A045899 -id:A045899 - 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

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A245031 Numbers m such that 3m+1 and 8m+1 are both squares.

Original entry on oeis.org

0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600

Offset: 1

Bruno Berselli, Jul 15 2014

Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=1: A006454;
k=3: this sequence;
k=4: A029549;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
k=6: A200999;
k=7: A157879.
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=1: A045899;
k=2: A045502;
k=4: A059989;
k=5: A159683;
k=6: 8*A029546;
k=7: A160695;
k=8: this sequence.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000217.

Cf. A006454, A029546, A029549, A045502, A045899, A059989, A157879, A159683, A160695, A200999.

I:=[0,1,21,120,2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];

LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]

a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);

a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a

G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1):   40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).

Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016

A046184 Indices of octagonal numbers which are also squares.

Original entry on oeis.org

1, 9, 121, 1681, 23409, 326041, 4541161, 63250209, 880961761, 12270214441, 170902040409, 2380358351281, 33154114877521, 461777249934009, 6431727384198601, 89582406128846401, 1247721958419651009, 17378525011746267721, 242051628206028097081

Offset: 1

Eric W. Weisstein

The equation a(t)*(3*a(t)-2) = m*m is equivalent to the Pell equation (3*a(t)-1)*(3*a(t)-1) - 3*m*m = 1. - Paul Weisenhorn, May 12 2009
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 16 2011
Also numbers n such that the octagonal number N(n) is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Also nonnegative integers y in the solutions to 2*x^2 - 6*y^2 + 4*x + 4*y + 2 + 2 = 0, the corresponding values of x being A251963. - Colin Barker, Dec 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Octagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A028230, A036428, A251963.

Cf. A006253.

I:=[1, 9, 121]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Nov 17 2011

LinearRecurrence[ {15, -15, 1}, {1, 9, 121}, 17 ] (* Ant King, Nov 16 2011 *)
CoefficientList[Series[x (1-6x+x^2)/((1-x)(1-14x+x^2)),{x,0,30}],x] (* Harvey P. Dale, Sep 01 2021 *)

Vec(x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)) + O(x^100)) \\ Colin Barker, Dec 11 2014

{n: A000567(n) in A000290}.
Nearest integer to (1/6) * (2+sqrt(3))^(2n-1). - Ralf Stephan, Feb 24 2004
a(n) = A045899(n-1) + 1 = A051047(n+1) + 1 = A003697(2n-2). - N. J. A. Sloane, Jun 12 2004
a(n) = A001835(n)^2. - Lekraj Beedassy, Jul 21 2006
From Paul Weisenhorn, May 12 2009: (Start)
With A=(2+sqrt(3))^2=7+4*sqrt(3) the equation x*x-3*m*m=1 has solutions
x(t) + sqrt(3)*m(t) = (2+sqrt(3))*A^t and the recurrences
x(t+2) = 14*x(t+1) - x(t) with  = 2, 26, 362, 5042
m(t+2) = 14*m(t+1) - m(t) with  = 1, 15, 209, 2911
a(t+2) = 14*a(t+1) - a(t) - 4 with  = 1, 9, 121, as above. (End)
From Ant King, Nov 15 2011: (Start)
a(n) = 14*a(n-1) - a(n-2) - 4.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (1/6)*( (2+sqrt(3))^(2n-1) + (2-sqrt(3))^(2n-1) + 2 ).
a(n) = ceiling( (1/6)*(2 + sqrt(3))^(2n-1) ).
a(n) = (1/6)*( (tan(5*Pi/12))^(2n-1) + (tan(Pi/12))^(2n-1) + 2 ).
a(n) = ceiling ( (1/6)*(tan(5*Pi/12))^(2n-1) ).
G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)
a(n) = A006253(2n-2). - Andrey Goder, Oct 17 2021

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

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

A067900 a(n) = 14*a(n-1) - a(n-2); a(0) = 0, a(1) = 8.

