A053141 -id:A053141 - 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)

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

Original entry on oeis.org

0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895

Offset: 0

Charlie Marion, Feb 15 2012

For n > 1, a(n) = 98*a(n-1) - a(n-2) + 15. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1) = A014105(m) and for n > 1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m). Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

			2*0 = 3*0.
2*15 = 3*10.
2*1485 = 3*990.
2*145530 = 3*97020.

Colin Barker, Table of n, a(n) for n = 0..500
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

A309507 Number of ways the n-th triangular number T(n) = A000217(n) can be written as the difference of two positive triangular numbers.

Original entry on oeis.org

0, 1, 1, 1, 3, 3, 1, 2, 5, 3, 3, 3, 3, 7, 3, 1, 5, 5, 3, 7, 7, 3, 3, 5, 5, 7, 7, 3, 7, 7, 1, 3, 7, 7, 11, 5, 3, 7, 7, 3, 7, 7, 3, 11, 11, 3, 3, 5, 8, 11, 7, 3, 7, 15, 7, 7, 7, 3, 7, 7, 3, 11, 5, 3, 15, 7, 3, 7, 15, 7, 5, 5, 3, 11, 11, 7, 15, 7, 3, 9, 9, 3, 7

Offset: 1

Alois P. Heinz, Aug 05 2019

nonn

Equivalently, a(n) is the number of triples [n,k,m] with k>0 satisfying the Diophantine equation n*(n+1) + k*(k+1) - m*(m+1) = 0. Any such triple satisfies a triangle inequality, n+k > m. The n for which there is a triple [n,n,m] are listed in A053141. - Bradley Klee, Mar 01 2020; edited by N. J. A. Sloane, Mar 31 2020

			a(5) = 3: T(5) = T(6)-T(3) = T(8)-T(6) = T(15)-T(14).
a(7) = 1: T(7) = T(28)-T(27).
a(8) = 2: T(8) = T(13)-T(10) = T(36)-T(35).
a(9) = 5: T(9) = T(10)-T(4) = T(11)-T(6) = T(16)-T(13) = T(23)-T(21) = T(45)-T(44).
a(49) = 8: T(49) = T(52)-T(17) = T(61)-T(36) = T(94)-T(80) = T(127)-T(117) = T(178)-T(171) = T(247)-T(242) = T(613)-T(611) = T(1225)-T(1224).
The triples with n <= 16 are:
2, 2, 3
3, 5, 6
4, 9, 10
5, 3, 6
5, 6, 8
5, 14, 15
6, 5, 8
6, 9, 11
6, 20, 21
7, 27, 28
8, 10, 13
8, 35, 36
9, 4, 10
9, 6, 11
9, 13, 16
9, 21, 23
9, 44, 45
10, 8, 13
10, 26, 28
10, 54, 55
11, 14, 18
11, 20, 23
11, 65, 66
12, 17, 21
12, 24, 27
12, 77, 78
13, 9, 16
13, 44, 46
13, 90, 91
14, 5, 15
14, 11, 18
14, 14, 20
14, 18, 23
14, 33, 36
14, 51, 53
14, 104, 105
15, 21, 26
15, 38, 41
15, 119, 120
16, 135, 136. - _N. J. A. Sloane_, Mar 31 2020

Alois P. Heinz, Table of n, a(n) for n = 1..20000
J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000217, A001108, A046079 (the same for squares), A068194, A100821 (the same for primes for n>1), A309332.
See also A053141. The monotonic triples [n,k,m] with n <= k <= m are counted in A333529.
Cf. A092517, A063440.

with(numtheory): seq(tau(n*(n+1))-tau(n*(n+1)/2)-1, n=1..80); # Ridouane Oudra, Dec 08 2023

TriTriples[TNn_] := Sort[Select[{TNn, (TNn + TNn^2 - # - #^2)/(2 #),
      (TNn + TNn^2 - # + #^2)/(2 #)} & /@
    Complement[Divisors[TNn (TNn + 1)], {TNn}],
   And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]]
Length[TriTriples[#]] & /@ Range[100]
(* Bradley Klee, Mar 01 2020 *)

a(n) = 1 <=> n in { A068194 } \ { 1 }.
a(n) is even <=> n in { A001108 } \ { 0 }.
a(n) = number of odd divisors of n*(n+1) (or, equally, of T(n)) that are greater than 1. - N. J. A. Sloane, Apr 03 2020
a(n) = A092517(n) - A063440(n) - 1. - Ridouane Oudra, Dec 08 2023

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)

A098602 a(n) = A001652(n) * A046090(n).

Original entry on oeis.org

0, 12, 420, 14280, 485112, 16479540, 559819260, 19017375312, 646030941360, 21946034630940, 745519146510612, 25325704946729880, 860328449042305320, 29225841562491651012, 992818284675673829100, 33726595837410418538400, 1145711440187278556476512

Offset: 0

Charlie Marion, Oct 26 2004

From Ron Knott, Nov 25 2013: (Start)
a(n) = 2*r*(r+1) which is also of form s(s+1) where the s is in A053141.
a(n) is an oblong number (A002378) which is twice another oblong number. (End)
2*a(n)+1 and 4*a(n)+1 are both square. - Paul Cleary, Jun 23 2014

			a(1) = 12 = 2(2*3) = 3*4, a(2) = 420 = 2(14*15) = 20*21.

