_] := a[n] = 14*a[n-1] - a[n-2] + 1; a[0] = 0; a[1] = 1; Table[ a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 15 2011, after given formula *)

{a(n) = polchebyshev( n, 2, 7) / 14 + polchebyshev( n, 1, 7)/ 84 - 1 / 12}; /* Michael Somos, Jun 16 2011 */

concat(0, Vec(-x/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

[(chebyshev_U(n,7) - chebyshev_U(n-1,7) - 1)/12 for n in (0..30)] # G. C. Greubel, Feb 03 2022

G.f.: x / ((1 - x) * (1 - 14*x +x^2)).
a(n+1) = Sum_{k=0..n} S(k, 14), n >= 0, where S(k, 14) = U(k, 7) = A007655(k+2).
a(n+1) = (S(n+1, 14) - S(n, 14) - 1)/12, n >= 0.
a(n) = 14 * a(n-1) - a(n-2) + 1. a(0)=0, a(1)=1.
a(-n) = a(n-1).
a(n) = A061278(n)*(A061278(n)+1)/2.
a(n) = (1/288)*(-24 + (12-6*sqrt(3))*(7-4*sqrt(3))^n + (12+6*sqrt(3))*(7+4*sqrt(3))^n).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) with a(0)=0, a(1)=1, a(2)=15. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002
a(2*n) = A108281(n + 1). a(2*n + 1) = A014979(n + 2). - Michael Somos, Jun 16 2011
a(n) = (1/2)*A217855(n) = (1/3)*A076140(n) = (1/4)*A123480(n) = (1/8)*A045899(n). - Peter Bala, Dec 31 2012
a(n) = A001353(n) * A001353(n-1) / 4. - Richard R. Forberg, Aug 26 2013
a(n) = ((2+sqrt(3))^(2*n+1) + (2-sqrt(3))^(2*n+1))/48 - 1/12. - Vladimir Pletser, Jan 15 2021

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002

Chebyshev comments from Wolfdieter Lang, Aug 31 2004

A076140 Triangular numbers T(k) that are three times another triangular number: T(k) such that T(k) = 3*T(m) for some m.

Original entry on oeis.org

0, 3, 45, 630, 8778, 122265, 1702935, 23718828, 330360660, 4601330415, 64088265153, 892634381730, 12432793079070, 173166468725253, 2411897769074475, 33593402298317400, 467895734407369128, 6516946879404850395, 90769360577260536405, 1264254101202242659278

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

This is a subsequence of A045943. - Michel Marcus, Apr 26 2014

			a(3) = 630 because 630 = T(35) and 630/3 = 210 = T(20).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Subsequence of A000217.
The m values are in A061278 and the k values are in A001571.
Cf. A076139, A045899, A123480, A217855.
Cf. A045943.

Join[{0}, CoefficientList[Series[3/(1 - 15x + 15x^2 - x^3), {x, 0, 20}], x]]  (* Harvey P. Dale, Apr 02 2011 *)
triNums = Accumulate[Range[0, 9999]]; Select[triNums, MemberQ[triNums, #/3] &] (* Alonso del Arte, Mar 24 2020 *)

concat(0, Vec(-3*x/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

a(n) = (3/288)*(-24 + (12 - 6*sqrt(3))*(7 - 4*sqrt(3))^n + (12 + 6*sqrt(3))*(7 + 4*sqrt(3))^n).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002: (Start)
a(0) = 0, a(1) = 3, a(2) = 45; a(n) = 15*(a(n-1) -a (n-2)) + a(n-3) for n >= 3.
G.f.: (3*x)/(1 - 15*x + 15*x^2 - x^3). (End)
a(n) = 3*A076139(n) = 3/2*A217855(n) = 3/4*A123480(n) = 3/8*A045899(n). - Peter Bala, Dec 31 2012
a(0) = 0, a(n) = 14 * a(n - 1) - a(n - 2) + 3 for n > 0. - Vladimir Pletser, Mar 23 2020
a(n) = ((2+sqrt(3))*(7+4*sqrt(3))^n + ((2-sqrt(3))*(7-4*sqrt(3))^n))/16 - 1/4 = ((2+sqrt(3))^(2n+1) + ((2-sqrt(3))^(2n+1)))/16 - 1/4. - Vladimir Pletser, Jan 15 2021

