A076139 -id:A076139 - cp's OEIS frontend

A001570 Numbers k such that k^2 is centered hexagonal.

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621

Offset: 1

N. J. A. Sloane

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 1..870 (terms 1..101 from T. D. Noe)
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes, part C (in French).
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics, Hex Number
Wikipedia, Beal's conjecture
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
Cf. A076139, A076140, A102871.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

[((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // G. C. Greubel, Nov 04 2017

A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
  ] (* Complement of A041017 *)
a[12, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)

{a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

Original entry on oeis.org

1, 23, 506, 11110, 243915, 5355021, 117566548, 2581109036, 56666832245, 1244089200355, 27313295575566, 599648413462098, 13164951800590591, 289029291199530905, 6345479454589089320, 139311518709760434136, 3058507932160140461673

Offset: 0

Bruno Berselli, Jun 08 2012

Partial sums of A077421.

Bruno Berselli, Table of n, a(n) for n = 0..200
Robert K. Moniot, A Family of Integer Sequences, 2021.
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3): A334673 (k=24), A212336 (k=23), A212335 (k=22), A097833 (k=21), A097832 (k=20), A049664 (k=19), A097831-A097829 (k=18,17,16), A076139 (k=15), A097828-A097826 (k=14,13,12), A097784 (k=11), A092420 (k=10), A076765 (k=9), A092521 (k=8), A053142 (k=7), A089817(k=6), A061278 (k=5), A027941 (k=4), A000217 (k=3), A021823 (k=2), A133872 (k=1), A079978 (k=0).

m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-23*x+23*x^2-x^3)));

I:=[1,23,506]; [n le 3 select I[n] else 23*Self(n-1)-23*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Aug 18 2013

a:= n-> (<<0|1|0>, <0|0|1>, <1|-23|23>>^n. <<1, 23, 506>>)[1, 1]:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 15 2012

CoefficientList[Series[1/(1 - 23 x + 23 x^2 - x^3), {x, 0, 16}], x]
LinearRecurrence[{23, -23, 1}, {1, 23, 506}, 20] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(coeff(taylor(1/(1-23*x+23*x^2-x^3), x, 0, n), x, n), n, 0, 16);

PARI
```
Vec(1/(1-23*x+23*x^2-x^3)+O(x^17))
```

[(1/20)*(-1 +21*chebyshev_U(n, 11) -chebyshev_U(n-1, 11)) for n in (0..30)] # G. C. Greubel, Feb 07 2022

G.f.: 1/((1-x)*(1 - 22*x + x^2)).
a(n) = (((6+sqrt(30))^(2*n+3) + (6-sqrt(30))^(2*n+3))/6^(n+1) - 12)/240.
a(n) = a(-n-3) = 23*a(n-1) - 23*a(n-2) + a(n-3).
a(n)*a(n+2) = a(n+1)*(a(n+1)-1).
a(n+1) - 11*a(n) = A133285(n+2).
11*a(n+1) - a(n) = (1/5)*A157096(n+2).
a(n) = (1/20)*(-1 + 21*ChebyshevU(n, 11) - ChebyshevU(n-1, 11)). - G. C. Greubel, Feb 07 2022

A076140 Triangular numbers T(k) that are three times another triangular number: T(k) such that T(k) = 3*T(m) for some m.

Original entry on oeis.org

0, 3, 45, 630, 8778, 122265, 1702935, 23718828, 330360660, 4601330415, 64088265153, 892634381730, 12432793079070, 173166468725253, 2411897769074475, 33593402298317400, 467895734407369128, 6516946879404850395, 90769360577260536405, 1264254101202242659278

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

This is a subsequence of A045943. - Michel Marcus, Apr 26 2014

			a(3) = 630 because 630 = T(35) and 630/3 = 210 = T(20).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
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 (15,-15,1).

Subsequence of A000217.
The m values are in A061278 and the k values are in A001571.
Cf. A076139, A045899, A123480, A217855.
Cf. A045943.

Join[{0}, CoefficientList[Series[3/(1 - 15x + 15x^2 - x^3), {x, 0, 20}], x]]  (* Harvey P. Dale, Apr 02 2011 *)
triNums = Accumulate[Range[0, 9999]]; Select[triNums, MemberQ[triNums, #/3] &] (* Alonso del Arte, Mar 24 2020 *)

concat(0, Vec(-3*x/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

a(n) = (3/288)*(-24 + (12 - 6*sqrt(3))*(7 - 4*sqrt(3))^n + (12 + 6*sqrt(3))*(7 + 4*sqrt(3))^n).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002: (Start)
a(0) = 0, a(1) = 3, a(2) = 45; a(n) = 15*(a(n-1) -a (n-2)) + a(n-3) for n >= 3.
G.f.: (3*x)/(1 - 15*x + 15*x^2 - x^3). (End)
a(n) = 3*A076139(n) = 3/2*A217855(n) = 3/4*A123480(n) = 3/8*A045899(n). - Peter Bala, Dec 31 2012
a(0) = 0, a(n) = 14 * a(n - 1) - a(n - 2) + 3 for n > 0. - Vladimir Pletser, Mar 23 2020
a(n) = ((2+sqrt(3))*(7+4*sqrt(3))^n + ((2-sqrt(3))*(7-4*sqrt(3))^n))/16 - 1/4 = ((2+sqrt(3))^(2n+1) + ((2-sqrt(3))^(2n+1)))/16 - 1/4. - Vladimir Pletser, Jan 15 2021

