A077290 -id:A077290 - cp's OEIS frontend

A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

0, 1, 3, 14, 34, 143, 341, 1420, 3380, 14061, 33463, 139194, 331254, 1377883, 3279081, 13639640, 32459560, 135018521, 321316523, 1336545574, 3180705674, 13230437223, 31485740221, 130967826660, 311676696540, 1296447829381, 3085281225183, 12833510467154

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - Gerald McGarvey, Apr 02 2005

			a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 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 (1,10,-10,-1,1).

The k values are in A077291
Cf. A077289, A077290, A077291.
Cf. A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // G. C. Greubel, Jul 15 2018

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - Vladimir Pletser, Jul 24 2020

CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}],x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ G. C. Greubel, Jul 15 2018

Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - Vladimir Pletser, Feb 29 2020
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
2*a(n) + 1 = A080806(n+1). - R. J. Mathar, Oct 01 2021

A077289 Triangular numbers that are 1/6 of another triangular number.

Original entry on oeis.org

0, 1, 6, 105, 595, 10296, 58311, 1008910, 5713890, 98862891, 559902916, 9687554415, 54864771885, 949281469786, 5376187741821, 93019896484620, 526811533926580, 9115000574022981, 51622154137063026, 893177036357767525, 5058444293898249975, 87522234562487194476

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

The triangular numbers multiplied by 6 are in A077290.

			b(3)=14 so a(3) = 14*15/2 = 105, etc.

Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A077288, A077290, A077291.

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 98*a(n-2)-a(n-4)+7}, a(n), remember); map(f, [`$`(0 .. 1000)])[]; # Vladimir Pletser, Feb 19 2021

tr6Q[n_]:= IntegerQ[1/2 (Sqrt[1+48n]-1)]; Select[Accumulate[ Range[0,1380000]],tr6Q]  (* Harvey P. Dale, Apr 21 2011 *)

T(n)=n*(n+1)\2;
istriang(n)=issquare(8*n+1);
for(n=0, 10^10, t=T(n); if ( t%6==0 && istriang(t\6), print1(t\6, ", ") ) );
\\ Joerg Arndt, Jul 03 2013

concat(0, Vec(-x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Let b(n) be A077288. Then a(n)=b(n)*(b(n)+1)/2.
G.f.: -x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)). - Colin Barker, Jul 02 2013
a(n) = 98*a(n-2) - a(n-4) + 7. - Vladimir Pletser, Feb 19 2021
96*a(n) = 9*A072256(n+1) -2*(-1)^n*A054320(n) -7. - R. J. Mathar, Oct 01 2021

A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k.

Original entry on oeis.org

0, 3, 8, 35, 84, 351, 836, 3479, 8280, 34443, 81968, 340955, 811404, 3375111, 8032076, 33410159, 79509360, 330726483, 787061528, 3273854675, 7791105924, 32407820271, 77123997716, 320804348039, 763448871240, 3175635660123, 7557364714688, 31435552253195

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

The corresponding m are given in A077288.

Numbers x such that (2*x^2 + 2*x + 3)/3 = y^2. The corresponding y are given by A080806. - Klaus Purath, Jul 30 2025

			b(3)=630 so a(3) = (-1 + sqrt(8*630 + 1))/2 = (-1 + sqrt(5041))/2 = (71 - 1)/2 = 35.

Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 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 (1,10,-10,-1,1).

Cf. A077288, A077289, A077290, A080806.

f := gfun:-rectoproc({a(-2) = -4, a(-1) = -1, a(0) = 0, a(1) = 3, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 26 2020

LinearRecurrence[{1,10,-10,-1,1},{0,3,8,35,84},30] (* Harvey P. Dale, Oct 11 2019 *)

concat(0, Vec(x*(x^3+3*x^2-5*x-3)/((x-1)*(x^4-10*x^2+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Let b(n) be A077290. Then a(n) = (-1 + sqrt(8*b(n) + 1))/2.
G.f.: x*(x^3+3*x^2-5*x-3) / ((x-1)*(x^4-10*x^2+1)). - Colin Barker, Mar 09 2012
From Vladimir Pletser, Jul 26 2020: (Start)
a(n) = 10*a(n-2) - a(n-4) + 4 with a(-2)=-4, a(-1)=-1, a(0)=0, a(1)=3.
Can be defined for negative n by setting a(-n) = - a(n-1) - 1 for all n in Z.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). (End)

