A217855 -id:A217855 - cp's OEIS frontend

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

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

A123480 Coefficients of the series giving the best rational approximations to sqrt(3).

Original entry on oeis.org

4, 60, 840, 11704, 163020, 2270580, 31625104, 440480880, 6135107220, 85451020204, 1190179175640, 16577057438760, 230888624967004, 3215863692099300, 44791203064423200, 623860979209825504, 8689262505873133860, 121025814103014048540, 1685672134936323545704

Offset: 1

Gene Ward Smith, Sep 28 2006

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

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A123478, A123479, A029549, A123482. A045899, A076139, A076140, A217855.

CoefficientList[Series[-4*x/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)

my(x='x+O('x^50)); Vec(-4*x/((x-1)*(x^2-14*x+1))) \\ G. C. Greubel, Oct 13 2017

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
a(n) = -1/3 + (1/6 + 1/12*3^(1/2))*(7 + 4*3^(1/2))^n + (1/6 - 1/12*3^(1/2))*(7 - 4*3^(1/2))^n.
a(n) = 4*A076139(n) = 2*A217855(n) = 1/2*A045899(n) = 4/3*A076140(n). - Peter Bala, Dec 31 2012
G.f.: -4*x/((x-1)*(x^2-14*x+1)). - Colin Barker, Jan 20 2013
a(n) = A001353(n)*A001353(n+1). - Antonio Alberto Olivares, Apr 06 2020

A084232 RMS values associated with A084231.

Original entry on oeis.org

1, 195, 37829, 7338631, 1423656585, 276182038859, 53577891882061, 10393834843080975, 2016350381665827089, 391161580208327374291, 75883330210033844785365, 14720974899166357560986519, 2855793247108063332986599321, 554009168964065120241839281755

Offset: 0

Ignacio Larrosa Cañestro, May 20 2003

From Klaus Purath, Aug 20 2025: (Start)
Solutions to the Pell equation (7*b(n))^2 - 3*(4*a(n))^2 = 1. The corresponding b(n) are given by A302332.
For any two consecutive terms (x,y), x^2 - 194*x*y + y^2 - 196 = 0. By analogy to this, for three consecutive terms (x, y, z), y^2 - x*z - 196 = 0. (End)

			a(1)=195 because 195 = sqrt((Sum_{k=1..337}k^2)/337) and 337 = A084231(1).

Indranil Ghosh, Table of n, a(n) for n = 0..436
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (194,-1).

Cf. A084231, A217855, A302332.

LinearRecurrence[{194,-1},{1,195},20] (* Harvey P. Dale, Nov 10 2021 *)

a(n) = ((7+4*sqrt(3))^(2*n+1)-(7-4*sqrt(3))^(2*n+1))/(8*sqrt(3)). [simplified by Bruno Berselli, Oct 19 2012]
a(n) = floor(((7*sqrt(3) + 12)/24)*(56*sqrt(3) + 97)^n).
a(n+2) = 194*a(n+1) - a(n).
G.f.: (1-x)/(1-194*x+x^2). - Philippe Deléham, Nov 18 2008
a(n)^2 = (Sum_{i=1..A084231(n+1)}i^2)/A084231(n+1). - Bruno Berselli, Oct 17 2012

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

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

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

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

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A123480 Coefficients of the series giving the best rational approximations to sqrt(3).

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A084232 RMS values associated with A084231.

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A221075 Simple continued fraction expansion of an infinite product.

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