Reinhard Zumkeller, Table of n, a(n) for n = 0..255
Nikola Adžaga, Andrej Dujella, Dijana Kreso, Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002378, A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(12*x/((1-x)*(x^2-34*x+1)))); // G. C. Greubel, Jul 15 2018

2*Table[ Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Ray Chandler, Nov 10 2004, copied incorrect program from A029549, revised Jul 09 2015 *)
RecurrenceTable[{a[n+3] == 35 a[n+2] - 35 a[n+1] + a[n], a[1] == 0, a[2] == 12, a[3] == 420}, a, {n, 1, 10}] (* Ron Knott, Nov 25 2013 *)
LinearRecurrence[{35, -35, 1}, {0, 12, 420}, 25] (* T. D. Noe, Nov 25 2013 *)
Table[(LucasL[4*n+2, 2] - 6)/16, {n,0,30}] (* G. C. Greubel, Jul 15 2018 *)

concat(0, Vec(12*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

{a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(4*n+2) - 6)/16), ", ")) \\ G. C. Greubel, Jul 15 2018

a(n) = 2*A029549(n) = 2*A001109(n)*A001109(n+1).
a(n) = (A001653(n)^2 - 1)/2.
a(n) = A053141(n)^2 + A011900(n)^2 - 1.
For n>0, a(n) = A053141(2n) - 2*A001109(n-1)^2.
For n>0, a(n) = 3*(A001542(n)^2 - A001542(n-1)^2).
For n>0, a(n) = A053141(2n-1) + 2*(A001653(2n-1) - A001109(n-1)^2).
a(n+1) + a(n) = 3*A001542(n+1)^2.
a(n+1) - a(n) = A001542(2*n).
a(n+1)*a(n) = 4*(A001109(n)^4 - A001109(n)^2) = 4*A001110(n)*(A001110(n) - 1).
From Ron Knott, Nov 25 2013: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: 12*x / ((1-x)*(x^2-34*x+1)). (End)
a(n) = (-6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n)+(3+2*sqrt(2))*(17+12*sqrt(2))^n)/16. -  Colin Barker, Mar 02 2016

More terms from Ray Chandler, Nov 10 2004

Corrected by Bill Lam (bill_lam(AT)myrealbox.com), Feb 27 2006

A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.

Original entry on oeis.org

0, 55, 26565, 12804330, 6171660550, 2974727580825, 1433812522297155, 691094661019647940, 333106192798948009980, 160556493834431921162475, 77387896922003387052303025, 37300805759911798127288895630

Offset: 0

Charlie Marion, Dec 20 2011

			6*0 = 5*0;
6*55 = 5*66;
6*26565 = 5*31878;
6*12804330 = 5*15365196.

Vincenzo Librandi, Table of n, a(n) for n = 0..229
Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

I:=[0, 55, 26565]; [n le 3 select I[n] else 483*Self(n-1)-483*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Dec 22 2011

LinearRecurrence[{483,-483,1},{0,55,26565},30] (* Vincenzo Librandi, Dec 22 2011 *)

makelist(expand(((11-2*sqrt(30))^(2*n+1)+(11+2*sqrt(30))^(2*n+1)-22)/192), n, 0, 11); /* Bruno Berselli, Dec 21 2011 */

concat(0,Vec(55/(1-x)/(1-482*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 23 2011

For n > 1, a(n) = 482*a(n-1) - a(n-2) + 55. See A200993 for generalization.
From Bruno Berselli, Dec 21 2011: (Start)
G.f.: 55*x/((1-x)*(1-482*x+x^2)).
a(n) = a(-n-1) = 483*a(n-1)-483*a(n-2)+a(n-3).
a(n) = ((11-2*r)^(2*n+1)+(11+2*r)^(2*n+1)-22)/192, where r=sqrt(30). (End)

a(11) corrected by Bruno Berselli, Dec 21 2011

a(6) corrected by Vincenzo Librandi, Dec 22 2011

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

A221073 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 4, 1, 8, 1, 32, 1, 56, 1, 196, 1, 336, 1, 1152, 1, 1968, 1, 6724, 1, 11480, 1, 39200, 1, 66920, 1, 228484, 1, 390048, 1, 1331712, 1, 2273376, 1, 7761796, 1, 13250216, 1, 45239072, 1, 77227928, 1, 263672644, 1, 450117360, 1, 1536796800, 1, 2623476240, 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 = 2. For other cases see A221074 (m = 3), A221075 (m = 4) and A221076 (m = 5).

If we denote the present sequence by [2; 4, 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-sqrt(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+3)]/[1 - sqrt(2)*{(sqrt(2)-1)^(2*k+1)}^(4*n+1)]. An example is given below.

