A165518 -id:A165518 - 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

A075870 Numbers k such that 2*k^2 - 4 is a square.

Original entry on oeis.org

2, 10, 58, 338, 1970, 11482, 66922, 390050, 2273378, 13250218, 77227930, 450117362, 2623476242, 15290740090, 89120964298, 519435045698, 3027489309890, 17645500813642, 102845515571962, 599427592618130, 3493720040136818, 20362892648202778, 118683635849079850

Offset: 1

Gregory V. Richardson, Oct 16 2002

Lim_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2).
Also gives solutions to the equation x^2-2 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators = A075870, denominators = A002315. - Clark Kimberling, Aug 27 2008
Numbers n such that sqrt(floor(n^2/2 - 1)) is an integer. The integer square roots are given by A002315. - Richard R. Forberg, Aug 01 2013
a(n) are the integer square roots of m^2 + (m+2)^2. The values of m are given by A065113 (except for m = 0). The values of this expression are given by A165518. - Richard R. Forberg, Aug 15 2013
Values of x (or y) in the solutions to x^2 - 6*x*y + y^2 + 16 = 0. - Colin Barker, Feb 04 2014
Also integers k such that k^2 is equal to the sum of four consecutive triangular numbers. - Colin Barker, Dec 20 2014
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = (y-1)^2 + (y+1)^2. y-values are listed in A002315. Example: for x=58 and y=41, 57*58/2 + 58*59/2 = 40^2 + 42^2. - Bruno Berselli, Mar 19 2018

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=2, 2*2^2 - 4 = 8 - 4 = 4 = 2^2.
For k=10, 2*10^2 - 4 = 200 - 4 =  196 = 14^2.
For k=58, 2*58^2 - 4 = 6728 - 4 =  6724 = 82^2.
... (End)

A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Colin Barker, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A000217, A000290, A002315.

Twice A001653.

a:=[2,10];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

a:= proc(n) option remember: if n = 1 then 2 elif n = 2 then 10 elif  n >= 3 then 6*procname(n-1) - procname(n-2) fi; end: seq(a(n), n = 0..25); # Muniru A Asiru, Mar 19 2018

LinearRecurrence[{6,-1},{2,10},30] (* Harvey P. Dale, Sep 27 2018 *)

Vec(2*x*(1-x)/(1-6*x+x^2) + O(x^100)) \\ Colin Barker, Dec 20 2014

a(n) = 2 * A001653(n).
a(n) = (1/sqrt(2))*((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1)) = 6*a(n-1) - a(n-2).
G.f.: 2*x*(1-x)/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/2). - Paul Weisenhorn, Jun 11 2020
From Peter Bala, Aug 21 2022: (Start)
a(n) = 2*Pell(2*n-1).
1/a(n) - 1/a(n+1) = 1/(Pell(2*n) + 1/Pell(2*n)), where Pell(n) = A000129(n). (End)

More terms from Colin Barker, Dec 20 2014

A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

Original entry on oeis.org

6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192

Offset: 1

Robert G. Wilson v, Nov 12 2001

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

			T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001652, A002315, A003499 (first differences), A065651.

CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]
LinearRecurrence[{7,-7,1},{6,40,238},41] (* Harvey P. Dale, Dec 27 2011 *)

a(n)=-1+subst(poltchebi(abs(n+1))-poltchebi(abs(n)),x,3)/2

Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

a(n) = 2*A001652(n) = -1 + A002315(n).
a(n) - a(n-1) = A003499(n).
From Michael Somos, Apr 07 2003: (Start)
G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).
a(n) = 6*a(n-1) - a(n-2) + 4.
a(-1-n) = -a(n) - 2. (End)
a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011
a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012
a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016
From Klaus Purath, Sep 05 2021: (Start)
(a(n+1) - a(n) - a(n-1) + a(n-2))/8 = A005319(n), for n >= 3.
((a(n) - a(n-1))^2)/2 - 2 = A005319(n)^2 = 2*A132592(n), for n>= 2.
a(n) = A265278(2*n+1).
a(n) = A293004(2*n+1).
a(n) = A213667(2*n).
a(n) = Sum_{k=1..n} A003499(k). (End)

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.

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A075870 Numbers k such that 2*k^2 - 4 is a square.

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A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

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