A001571 -id:A001571 - 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)

A102871 a(n) = a(n-3) - 5a(n-2) + 5a(n-1), a(0) = 1, a(1) = 3, a(2) = 10.

Original entry on oeis.org

1, 3, 10, 36, 133, 495, 1846, 6888, 25705, 95931, 358018, 1336140, 4986541, 18610023, 69453550, 259204176, 967363153, 3610248435, 13473630586, 50284273908, 187663465045, 700369586271, 2613814880038, 9754889933880, 36405744855481, 135868089488043, 507066613096690

Offset: 0

Creighton Dement, Mar 01 2005

A floretion-generated sequence resulting from a particular transform of the periodic sequence (-1,1).
Floretion Algebra Multiplication Program, FAMP Code: .5em[J* ]forseq[ .25( 'i + 'j + 'k + i' + j' + k' + 'ii' + 'jj' + 'kk' + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + e ) ], em[J]forseq = A001834, vesforseq = (1,-1,1,-1). ForType 1A. Identity used: em[J]forseq + em[J* ]forseq = vesforseq.
Also indices of the centered triangular numbers which are triangular numbers - Richard Choulet, Oct 09 2007
Place a(n) red and b(n) blue balls in an urn; draw 2 balls without replacement. Probability(2 red balls) = 3*Probability(2 blue balls); b(n)=A101265(n). - Paul Weisenhorn, Aug 02 2010

			For n=5, a(5)=495; b(5)=286; binomial(495,2) = 122265 = 3*binomial(286,2). - _Paul Weisenhorn_, Aug 02 2010

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075 (first differences), A001834, A082841, A101265.

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=4*a[n-1]-a[n-2]-1 od: seq(a[n], n=1..23); # Zerinvary Lajos, Mar 08 2008

LinearRecurrence[{5,-5,1},{1,3,10},30] (* Harvey P. Dale, Oct 04 2011 *)

2*a(n) - A001834(n) = (-1)^(n+1); a(n) = 4*a(n-1) - a(n) - 1;
G.f.: (2*x-1)/((x-1)*(x^2-4*x+1)).
Superseeker results: a(n+2) - 2a(n+1) + a(n) = A001834(n+1) (from this and the first relation involving A001834 it follows that 4a(n+1) - a(n+2) - a(n) = (-1)^n as well as the recurrence relation given for A001834 ); a(n+1) - a(n) = A001075(n+1); a(n+2) - a(n) = A082841(n+1).
a(j+3) - 3*a(j+2) - 3*a(j+1) + a(j) = -2 for all j.
a(n+1) = 2*a(n) - 1/2 + (1/2)*(12*a(n)^2 - 12*a(n) + 9)^(1/2). - Richard Choulet, Oct 09 2007
a(n) = (sqrt(12*b(n)*(b(n)-1) + 1) + 1)/2; b(n) = A101265(n). - Paul Weisenhorn, Aug 02 2010
a(n) = A001571(n) + 1. - Johannes Boot, Jun 17 2011
E.g.f.: (exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) + cosh(x) + sinh(x))/2. - Stefano Spezia, Sep 19 2023

A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

Original entry on oeis.org

-1, 1, 8, 34, 131, 493, 1844, 6886, 25703, 95929, 358016, 1336138, 4986539, 18610021, 69453548, 259204174, 967363151, 3610248433, 13473630584, 50284273906, 187663465043, 700369586269, 2613814880036, 9754889933878, 36405744855479, 135868089488041

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Apr 14 2003

Apart from the initial -1, these are the numbers k such that the triangular number k*(k + 1)/2 is the sum of three consecutive triangular numbers - see A129803. - Brian Nowell, Nov 03 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001571, A071954.

a:=[-1,1,8];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( -(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2)) )); // G. C. Greubel, Feb 25 2019

CoefficientList[Series[(-1+6x-2x^2)/((1-x)(1-4x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 15 2014 *)
LinearRecurrence[{5,-5,1}, {-1,1,8}, 30] (* G. C. Greubel, Feb 25 2019 *)

is(n)=ispolygonal(3/2*n*(n+1)+4,3) || n==-1 \\ Charles R Greathouse IV, Apr 14 2014

my(x='x+O('x^30)); Vec(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

a(n) = A001571(n) - 1. - N. J. A. Sloane, Nov 03 2009
G.f.: -(1 -6*x +2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = -3/2 + (1/12)*( (a -2*b +5)*a^n + (b -2*a +5)*b^n ), with a = 2 + sqrt(3), b = 2 - sqrt(3):.
a(n) = -3/2 + (3/4)*A003500(n) - (1/4)*A003500(n-1).
a(n) = (1/2)*(A001834(n) - 3).
E.g.f.: ((1 + sqrt(3))*exp((2 + sqrt(3))*x) + (1 - sqrt(3))*exp((2 - sqrt(3))*x) - 6*exp(x))/4. - Franck Maminirina Ramaharo, Nov 12 2018

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

A093446 Largest member of the n-th row of the triangular triangle (A093445).