A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

Original entry on oeis.org

0, 15, 66, 17391, 76245, 20069280, 87986745, 23159931810, 101536627566, 26726541239541, 117173180224500, 30842405430498585, 135217748442445515, 35592109140254127630, 156041164529401899891, 41073263105447832786516, 180071368649181350028780, 47398510031577658781511915

Offset: 0

Vladimir Pletser, Aug 07 2020

The triangular numbers T(t) that are one-eighth of other triangular numbers T(u) : T(t)=T(u)/8. The t's are in A336623, the T(u)'s are in A336626 and the u's are in A336625.

Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(1)= 15 is a term because it is triangular and 8*15 = 120 is also triangular.
a(2) = 1154*a(0) - a(-2) + 81 = 0 - 15 + 81 = 66;
a(3) = 1154*a(1) - a(-1) + 81 = 1154*15 - 0 + 81 = 17391, etc.

Vladimir Pletser, Table of n, a(n) for n = 0..650
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 (1,1154,-1154,-1,1).

Subsequence of A000217.
Cf. A336623, A336626, A336625.
Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400.

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 81, a(1) = 15, a(0) = 0, a(-1) = 0, a(-2) = 15}, a(n), remember): map(f, [$ (0 .. 40)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 15, 66, 17391, 76245}, 18] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(4*n + 2)*(11 - 6*(-1)^n*Sqrt[2]) + (Sqrt[2] - 1)^(4*n + 2)*(11 + 6*(-1)^n*Sqrt[2]) - 18)/256, {n, 0, 17}]] (* Vaclav Kotesovec, Sep 08 2020 *)
Select[Accumulate[Range[0, 10^6]]/8, OddQ[Sqrt[8 # + 1]] &] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Jan 15 2024 *)

concat(0, Vec(3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)) + O(x^40))) \\ Colin Barker, Aug 08 2020

a(n) = 1154*a(n-2) - a(n-4) + 81, for n>=2 with a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(2)=66, a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A336623(n).
G.f.: 3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((sqrt(2) + 1)^(4*n + 2) * (11 - 6*(-1)^n*sqrt(2)) + (sqrt(2) - 1)^(4*n + 2) * (11 + 6*(-1)^n*sqrt(2)) - 18)/256. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11 - 6*sqrt(2))*(1 + sqrt(2))^(4n + 2) + (11 + 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for even n.
a(n) = ((11 + 6*sqrt(2))*(1 + sqrt(2) )^(4n + 2) + (11 - 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for odd n. (End)
128*a(n) = -9+33*A077420(n)-24*(-1)^n*A046176(n+1). - R. J. Mathar, May 05 2023

A336625 Indices of triangular numbers that are eight times other triangular numbers.

Original entry on oeis.org

0, 15, 32, 527, 1104, 17919, 37520, 608735, 1274592, 20679087, 43298624, 702480239, 1470878640, 23863649055, 49966575152, 810661587647, 1697392676544, 27538630330959, 57661384427360, 935502769664975, 1958789677853712, 31779555538278207, 66541187662598864, 1079569385531794079, 2260441590850507680

Offset: 1

Vladimir Pletser, Aug 13 2020

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 8*(b(n)^2 + b(n)) or T(a(n)) = 8*T(b(n)) where T(x) is the triangular number of x. The T(a)'s are in A336626, the T(b)'s are in A336624 and the b's are in A336623.

Can be defined for negative n by setting a(-n) = -a(n+1) - 1 for all n in Z.

			a(3) = 34*a(1) - a(-1) + 16 = 0 - (-16) + 16 = 32,
a(4) = 34*a(2) - a(0) + 16 = 34*15 - (-1) + 16 = 527, etc.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
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, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A336623, A336624, A336626, A166477 (at n=8).

Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.

f := gfun:-rectoproc({a(n) = 34*a(n - 2) - a(n - 4) + 16, a(2) = 15, a(1) = 0, a(0) = -1, a(-1) = -16}, a(n), remember); map(f, [$ (0 .. 1000)]); #

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 15, 32, 527, 1104, 17919}, 29] (* Amiram Eldar, Aug 18 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(2*n + 1) * (3 - Sqrt[2]*(-1)^n) - (Sqrt[2] - 1)^(2*n + 1) * (3 + Sqrt[2]*(-1)^n) - 2)/4, {n, 0, 20}]] (* Vaclav Kotesovec, Sep 08 2020 *)

concat(0, Vec(x*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^22))) \\ Colin Barker, Aug 14 2020

a(n) = 34*a(n-2) - a(n-4) + 16, for n>=2 with a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=32, a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = (-1 + sqrt(8*b(n) + 1))/2, where b(n) is A336626(n).
G.f.: x^2*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)). - Colin Barker, Aug 14 2020
a(n) = ((sqrt(2) + 1)^(2*n+1) * (3 - sqrt(2)*(-1)^n) - (sqrt(2) - 1)^(2*n+1) * (3 + sqrt(2)*(-1)^n) - 2)/4. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((3 - sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 + sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for even n.
a(n) = ((3 + sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 - sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for odd n. (End)

A336623 First member of the Diophantine pair (m, k) that satisfies 8*(m^2 + m) = k^2 + k; a(n) = m.

Original entry on oeis.org

0, 5, 11, 186, 390, 6335, 13265, 215220, 450636, 7311161, 15308375, 248364270, 520034130, 8437074035, 17665852061, 286612152936, 600118935960, 9736376125805, 20386377970595, 330750176124450, 692536732064286, 11235769612105511, 23525862512215145, 381685416635462940

Offset: 0

Vladimir Pletser, Aug 07 2020

The indices of triangular numbers that are one-eighth of other triangular numbers [m of T(m) such that T(m)=T(k)/8]. The T(m)'s are in A336624, the T(k)'s are in A336626 and the k's are in A336625.
Also, nonnegative m such that 32*m^2 + 32*m + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(2) = 34 a(0) - a(-2)+16=0 -5 +16 = 11 ; a(3) = 34 a(1) - a(-1)+16 = 34*5 -0 +16 = 186, etc.

Vladimir Pletser, Table of n, a(n) for n = 0..1000
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 (1,34,-34,-1,1).

Cf. A336624, A336626, A336625.

Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289 , A077290, A077398, A077401, A077399, A077400, A000217.

f := gfun:-rectoproc({a(n) = 34*a(n - 2) - a(n - 4) + 16, a(1) = 5, a(0) = 0, a(-1) = 0,  a(-2) = 5}, a(n), remember); map(f, [$ (0 .. 50)]); #

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 5, 11, 186, 390}, 24] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((3*Sqrt[2] - 2*(-1)^n)*(1 + Sqrt[2])^(2*n + 1) + (3*Sqrt[2] + 2*(-1)^n)*(Sqrt[2] - 1)^(2*n + 1) - 8)/16, {n, 0, 20}]] (* Vaclav Kotesovec, Sep 08 2020 *)

concat(0, Vec(x*(5 + 6*x + 5*x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^22))) \\ Colin Barker, Aug 08 2020