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Original entry on oeis.org

0, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080, 3371344269872647091408

Offset: 1

Andrej Dujella (duje(AT)math.hr)

Essentially the same as A051047.
It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - Jonathan Vos Post, Feb 28 2011
Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - Bruno Berselli, Feb 20 2018

G. C. Greubel, Table of n, a(n) for n = 1..870
A. Baker and H. Davenport, The Equations 3x^2-2=y^2 and 8x^2-7=z^2, Quart. J. Math. Oxford 20 (1969).
A. Dujella and A. Pethoe, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.
A. Dujella, The Problem of Diophantus and Davenport, References
A. Dujella, Publications of Andrej Dujella
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6
P. Gibbs, 1,3,8,120 ... A Diophantine Problem
P. Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A051047, A067900. A076139, A076140, A123480, A217855.

Cf. A046184, A245031.

f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)
Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* G. C. Greubel, Jun 07 2017 *)
LinearRecurrence[{15,-15,1},{0,8,120},20] (* Harvey P. Dale, Jul 14 2024 *)

x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ G. C. Greubel, Jun 07 2017

a(n) = A046184(n+1) - 1.
a(n) = 14*a(n-1) - a(n-2) + 8.
a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006
a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - Peter Bala, Dec 31 2012
From Colin Barker, Jul 30 2013: (Start)
G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)
E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - Ilya Gutkovskiy, Apr 28 2016

A217855 Numbers m such that 16m(3*m+1)+1 is a square.

Original entry on oeis.org

0, 2, 30, 420, 5852, 81510, 1135290, 15812552, 220240440, 3067553610, 42725510102, 595089587820, 8288528719380, 115444312483502, 1607931846049650, 22395601532211600, 311930489604912752, 4344631252936566930, 60512907051507024270, 842836067468161772852

Offset: 0

Arnaldo Vicentini, Oct 12 2012

If S(h) is the sum of the squares of the integers from 1 to h, then S(h)/h = (h+1)*(2*h+1)/6. If S(h)/h is a square, then h+1 is even and 2*h+1 is a multiple of 3, so that (h+1)/2 and (2*h+1)/3 must be odd squares. The numbers h that satisfy this condition are in A084231. Let h = 1+24*k+72*k^2; for any natural k we have (h+1)/2 = (6*k+1)^2, but only if 16*k*(3*k+1)+1 is a square also (2*h+1)/3 is a square. Searching what integer numbers k satisfy this condition leads to this sequence.

			a(0) = ((2+sqrt(3))^1+(2-sqrt(3))^1-4)/24 = 0,
a(1) = ((2+sqrt(3))^3+(2-sqrt(3))^3-4)/24 = 2,
a(2) = ((2+sqrt(3))^5+(2-sqrt(3))^5-4)/24 = 30;
a(2) = 14*a(1)-a(0)+2 = 30,
a(3) = 14*a(2)-a(1)+2 = 420,
a(4) = 14*a(3)-a(2)+2 = 5852;
a(3) = 15*a(2)-15*a(1)+a(0) = 420,
a(4) = 15*a(3)-15*a(2)+a(1) = 5852.
Regarding to the comment, a(3) = 420 and so 72*a(3)^2+24*a(3)+1 = A084231(4) = 12710881, therefore Sum_{i=1..12710881} i^2/12710881 = 7338631^2 = A084232(3)^2. - _Bruno Berselli_, Oct 17 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Forum Rudi Mathematici, Un po' di calcoli...un po' di logica, post #923 (in Italian)
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A028230, A076139, A084231, A084232. A045899, A076140, A123480.

