A084703
Squares k such that 2*k+1 is also a square.
Original entry on oeis.org
0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844
Offset: 0
- D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
- Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
- Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
- Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1),
A076218 (k=3),
A278310 (k=-5).
-
[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022
-
b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)
-
[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022
A029546
Expansion of 1/( (1-x)*(1-34*x+x^2) ).
Original entry on oeis.org
1, 35, 1190, 40426, 1373295, 46651605, 1584781276, 53835911780, 1828836219245, 62126595542551, 2110475412227490, 71694037420192110, 2435486796874304251, 82734857056306152425, 2810549653117534878200, 95475953348939879706376, 3243371864210838375138585
Offset: 0
-
List([0..20], n-> (Lucas(2,-1, 4*n+6)[2] -6)/192 ); # G. C. Greubel, Jan 13 2020
-
I:=[1,35,1190]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2) +Self(n-3): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011
-
seq(coeff(series(1/( (1-x)*(1-34*x+x^2) ), x, n+1), x, n), n = 0..20); # G. C. Greubel, Jan 13 2020
-
LinearRecurrence[{35,-35,1},{1,35,1190},20] (* Vincenzo Librandi, Nov 22 2011 *)
Table[(Fibonacci[2*n+3, 2]^2 -1)/24, {n,0,20}] (* G. C. Greubel, Jan 13 2020 *)
-
Vec(1/(1-35*x+35*x^2-x^3)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012
-
[(lucas_number2(4*n+6, 2,-1) -6)/192 for n in (0..20)] # G. C. Greubel, Jan 13 2020
A123480
Coefficients of the series giving the best rational approximations to sqrt(3).
Original entry on oeis.org
4, 60, 840, 11704, 163020, 2270580, 31625104, 440480880, 6135107220, 85451020204, 1190179175640, 16577057438760, 230888624967004, 3215863692099300, 44791203064423200, 623860979209825504, 8689262505873133860, 121025814103014048540, 1685672134936323545704
Offset: 1
-
CoefficientList[Series[-4*x/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)
-
my(x='x+O('x^50)); Vec(-4*x/((x-1)*(x^2-14*x+1))) \\ G. C. Greubel, Oct 13 2017
A200993
Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3*T(m) = 2*T(k) for some k.
Original entry on oeis.org
0, 10, 990, 97020, 9506980, 931587030, 91286021970, 8945098566040, 876528373449960, 85890835499530050, 8416425350580494950, 824723793521388975060, 80814515339745539060940, 7918997779501541438997070, 775980967875811315482651930, 76038215854050007375860892080
Offset: 0
3*0 = 2*0.
3*10 = 2*15.
3*990 = 2*1485.
3*97020 = 2*145530.
- 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).
-
m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(10*x/((1-x)*(x^2-98*x+1)))); // G. C. Greubel, Jul 15 2018
-
LinearRecurrence[{99,-99,1},{0,10,990},20] (* Harvey P. Dale, Feb 25 2018 *)
-
concat(0, Vec(10*x/((1-x)*(1-98*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 02 2016
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
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).
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
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
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.
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
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.
Original entry on oeis.org
0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895
Offset: 0
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).
-
m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(15*x/((1-x)*(1-98*x+x^2)))); // G. C. Greubel, Jul 15 2018
-
LinearRecurrence[{99, -99, 1}, {0, 15, 1485}, 20] (* T. D. Noe, Feb 15 2012 *)
-
concat(0, Vec(15*x/((1-x)*(1-98*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016
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
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.
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
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
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).
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]
Original entry on oeis.org
0, 12, 420, 14280, 485112, 16479540, 559819260, 19017375312, 646030941360, 21946034630940, 745519146510612, 25325704946729880, 860328449042305320, 29225841562491651012, 992818284675673829100, 33726595837410418538400, 1145711440187278556476512
Offset: 0
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).
-
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
Corrected by Bill Lam (bill_lam(AT)myrealbox.com), Feb 27 2006
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