a(n) = 34 a(n-2) - a(n-4) + 16 for n>=2, with a(1)=5, a(0)=0, a(-1)=0, a(-2)=5.
a(n) = a(n-1) + 34 a(n-2) - 34 a(n-3) - a(n-4)+ a(n-5) for n>=3 with a(2)=11, a(1)=5, a(0)=0, a(-1)=0, a(-2)=5.
a(n) = (C+((-1)^n)*D)*A^n + (E+((-1)^n)*F)*B^n -1/2 with A = (sqrt(2) + 1)^2 ; B = (sqrt(2) - 1)^2 ; C = 3*(2 + sqrt(2))/16 ; D = -(1 + sqrt(2))/8 ; E = 3*(2 - sqrt(2))/16 ; F = (sqrt(2) - 1)/8 and n>=0.
a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A336624(n).
G.f.: x*(5 + 6*x + 5*x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((3*sqrt(2) - 2*(-1)^n) * (1 + sqrt(2))^(2*n + 1) + (3*sqrt(2) + 2*(-1)^n) * (sqrt(2) - 1)^(2*n + 1) - 8)/16. - Vaclav Kotesovec, Sep 08 2020
Comment from _Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((4 + sqrt(2))(1 + sqrt(2))^(2n) + (4 - sqrt(2))(1 - sqrt(2))^(2n))/16  - 1/2 for even n.
a(n) = ((8 + 5 sqrt(2))(1 + sqrt(2))^(2n) + (8 - 5 sqrt(2))(1 - sqrt(2))^(2n))/16  - 1/2 for odd n. (End)

A336626 Triangular numbers that are eight times another triangular number.

Original entry on oeis.org

0, 120, 528, 139128, 609960, 160554240, 703893960, 185279454480, 812293020528, 213812329916328, 937385441796000, 246739243443988680, 1081741987539564120, 284736873122033021040, 1248329316235215199128, 328586104843582662292128, 1440570949193450800230240, 379188080252621270252095320

Offset: 1

Vladimir Pletser, Oct 04 2020

The triangular numbers T(t) that are eight times another triangular number T(u) : T(t) = 8*T(u). The t's are in A336625, the T(u)'s are in A336624 and the u's are in A336623.

Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(2) = 120 is a term because it is triangular and 120/8 = 15 is also triangular.
a(3) = 1154*a(1) - a(-1) + 648 = 0 - 120 + 648 = 528;
a(4) = 1154*a(2) - a(0) + 648 = 1154*120 - 0 + 648 = 139128, etc.
.
From _Peter Luschny_, Oct 19 2020: (Start)
Related sequences in context, as computed by the Julia function:
n   [A336623, A336624,        A336625,  A336626        ]
[0] [0,       0,              0,        0              ]
[1] [5,       15,             15,       120            ]
[2] [11,      66,             32,       528            ]
[3] [186,     17391,          527,      139128         ]
[4] [390,     76245,          1104,     609960         ]
[5] [6335,    20069280,       17919,    160554240      ]
[6] [13265,   87986745,       37520,    703893960      ]
[7] [215220,  23159931810,    608735,   185279454480   ]
[8] [450636,  101536627566,   1274592,  812293020528   ]
[9] [7311161, 26726541239541, 20679087, 213812329916328] (End)

Vladimir Pletser, Table of n, a(n) for n = 1..653
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.
V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Subsequence of A000217.
Cf. A336623, A336624, A336625.
Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077260, A077261, A077262, A077288, A077289, A077290, A077291, A077398, A077399, A077400, A077401.

function omnibus()
    println("[A336623, A336624, A336625, A336626]")
    println([0, 0, 0, 0])
    t, h = 1, 1
    for n in 1:999999999
        d, r = divrem(t, 8)
        if r == 0
            d2 = 2*d
            s = isqrt(d2)
            d2 == s * (s + 1) && println([s, d, n, t])
        end
        t, h = t + h + 1, h + 1
    end
end
omnibus() # Peter Luschny, Oct 19 2020

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 648, a(2) = 120, a(1) = 0, a(0) = 0, a(-1) = 120}, a(n), remember); map(f, [$ (1 .. 1000)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 120, 528, 139128, 609960}, 18]