Original entry on oeis.org

0, 8, 112, 1560, 21728, 302632, 4215120, 58709048, 817711552, 11389252680, 158631825968, 2209456310872, 30773756526240, 428623135056488, 5969950134264592, 83150678744647800, 1158139552290804608, 16130803053326616712, 224673103194281829360, 3129292641666618994328

Offset: 0

Lekraj Beedassy, May 13 2003

Solves for y in x^2 - 3*y^2 = 4. Quadruples (a=b-y, b, c=b+y, d), with b=y^2 + 1 and d=x*y, where (x, y) solves x^2 - 3*y^2 = 4, satisfy the triangle relation (a^2 + b^2 + c^2 + d^2)^2 = 3*(a^4 + b^4 + c^4 + d^4). Thus d corresponds to the distance sum of the Fermat (or Torriccelli) point from its vertices in a triangle whose sides are in A.P. with middle side b and common difference y.

G. C. Greubel, Table of n, a(n) for n = 0..870
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A067902.
First differences of A045899.
Equals 8 * A007655(n+1).

m:=7;; a:=[0,8];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

m:=7; I:=[0,8]; [n le 2 select I[n] else 2*m*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 23 2019

a := proc(n) option remember: if n=0 then RETURN(0) fi: if n=1 then RETURN(8) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
seq( simplify(8*ChebyshevU(n-1, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

LinearRecurrence[{14,-1}, {0,8}, 17] (* Jean-François Alcover, Sep 19 2017 *)
8*ChebyshevU[Range[21] -2, 7] (* G. C. Greubel, Dec 23 2019 *)

vector(21, n, 8*polchebyshev(n-2, 2, 7) ) \\ G. C. Greubel, Dec 23 2019

[8*chebyshev_U(n-1,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: 8*x/(1-14*x+x^2). - Philippe Deléham, Nov 17 2008

E.g.f.: 2*exp(7*x)*sinh(4*sqrt(3)*x)/sqrt(3). - Stefano Spezia, Dec 12 2022

A120892 a(n)=3a(n-1)+3a(n-2)-a(n-3);a(0)=1,a(1)=0,a(2)=3. a(n)=4*{a(n-1)+(-1)^n}-a(n-2);a(0)=1,a(1)=0.

Original entry on oeis.org

1, 0, 3, 8, 33, 120, 451, 1680, 6273, 23408, 87363, 326040, 1216801, 4541160, 16947843, 63250208, 236052993, 880961760, 3287794051, 12270214440, 45793063713, 170902040408, 637815097923, 2380358351280, 8883618307201

Offset: 0

Lekraj Beedassy, Jul 13 2006

For n>1, short leg of primitive Pythagorean triangles having an angle nearing pi/3 with larger values of sides.[Complete triple (X,Y,Z),XA001353(n),Z=A120893(n), with recurrence relations Y(i+1)=2*{Y(i)-(-1)^i} + 3*a(i) ; Z(i+1)=2*{2*Z(i)-a(i-1)} - 3*(-1)^i] A120893(n)=2*a(n)-(-1)^n.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
J. P. Chabert, Right Triangle Applet (Hypotenuse & angles computation, given legs<350)
Index entries for linear recurrences with constant coefficients, signature (3, 3, -1).

LinearRecurrence[{3,3,-1},{1,0,3},30] (* Harvey P. Dale, Mar 05 2014 *)

a(n)=([0,1,0; 0,0,1; -1,3,3]^n*[1;0;3])[1,1] \\ Charles R Greathouse IV, Oct 19 2022

Union of A045899 and A011922.

O.g.f.: -(-1+3*x)/((x+1)*(x^2-4*x+1)). - R. J. Mathar, Nov 23 2007

Corrected and extended by T. D. Noe, Nov 07 2006

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

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.

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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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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A245031 Numbers m such that 3*m+1 and 8*m+1 are both squares.

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A046184 Indices of octagonal numbers which are also squares.

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

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A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A245031 Numbers m such that 3m+1 and 8m+1 are both squares.

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

A120892 a(n)=3a(n-1)+3a(n-2)-a(n-3);a(0)=1,a(1)=0,a(2)=3. a(n)=4*{a(n-1)+(-1)^n}-a(n-2);a(0)=1,a(1)=0.