m:=19; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2/((1-x)*(1-14*x+x^2)))); // Bruno Berselli, Oct 16 2012

LinearRecurrence[{15, -15, 1}, {0, 2, 30}, 20] (* Bruno Berselli, Oct 16 2012 *)

makelist(expand(((2+sqrt(3))^(2*n+1)+(2-sqrt(3))^(2*n+1)-4)/24), n, 0, 19); /* Bruno Berselli, Oct 16 2012 */

a=vector(20); a[1]=0; a[2]=2; a[3]=30; for(i=4, #a, a[i]=15*a[i-1]-15*a[i-2]+a[i-3]); a \\ Bruno Berselli, Oct 16 2012

a(n) = ((2 + sqrt(3))^(2*n + 1) + (2 - sqrt(3))^(2*n + 1) - 4)/24.
a(n) = (cosh((2*n + 1)*log(2 + sqrt(3))) - 2)/12.
a(n) = a(-n-1) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = 14*a(n-1) - a(n-2) + 2.
a(n) = 2*A076139(n).
72*a(n)^2 + 24*a(n)+1 = A084231(n+1).
16*a(n)*(3*a(n) + 1) + 1 = A028230(n+1)^2.
G.f.: 2*x/((1 - x)*(1 - 14*x + x^2)). - Bruno Berselli, Oct 16 2012
a(n) = 2*A076139(n) = 1/2*A123480(n) = 1/4*A045899(n) = 2/3*A076140(n). - Peter Bala, Dec 31 2012

A123479 Coefficients of series giving the best rational approximations to sqrt(6).

Original entry on oeis.org

20, 1980, 194040, 19013960, 1863174060, 182572043940, 17890197132080, 1753056746899920, 171781670999060100, 16832850701160989900, 1649447587042777950120, 161629030679491078121880, 15837995559003082877994140, 1551961935751622630965303860

Offset: 1

Gene Ward Smith, Sep 28 2006

The partial sums of the series 5/2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(6), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2;2], [2;2,4,2], [2;2,4,2,4,2], [2;2,4,2,4,2,4,2] and so forth.

Sequence of numbers x=a(n) such 4*x+1 and 6*x+1 are both square, and their square roots are A138288(n) and A054320(n). - Paul Cleary, Jun 23 2014

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A123478, A123480, A029549, A123482.

LinearRecurrence[{99,-99,1},{0,20,1980},{2, 25}] (* Paul Cleary, Jun 23 2014 *)

Vec(-20*x/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 99*a(n+2) - 99*a(n+1) + a(n).
a(n) = -5/24 + (( + 2*6^(1/2))/48)*(49 + 20*6^(1/2))^n + ((5 - 2*6^(1/2))/48)*(49 - 20*6^(1/2))^n.
G.f.: -20*x / ((x-1)*(x^2-98*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A221075 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 12, 1, 24, 1, 192, 1, 360, 1, 2700, 1, 5040, 1, 37632, 1, 70224, 1, 524172, 1, 978120, 1, 7300800, 1, 13623480, 1, 101687052, 1, 189750624, 1, 1416317952, 1, 2642885280, 1, 19726764300, 1, 36810643320, 1

Offset: 0

Peter Bala, Jan 06 2013

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).

If we denote the present sequence by [2; 12, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below

			Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
= 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
product {n >= 0} {1 - 2*(26 - 15*sqrt(3))^(4*n+3)}/{1 - 2*(26 - 15*sqrt(3))^(4*n+1)} = 1.04000 05921 62729 43797 ... = 1 + 1/(24 + 1/(1 + 1/(2700 + 1/(1 + 1/(70224 + 1/(1 + 1/(7300800 + ...))))))).

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,14,0,-14,0,-1,0,1).

Cf. A001353, A007654, A045899, A076139, A076140, A098301, A123480, A174500, A217855, A221073 (m = 2), A221074 (m = 3), A221076 (m = 5).