a(n) = 8*A336624(n).
a(n) = 1154*a(n-2) - a(n-4) + 648, for n>=2 with a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=528, a(2)=120, a(1)=0, a(0)=0, a(-1)=120.
a(n) = ((10*sqrt(2))/17 + 15/17)*(17 + 12*sqrt(2))^n + (-(10*sqrt(2))/17 + 15/17)*(17 - 12*sqrt(2))^n + (-15/17 - (45*sqrt(2))/68)*(-17 - 12*sqrt(2))^n + (-15/17 + (45*sqrt(2))/68)*(-17 + 12*sqrt(2))^n - 27*(-4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 + 12*sqrt(2)))^n/(1088*(-17 + 12*sqrt(2))) - 27*(4 + 3*sqrt(2))*sqrt(2)*(-1/(-17 - 12*sqrt(2)))^n/(1088*(-17 - 12*sqrt(2))) - 9/16 - 9*(-3 + 2*sqrt(2))*sqrt(2)*(-1/(17 - 12*sqrt(2)))^n/(272*(17 - 12*sqrt(2))) - 9*(3 + 2*sqrt(2))*sqrt(2)*(-1/(17 + 12*sqrt(2)))^n/(272*(17 + 12*sqrt(2))).
Let b(n) be A336625(n). Then a(n) = b(n)*(b(n)+1)/2.
G.f.: 24*x^2*(5 + 17*x + 5*x^2)/(1 - x - 1154*x^2 + 1154*x^3 + x^4 - x^5). - Stefano Spezia, Oct 05 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11*(1 + sqrt(2))^2 - (-1)^n*6*(4 + 3*sqrt(2)))*(1 + sqrt(2))^(4n) + (11*(1 - sqrt(2))^2 - (-1)^n*6*(4 - 3*sqrt(2)))*(1 - sqrt(2))^(4n))/32 - 9/16.
a(n) = ((1 + 2*sqrt(2))^2*(1 + sqrt(2))^(4n) + (1 - 2*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for even n.
a(n) = ((5 + 4*sqrt(2))^2*(1 + sqrt(2))^(4n) + (5 - 4*sqrt(2))^2*(1 - sqrt(2))^(4n))/32 - 9/16 for odd n. (End)

A341895 Indices of triangular numbers that are ten times other triangular numbers.

Original entry on oeis.org

0, 4, 20, 39, 175, 779, 1500, 6664, 29600, 56979, 253075, 1124039, 2163720, 9610204, 42683900, 82164399, 364934695, 1620864179, 3120083460, 13857908224, 61550154920, 118481007099, 526235577835, 2337285022799, 4499158186320, 19983094049524, 88755280711460, 170849530073079, 758831338304095, 3370363382012699

Offset: 1

Vladimir Pletser, Feb 23 2021

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 10*(b(n)^2 + b(n)) or T(a(n)) = 10*T(b(n)) where T(x) is the triangular number of x. The T(b)'s are in A068085 and the b's are in A341893.

Can be defined for negative n by setting a(-n) = -a(n+1) - 1 for all n in Z.

			a(2) = 4 is a term because its triangular number, T(a(2)) = 4*5 / 2 = 10 is ten times a triangular number.
a(4) = 38*a(1) - a(-2) + 18 = 38*0 - (-21) + 18 = 39, etc.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
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 (1,38,-38,-1,1).

Cf. A341893, A068085, A166477 (n=10).

Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.

f := gfun:-rectoproc({a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)]) ; #

Rest@ CoefficientList[Series[x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 30}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = 38*a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
G.f.: x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7). - Stefano Spezia, Feb 24 2021
a(n) = (A198943(n) + 1)/2 - 1. - Hugo Pfoertner, Feb 26 2021

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

Original entry on oeis.org

0, 1, 6, 12, 55, 246, 474, 2107, 9360, 18018, 80029, 355452, 684228, 3039013, 13497834, 25982664, 115402483, 512562258, 986657022, 4382255359, 19463867988, 37466984190, 166410301177, 739114421304, 1422758742216, 6319209189385, 28066884141582, 54027365220036, 239963538895471, 1065802482958830, 2051617119619170

Offset: 1

Vladimir Pletser, Feb 23 2021

The indices of triangular numbers that are one-tenth of other triangular numbers [t of T(t) such that T(t)=T(u)/10].
First member of the Diophantine pair (t, u) that satisfies 10*(t^2 + t) = u^2 + u; a(n) = t.
The T(t)'s are in A068085 and the u's are in A341895.
Also, nonnegative t such that 40*t^2 + 40*t + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(4) = 12 is a term because its triangular number, (12*13) / 2 = 78 is one-tenth of 780, the triangular number of 39.
a(4) = 38 a(1) - a(-2) +18 = 0 - 6 +18 = 12 ;
a(5) = 38 a(2) - a(-1) + 18 = 38*1 - 1 +18 = 55.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A068085, A341895.
Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.
Cf. A180003.