			Product {n >= 0} {1 - sqrt(2)*(sqrt(2) - 1)^(4*n+3)}/{1 - sqrt(2)*(sqrt(2) - 1)^(4*n+1)} = 2.20409 39255 78752 05766 ...
= 2 + 1/(4 + 1/(1 + 1/(8 + 1/(1 + 1/(32 + 1/(1 + 1/(56 + ...))))))).
We have (sqrt(2) - 1)^3 = 5*sqrt(2) - 7 so product {n >= 0} {1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+3)}/{1 - sqrt(2)*(5*sqrt(2) - 7)^(4*n+1)} = 1.11117 34981 94843 98511 ... = 1 + 1/(8 + 1/(1 + 1/(196 + 1/(1 + 1/(1968 + 1/(1 + 1/(39200 + ...))))))).

G. C. Greubel, Table of n, a(n) for n = 0..1000
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,6,0,-6,0,-1,0,1).

Cf. A001108, A053141, A174500, A221074 (m = 3), A221075 (m = 4), A221076 (m = 5).

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)))); // G. C. Greubel, Jul 15 2018

NProduct[( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 3) - 1)/( Sqrt[2]*(Sqrt[2] - 1)^(4*n + 1) - 1), {n, 0, Infinity}, WorkingPrecision -> 200] // ContinuedFraction[#, 37] & (* Jean-François Alcover, Mar 06 2013 *)
Join[{2},LinearRecurrence[{0,1,0,6,0,-6,0,-1,0,1},{4,1,8,1,32,1,56,1,196,1},60]] (* Harvey P. Dale, Feb 16 2014 *)

x='x+O('x^30); Vec((x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2)/((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1))) \\ G. C. Greubel, Jul 15 2018

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(2) + 1)^(2*n) + (sqrt(2) - 1)^(2*n) - 2;
a(4*n - 1) = 1/sqrt(2)*{(sqrt(2) + 1)^(2*n + 1) + (sqrt(2) - 1)^(2*n + 1)} - 2.
a(4*n - 3) = 4*A001108(n); a(4*n - 1) = 4*A053141(n).
O.g.f.: 2 + x^2/(1 - x^2) + 4*x*(1 + x^2)^2/(1 - 7*x^4 + 7*x^8 - x^12) = 2 + 4*x + x^2 + 8*x^3 + x^4 + 32*x^5 + ....
O.g.f.: (x^10-2*x^8-6*x^6+12*x^4-4*x^3+x^2-4*x-2) / ((x-1)*(x+1)*(x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Jan 10 2014

More terms from Harvey P. Dale, Feb 16 2014

A222390 Nonnegative integers m such that 10m(m+1)+1 is a square.

Original entry on oeis.org

0, 3, 15, 132, 588, 5031, 22347, 191064, 848616, 7255419, 32225079, 275514876, 1223704404, 10462309887, 46468542291, 397292260848, 1764580902672, 15086643602355, 67007605759263, 572895164628660, 2544524437949340, 21754929612286743, 96624921036315675

Offset: 1

Bruno Berselli, Feb 18 2013

a(n+1)/a(n) tends alternately to (7+2*sqrt(10))/3 and (13+4*sqrt(10))/3; a(n+2)/a(n) tends to A176398^2.

Subsequence of A014601.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. nonnegative integers m such that k*m*(m+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), this sequence (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), A222393 (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].

Cf. A221875.

m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(3*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2))));

I:=[0,3,15,132,588]; [n le 5 select I[n] else Self(n-1) +38*Self(n-2)-38*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013

LinearRecurrence[{1, 38, -38, -1, 1}, {0, 3, 15, 132, 588}, 23]
CoefficientList[Series[3 x (1 + 4 x + x^2)/((1 - x) (1 - 6 x - x^2) (1 + 6 x - x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(-1/2+((5+(-1)^n*sqrt(10))*(3-sqrt(10))^(2*floor(n/2))+(5-(-1)^n*sqrt(10))*(3+sqrt(10))^(2*floor(n/2)))/20), n, 1, 23);

x='x+O('x^30); concat([0], Vec(3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)))) \\ G. C. Greubel, Jul 15 2018

G.f.: 3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)).
a(n) = a(-n+1) = a(n-1)+38*a(n-2)-38*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((5+t*(-1)^n)*(3-t)^(2*floor(n/2))+(5-t*(-1)^n)*(3+t)^(2*floor(n/2)))/20, where t=sqrt(10).
2*a(n)+1 = A221875(n).

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

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A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2*T(m) = 3*T(k) for some k.

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A309507 Number of ways the n-th triangular number T(n) = A000217(n) can be written as the difference of two positive 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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A336626 Triangular numbers that are eight times another triangular number.

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A098602 a(n) = A001652(n) * A046090(n).

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Extensions

A200994 Triangular numbers, T(m), that are three-halves of another triangular number; T(m) such that 2T(m) = 3T(k) for some k.

A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.

A222390 Nonnegative integers m such that 10m(m+1)+1 is a square.