Original entry on oeis.org

1, 3, 9, 18, 33, 54, 82, 120, 165, 225, 294, 378, 476, 588, 720, 865, 1035, 1221, 1430, 1662, 1914, 2197, 2499, 2835, 3195, 3585, 4008, 4456, 4947, 5463, 6021, 6612, 7239, 7910, 8610, 9366, 10153, 10989, 11868, 12788, 13764, 14775, 15850, 16965, 18135

Offset: 1

Amarnath Murthy, Apr 02 2004

nonn

The largest terms is near the middle term at about 42.265%. Lim a(n) -> 0.19245*n^3

			The row for n = 4 is (1+2+3+4), (5+6+7), (8+9), 10 => 10 18 17 10. The largest member is 18 hence a(4) = 18.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A093445, A001571.

a093446 = maximum . a093445_row  -- Reinhard Zumkeller, Oct 03 2012

T[n_] := n(n + 1)/2; TT[n_, k_] := T[k*n - T[k - 1]] - T[(k - 1)*n - T[k - 2]]; Max[ # ] & /@ Table[ TT[n, k], {n, 45}, {k, n}]

Edited, corrected and extended by Robert G. Wilson v, Apr 24 2004

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

A166477 Minimum positive integer solution x of equation n=x(x+1)/(t(t+1)); that is, ratio of product of two consecutive integers divided by product of two consecutive integers. Here n is a nonsquare integer (see A000037).

Original entry on oeis.org

3, 2, 5, 3, 6, 15, 4, 11, 8, 12, 20, 5, 51, 27, 19, 15, 6, 11, 45, 95, 12, 54, 7, 29, 24, 30, 1343, 54, 84, 14, 185, 95, 65, 15, 41, 35, 42, 560, 9, 23, 140, 287, 24, 17, 39, 105, 1539, 10, 48, 18, 87, 1770, 104, 183, 216, 27, 455, 11, 200, 119, 45, 20, 71, 63, 72, 14060, 99

Offset: 2

Carmine Suriano, Oct 14 2009

nonn

From R. J. Mathar, Oct 23 2010: (Start)
Writing x = (-1 + sqrt(1 + 4*n*t*(t+1)))/2, each solution is associated with a Diophantine equation 1 + 4*n*t*(t+1) = s^2. The sequence entries are the leading column if all solutions are presented in rows for a given n:
   n  Seq #  solutions
  -- ------- ------------------------------------------------
A001652 3, 20, 119, 696, 4059
A001571 2, 9, 35, 132, 494, 1845, 6887
       ...
A077262 5, 14, 99, 260, 1785, 4674
A077291 3, 8, 35, 84, 351, 836, 3479, 8280
A077401 6, 14, 104, 231, 1665, 3689
A336625 15, 32, 527, 1104, 17919
       ...
A341895 4, 20, 39, 175, 779, 1500, 6664, 29600
       11, 21, 230, 429, 4598, 8568
       8, 15, 119, 216, 1664, 3015, 23183
       12, 77, 845, 1494, 16302
       20, 35, 615, 1064, 18444, 31899
       5, 9, 44, 75, 350, 594, 2759, 4680, 21725, 36849
       ...
       51, 84, 3399, 5576
       27, 44, 935, 1512, 31779
       19, 285, 455, 6649
       15, 24, 279, 440, 5015, 7904
(End) [table reformatted by Jon E. Schoenfield, Apr 01 2018]

			For n=14, x=20; corresponding value of t is 5 since 14 = 20*21/(5*6).

Cf. A000037.

Cf. A166478 (associated t). - R. J. Mathar, Oct 23 2010

Deleted an 8 between 14 and 185. - R. J. Mathar, Oct 23 2010

A133161 Indices of the triangular numbers which are also centered triangular number.

Original entry on oeis.org

1, 4, 16, 61, 229, 856, 3196, 11929, 44521, 166156, 620104, 2314261, 8636941, 32233504, 120297076, 448954801, 1675522129, 6253133716, 23337012736, 87094917229, 325042656181, 1213075707496, 4527260173804, 16895964987721

Offset: 1

Richard Choulet, Oct 09 2007

Also, indices of the triangular numbers which are sums of three consecutive triangular numbers (see A129803).

A. Kozikowska, Topological classes of statically determinate beams with arbitrary number of supports, JOURNAL OF THEORETICAL AND APPLIED MECHANICS 49, 4, pp. 1079-1100, Warsaw 2011; (see Eq. 5.18). - _N. J. A. Sloane_, Dec 17 2011.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001834, A102871, A128862, A129803, A001571.

LinearRecurrence[{5,-5,1},{1,4,16},30] (* Harvey P. Dale, Aug 29 2017 *)

a(n+2)=4*a(n+1)-a(n)+1.
a(n+1)=2*a(n)+0.5+0.5*(12*a(n)^2+12*a(n)-15)^0.5.
G.f.: x*(1-x+x^2)/(1-x)/(1-4*x+x^2). - R. J. Mathar, Oct 24 2007
a(n)-a(n-1)= A005320(n-1). - R. J. Mathar, Mar 14 2016

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A336626 Triangular numbers that are eight times another triangular number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Maple

Mathematica

Formula

A102871 a(n) = a(n-3) - 5*a(n-2) + 5*a(n-1), a(0) = 1, a(1) = 3, a(2) = 10.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

PARI

Sage

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A093446 Largest member of the n-th row of the triangular triangle (A093445).

A102871 a(n) = a(n-3) - 5a(n-2) + 5a(n-1), a(0) = 1, a(1) = 3, a(2) = 10.

A166477 Minimum positive integer solution x of equation n=x(x+1)/(t(t+1)); that is, ratio of product of two consecutive integers divided by product of two consecutive integers. Here n is a nonsquare integer (see A000037).