f := gfun:-rectoproc({a(-3) = 6, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 6, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ;

Rest@ CoefficientList[Series[(x^2*(1 + 5*x + 6*x^2 + 5*x^3 + x^4))/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 31}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A068085(n).
a(n) = 38 a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
G.f.: x^2*(1 + 4*x+x^2)*(1+x+x^2)/ ((1-x)*(1-38*x^3+x^6)). - Stefano Spezia, Feb 24 2021
a(n) = A180003(n) - 1. - Hugo Pfoertner, Feb 28 2021

A341894 For square n > 0, a(n) = 0; for nonsquare n > 0, a(n) is the rank r such that t(r) + t(r-1) = u(r) - u(r-1) - 1, where u(r) and t(r) are indices of some triangular numbers in the Diophantine relation T(u(r)) = n*T(t(r)).

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 2, 2, 0, 3, 2, 2, 4, 2, 2, 0, 2, 2, 3, 2, 4, 4, 2, 2, 0, 3, 2, 4, 4, 2, 4, 2, 2, 2, 2, 0, 2, 2, 2, 4, 4, 2, 4, 2, 4, 6, 2, 2, 0, 3, 3, 4, 4, 2, 4, 2, 4, 4, 2, 2, 8, 2, 2, 0, 2, 4, 4, 2, 4, 4, 4, 2, 6, 2, 2, 6, 4, 4, 2, 2, 0, 3, 2, 2, 8, 4, 2, 4, 4, 2, 6, 4, 4, 4, 2, 4, 4, 2, 2, 0, 2, 2, 4, 2, 4, 8, 2, 2, 8, 2

Offset: 1

Vladimir Pletser, Mar 06 2021

nonn

Let t(i) and u(j) be the indices of triangular numbers that satisfy the Diophantine relation T(u(j)) = n*T(t(i)) for some integers i and j. The number of solutions (t(i), u(j)) of T(u(j)) = n*T(t(i)) is 0 or 1 for square n, and an infinity for nonsquare n.
For square n, a(n) is arbitrarily set to 0.
For nonsquare n, a(n) is the index r in the sequence of t(i) and u(j) such that t(r) + t(r-1) = u(r) - u(r-1) - 1.
Alternatively, for nonsquare n, a(n) is the index r such that the ratio t(i)/t(i-r) is decreasing monotonically without jumps for increasing values of i.
Alternatively, for n > 4, a(n) is the index r such that the ratio t(r)/t(r-1) varies between (s+1)/(s-1) and (s+2)/s, with s = [sqrt(n)], where [x] = floor(x).
Alternatively, for nonsquare n, a(n) is the number of fundamental solutions (X_f, Y_f) of the generalized Pell equation X^2 - n*Y^2 = 1 - n providing odd solutions, i.e., with X_f odd and Y_f odd (or Y_f even if y_f is odd, where y_f is the fundamental solution of the associated simple Pell equation x^2 - n*y^2 = 1).

