A336624 -id:A336624 - cp's OEIS frontend

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

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

A068085 Numbers k such that k and 10*k are both triangular numbers.

Original entry on oeis.org

0, 1, 21, 78, 1540, 30381, 112575, 2220778, 43809480, 162333171, 3202360435, 63173239878, 234084320106, 4617801526591, 91095768094695, 337549427259780, 6658866598983886, 131360034419310411, 486746040024282753, 9602081017933237120, 189421078536877518066, 701887452165588470145

Offset: 1

Amarnath Murthy, Feb 18 2002

Let y=sqrt(8*k+1) and x=sqrt(80*k+1), which must be integers if k and 10*k are triangular. These quantities satisfy the Pell-like equation x^2 - 10*y^2 = -9. All solutions x+y*sqrt(10) are obtained from 1+sqrt(10), 9+3*sqrt(10) and 41+13*sqrt(10) by multiplying by powers of the fundamental unit 19+6*sqrt(10).
Conjecture: satisfies a linear recurrence having signature (1, 0, 1442, -1442, 0, -1, 1). - Harvey P. Dale, Sep 03 2020
This conjecture is true because of the connection between (generalized) Pell equations and continued fractions of quadratic irrationals. - Georg Fischer, Feb 23 2021
From Vladimir Pletser, Feb 26 2021: (Start)
The triangular numbers T(t) that are one-tenth of other triangular numbers T(u) : T(t)=T(u)/10. The t's are in A341893, and the u's are in A341895.
Can be defined for negative n by setting a(n) = a(1-n) for all n in Z. (End)

			21 and 210 are both triangular numbers.

Georg Fischer, 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 (1,0,1442,-1442,0,-1,1).

Cf. for k and m*k both triangular: A075528 (m=2), A076139 (m=3), 0 (m=4), A077260 (m=5), A077289 (m=6), A077399 (m=7), A336624 (m=8), 0 (m=9), this sequence (m=10).

f := gfun:-rectoproc({a(-3) = 21, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 21, a(n) = 1442*a(n-3)-a(n-6)+99}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ; # Vladimir Pletser, Feb 26 2021

a[0]=0; a[1]=1; a[2]=21; a[n_] := a[n]=(99+1442a[n-3]+57Sqrt[(1+8a[n-3])(1+80a[n-3])])/2

a(n) = (99 + 1442*a(n-3) + 57*sqrt((1 + 8*a(n-3))*(1 + 88*a(n-3))))/2.
G.f.: -x^2*(x^4+20*x^3+57*x^2+20*x+1) / ((x-1)*(x^6-1442*x^3+1)). - Colin Barker, Jun 24 2014
From _Vladimir Pletser, Feb 26 2021: (Start)
a(n) = 1442 *a(n-3) - a(n-6) + 99, for n > 3, with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.
a(n) = a(n - 1) + 1442 ( a(n - 3) - a(n - 4) ) - ( a(n - 6) - a(n - 7) ) for n >= 4 with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A341893(n). (End)

Edited by Dean Hickerson, Feb 20 2002

More terms from Georg Fischer, Feb 23 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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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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A336626 Triangular numbers that are eight times another triangular number.

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

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A068085 Numbers k such that k and 10*k are both triangular numbers.

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

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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)).

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