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
a(4*n - 3) = 12*A098301(n) = 12*A001353(n)^2 = 4*A007654(n);
a(4*n - 1) = 24*A076139(n) = 12*A217855 = 8*A076140(n) = 6*A123480(n) = 3*A045899(n).
O.g.f.: 2 + x^2/(1 - x^2) + 12*x*(1 + x^2)^2/(1 - 15*x^4 + 15*x^8 - x^12) = 2 + 12*x + x^2 + 24*x^3 + x^4 + 192*x^5 + ....
O.g.f.: (x^10-2*x^8-14*x^6+28*x^4-12*x^3+x^2-12*x-2) / ((x-1)*(x+1)*(x^4-4*x^2+1)*(x^4+4*x^2+1)). - Colin Barker, Jan 10 2014

A123478 Coefficients of series giving the best rational approximations to sqrt(7).

Original entry on oeis.org

48, 12240, 3108960, 789663648, 200571457680, 50944360587120, 12939667017670848, 3286624478127808320, 834789677777445642480, 212033291530993065381648, 53855621259194461161296160, 13679115766543862141903843040, 3474441549080881789582414836048

Offset: 1

Gene Ward Smith, Sep 28 2006

The partial sums of the series 8/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(7), which constitute every fourth convergent of the continued fraction. The corresponding continued fractions are [2;1,1,1], [2;1,1,1,4,1,1,1], [2;1,1,1,4,1,1,1,4,1,1,1] and so forth.

Harvey P. Dale, Table of n, a(n) for n = 1..416
Index entries for linear recurrences with constant coefficients, signature (255,-255,1).

Cf. A123479, A123480, A029549, A123482.

LinearRecurrence[{255,-255,1},{48,12240,3108960},30] (* Harvey P. Dale, Nov 20 2016 *)

Vec(-48*x/((x-1)*(x^2-254*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 255 a(n+2) - 255 a(n+1) + a(n).
a(n) = -4/21 + (2/21+1/28*7^(1/2))*(127+48*7^(1/2))^n + (2/21-1/28*7^(1/2))*(127-48*7^(1/2))^n.
G.f.: -48*x / ((x-1)*(x^2-254*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A123482 Coefficients of the series giving the best rational approximations to sqrt(11).

Original entry on oeis.org

60, 23940, 9528120, 3792167880, 1509273288180, 600686976527820, 239071907384784240, 95150018452167599760, 37869468272055319920300, 15071953222259565160679700, 5998599512991034878630600360, 2387427534217209622129818263640, 950190160018936438572789038328420

Offset: 1

Gene Ward Smith, Oct 02 2006

The partial sums of the series 10/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(11), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [3;3,6,3], [3;3,6,3,6,3], [3;3,6,3,6,3,6,3] and so forth.

G. C. Greubel, Table of n, a(n) for n = 1..380
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A010468, A041014, A041015.

Cf. A123478, A123479, A123480, A029549.

CoefficientList[Series[-60*x/((x - 1)*(x^2 - 398*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)

Vec(-60*x/((x-1)*(x^2-398*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 399*a(n+2) - 399*a(n+1) + a(n).
a(n) = -5/33 + (5/66 + 1/44*11^(1/2))*(199 + 60*11^(1/2))^n + (5/66 - 1/44*11^(1/2))*(199 - 60*11^(1/2))^n.
G.f.: -60*x / ((x-1)*(x^2-398*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

cp's OEIS Frontend

A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Haskell

Macsyma

Magma

Maple

Mathematica

PARI

Sage

Scala

Formula

Extensions

A076139 Triangular numbers that are one-third of another triangular number: T(m) such that 3*T(m) = T(k) for some k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Sage

Formula

Extensions

A076140 Triangular numbers T(k) that are three times another triangular number: T(k) such that T(k) = 3*T(m) for some m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A217855 Numbers m such that 16*m*(3*m+1)+1 is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A123479 Coefficients of series giving the best rational approximations to sqrt(6).

Views

Author

Keywords

Comments

Links

A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

A217855 Numbers m such that 16m(3*m+1)+1 is a square.