			The following table gives the first values of nonsquare n and a(n) and the sequences yielding the values of t, u, T(t) and T(u) such that T(u) = n*T(t).
n       2       3       5       6       7       8      10
a(n)    1       1       2       2       2       2       3
t    A053141 A061278 A077259 A077288 A077398 A336623  A341893*
u    A001652 A001571 A077262 A077291 A077401 A336625* A341895*
T(t) A075528 A076139 A077260 A077289 A077399 A336624  A068085*
T(u) A029549 A076140 A077261 A077290 A077400 A336626*   -
With a(n) = r, the definition t(r) + t(r-1) = u(r) - u(r-1) - 1 yields:
- For n = 2, a(n) = 1: A053141(1) + A053141(0) = A001652(1) - A001652(0) - 1, i.e., 2 + 0 = 3 + 0 - 1 = 2.
- For n = 5, a(n) = 2: A077259(2) + A077259(1) = A077262(2) - A077262(1) - 1, i.e., 6 + 2 = 14 - 5 - 1 = 8.
- For n = 10, a(n) = 3: A341893(3+1*) + A341893(2+1*) = A341895(3+1*) - A341895(2+1*) - 1, i.e., 12 + 6 = 39 - 20 - 1 = 18.
Note that for those sequences marked with an *, the first term 0 appears for n = 1, contrary to all the other sequences, where the first term 0 appears for n = 0; the numbering must therefore be adapted and 1 must be added to compensate for this shift in indices.
The monotonic decrease of t(i)/t(i-r) can be seen also as:
- For n = 2, a(n) = 1: for 1 <= i <= 6, A053141(i)/A053141(i-1) decreases monotonically from 7 to 5.829.
- For n = 5, a(n) = 2: for 3 <= i <= 8, A077259(i)/A077259(i-2) decreases monotonically from 22 to 17.948, while A077259(i)/A077259(i-1) takes values alternatively varying between 3 and 2.618 and between 7.333 and 6.855.
- For n = 10, a(n) = 3: for 4 <= i <= 10, A341893(i)/A341893(i-3) decreases monotonically from 55 to 38, while A077259(i) / A077259(i-1) takes values alternatively varying between 6 and 4.44 and between 2 and 1.925.
For n > 4, the relation (s+1)/(s-1) <=  t(r)/t(r-1) <= (s+2)/s, with s = [sqrt(n)], yields:
- For n = 5, a(n) = 2: A077259(2)/A077259(1) = 6/2 = 3, and s = [sqrt(5)] = 2, (s+1)/(s-1) = 3 and (s+2)/s = 2.
- For n = 10, a(n) = 3: A077259(3+1*)/A077259(2+1*) = 12/6 = 2, and s = [sqrt(10)] = 3, (s+1)/(s-1) = 2 and (s+2)/s = 5/3 = 1.667.
Finally, the number of fundamental solutions of the generalized Pell equation is as follows.
- For n = 2, X^2 - 2*Y^2 = -1 has a single fundamental solution, (X_f, Y_f) = (1, 1), and the rank a(n) is 1.
- For n = 5, X^2 - 5*Y^2 = -4 has two fundamental solutions, (X_f, Y_f) = (1, 1) and (-1, 1), and the rank a(n) is 2.
- For n = 10, X^2 - 10*Y^2 = -9 has three fundamental solutions, (X_f, Y_f) = (1, 1), (-1, 1), and (9, 3), and the rank a(n) is 3.

J. S. Chahal and H. D'Souza, "Some remarks on triangular numbers", in A.D. Pollington and W. Mean, eds., Number Theory with an Emphasis on the Markov Spectrum, Lecture Notes in Pure Math, Dekker, New York, 1993, 61-67.

Vladimir Pletser, Table of n, a(n) for n = 1..256
T. Breiteig, Quotients of triangular numbers, The Mathematical Gazette, 99, 2015, 243-255.
Keith Matthews, The Diophantine Equation x^2 - Dy^2 = N, D > 0, in integers, Expositiones Mathematicae, 18, 2000, 323-331.
Keith Matthews, Quadratic Diophantine equations BCMATH programs, 2020.
Vladimir Pletser, Searching for Multiple of Triangular Numbers being Triangular Numbers, ResearchGate, DOI: 10.13140/RG.2.2.35428.91527, 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv: 2102.13494 [math.NT], 2021.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv: 2101.00998 [math.NT], 2021.

Cf. A068085, A053141, A061278, A077259, A077288, A077398, A336623, A341893, A001652, A001571, A077262, A077291, A077401, A336625, A341895, A075528, A076139, A077260, A077289, A077399, A336624, A068085, A029549, A076140, A077261, A077290, A077400, A336626, A000217, A341895, A341896.

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A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

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A077289 Triangular numbers that are 1/6 of another triangular number.

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A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k.

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A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

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A336625 Indices of triangular numbers that are eight times other triangular numbers.

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A336623 First member of the Diophantine pair (m, k) that satisfies 8*(m^2 + m) = k^2 + k; a(n) = m.

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