A061278 -id:A061278 - cp's OEIS frontend

A089817 a(n) = 5*a(n-1) - a(n-2) + 1 with a(0)=1, a(1)=6.

1, 6, 30, 145, 696, 3336, 15985, 76590, 366966, 1758241, 8424240, 40362960, 193390561, 926589846, 4439558670, 21271203505, 101916458856, 488311090776, 2339638995025, 11209883884350, 53709780426726, 257339018249281

Offset: 0

Paul Barry, Nov 14 2003

Partial sums of Chebyshev sequence S(n,5) = U(n,5/2) = A004254(n) (Chebyshev's polynomials of the second kind, see A049310). - Wolfdieter Lang, Aug 31 2004

In this sequence 4*a(n)*a(n+2)+1 is a square. - Bruno Berselli, Jun 19 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
F. M. van Lamoen, Square wreaths around hexagons, Forum Geometricorum, 6 (2006) 311-325.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-6,1).

Cf. A061278, A053142, A101368.

See. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

[Round((2/3 - Sqrt(21)/7)*(5/2 - Sqrt(21)/2)^n + (2/3 + Sqrt(21)/7)*(5/2 + Sqrt(21)/2)^n - 1/3): n in [0..30]]; // G. C. Greubel, Nov 20 2017

Join[{a=1,b=6},Table[c=5*b-a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011*)
CoefficientList[Series[1/(1 - 6*x + 6*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Nov 20 2017 *)

a(n)=([0,1,0; 0,0,1; 1,-6,6]^n*[1;6;30])[1,1] \\ Charles R Greathouse IV, Nov 29 2016

x='x+O('x^50); Vec(1/(1-6*x+6*x^2-x^3)) \\ G. C. Greubel, Nov 20 2017

For n > 0, a(n-1) = Sum_{i=1..n} Sum_{j=1..i} b(n) with b(n) as in A004253.
a(n) = (2/3 - sqrt(21)/7)*(5/2 - sqrt(21)/2)^n + (2/3 + sqrt(21)/7)*(5/2 + sqrt(21)/2)^n - 1/3.
G.f.: 1/((1-x)*(1 - 5*x + x^2)) = 1/(1 - 6*x + 6*x^2 - x^3).
a(n) = 6*a(n-1) - 6*a(n-2) + a(n-3) for n >= 2, a(-1):=0, a(0)=1, a(1)=6.
a(n) = (S(n+1, 5) - S(n, 5) - 1)/3 for n >= 0.
a(n)*a(n-2) = a(n-1)*(a(n-1)-1) for n > 1. - Bruno Berselli, Nov 29 2016

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

Original entry on oeis.org

1, 1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 28 2005

Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is an integer. For this sequence K=4. - Andrey Vyshnevyy, Sep 18 2015
The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - N. J. A. Sloane, Sep 08 2018

Seiichi Manyama, Table of n, a(n) for n = 0..1750
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A061278, A001353, A001835, A005246, A276122, A276271.

I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* Vincenzo Librandi, Sep 18 2015 *)
CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 11 2016 *)
a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* Michael Somos, Jul 09 2017 *)

M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))

{a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* Michael Somos, Jul 09 2017 */

a(n) = A101265(n), n>0. - R. J. Mathar, Aug 30 2008
a(n) = A079935(n+1) - A001571(n). - Gerry Martens, Jun 05 2015
a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - Seiichi Manyama, Aug 11 2016
From Ilya Gutkovskiy, Aug 11 2016: (Start)
G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
a(n) = 4*a(n-1) - a(n-2) - 1. - Seiichi Manyama, Aug 26 2016
From Seiichi Manyama, Sep 03 2016: (Start)
a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
a(n) = A005246(n)*A005246(n+1). (End)
From Michael Somos, Jul 09 2017: (Start)
0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)

a(26)-a(27) from Vincenzo Librandi, Sep 18 2015

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

Original entry on oeis.org

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

Offset: 0

Vladimir Pletser, Aug 07 2020

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

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

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

Vladimir Pletser, Table of n, a(n) for n = 0..650
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

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

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

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

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

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

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

Original entry on oeis.org

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

Offset: 1

Vladimir Pletser, Aug 13 2020

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

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

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

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

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

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

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

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

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

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

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

Original entry on oeis.org

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

Offset: 0

Vladimir Pletser, Aug 07 2020

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

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

Vladimir Pletser, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A336624, A336626, A336625.

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

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

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

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

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

A336626 Triangular numbers that are eight times another triangular number.

Original entry on oeis.org

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

Offset: 1

Vladimir Pletser, Oct 04 2020

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

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

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

Vladimir Pletser, Table of n, a(n) for n = 1..653
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

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

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

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

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

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

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

Original entry on oeis.org

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

Offset: 1

Vladimir Pletser, Feb 23 2021

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

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

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

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

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

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

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

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

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

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).

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

A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

Original entry on oeis.org

2, -1, 1, 3, 2, 7, 15, 1, 16, 8, 14, 4, 5, 15, 1, 2, 5, -1, 6, 3, 2, 39, 6, 1, 21, 7, 110, 3, 15, 7, 15, -1, 2, 8, 1, 4, 989, 8, 14, 2, 45, 15, 9, 4, 5, 335, 9, 1, 29, -1, 30, 15, 10, 415, 6, 2, 10, 32, 54, 3, 77, 55, 1, 5, 2, 7, 47750, 11, 15, 23, 47, -1, 48, 24, 16, 12, 5, 8, 2639, 1, 6720, 704, 38, 4, 2, 39, 505, 3, 13, 56, 9, 20, 13, 1631, 41

Offset: 1

Alexander Adamchuk, Apr 26 2007

sign

			a(5) = 2 because A129556(2) = 2>1 and A129556(1) = 0<1.

Max Alekseyev, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Centered Polygonal Number

Cf. A001542, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

a(n) = -1 for n in A166259.

a(n) = 1 for n = k^2-1.

Edited and b-file provided by Max Alekseyev, Jan 20 2010

cp's OEIS Frontend

A089817 a(n) = 5*a(n-1) - a(n-2) + 1 with a(0)=1, a(1)=6.

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A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).

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

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

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

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A336626 Triangular numbers that are eight times another triangular number.

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A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

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