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Original entry on oeis.org

0, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080, 3371344269872647091408

Offset: 1

Andrej Dujella (duje(AT)math.hr)

Essentially the same as A051047.
It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - Jonathan Vos Post, Feb 28 2011
Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - Bruno Berselli, Feb 20 2018

G. C. Greubel, Table of n, a(n) for n = 1..870
A. Baker and H. Davenport, The Equations 3x^2-2=y^2 and 8x^2-7=z^2, Quart. J. Math. Oxford 20 (1969).
A. Dujella and A. Pethoe, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.
A. Dujella, The Problem of Diophantus and Davenport, References
A. Dujella, Publications of Andrej Dujella
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6
P. Gibbs, 1,3,8,120 ... A Diophantine Problem
P. Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A051047, A067900. A076139, A076140, A123480, A217855.

Cf. A046184, A245031.

f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)
Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* G. C. Greubel, Jun 07 2017 *)
LinearRecurrence[{15,-15,1},{0,8,120},20] (* Harvey P. Dale, Jul 14 2024 *)

x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ G. C. Greubel, Jun 07 2017

a(n) = A046184(n+1) - 1.
a(n) = 14*a(n-1) - a(n-2) + 8.
a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006
a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - Peter Bala, Dec 31 2012
From Colin Barker, Jul 30 2013: (Start)
G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)
E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - Ilya Gutkovskiy, Apr 28 2016

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

Gene Ward Smith, Sep 28 2006

The partial sums of the series 2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(3), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [1;1,2,1], [1;1,2,1,2,1], [1;1,2,1,2,1,2,1], [1;1,2,1,2,1,2,1,2,1] and so forth.

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A123478, A123479, A029549, A123482. A045899, A076139, A076140, A217855.

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

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
a(n) = -1/3 + (1/6 + 1/12*3^(1/2))*(7 + 4*3^(1/2))^n + (1/6 - 1/12*3^(1/2))*(7 - 4*3^(1/2))^n.
a(n) = 4*A076139(n) = 2*A217855(n) = 1/2*A045899(n) = 4/3*A076140(n). - Peter Bala, Dec 31 2012
G.f.: -4*x/((x-1)*(x^2-14*x+1)). - Colin Barker, Jan 20 2013
a(n) = A001353(n)*A001353(n+1). - Antonio Alberto Olivares, Apr 06 2020

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)

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

Original entry on oeis.org

1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481

Offset: 0

N. J. A. Sloane

Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
From Colin Barker, Jan 06 2015: (Start)
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)

			8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - _Colin Barker_, Jan 07 2015

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..800
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A001570, A001844, A001921, A003215.

Cf. A005448, A006051, A016754, A076139, A156712.

I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012

seq(simplify((1 +ChebyshevU(n,7) +ChebyshevU(n-1,7))/2), n=0..30); # G. C. Greubel, Oct 07 2022

With[{s1=3+2Sqrt[3],s2=3-2Sqrt[3],t1=7+4Sqrt[3],t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12,{n,0,20}]]] (* or *) LinearRecurrence[ {15,-15,1},{1,8,105},21] (* Harvey P. Dale, Aug 14 2011 *)
CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3),{x,0,30}],x] (* Vincenzo Librandi, Apr 16 2012 *)

Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015

[(1+chebyshev_U(n,7) +chebyshev_U(n-1,7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = A001075(n)*A001353(n+1).
G.f.: (1-7*x)/((1-x)*(1-14*x+x^2)). - Simon Plouffe (in his 1992 dissertation) and Colin Barker, Jan 01 2012
a(n) = A076139(n+1) - 7*A076139(n). - R. J. Mathar, Jul 14 2015
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
a(n) = 1 - a(-1-n) = 1 + A001921(n) for all integers n. - Michael Somos, Jul 10 2025

Additional comments from James R. Buddenhagen, Mar 04 2001
Name improved by Colin Barker, Jan 07 2015
Edited by Robert Israel, Feb 20 2017

A217855 Numbers m such that 16m(3*m+1)+1 is a square.

Original entry on oeis.org

0, 2, 30, 420, 5852, 81510, 1135290, 15812552, 220240440, 3067553610, 42725510102, 595089587820, 8288528719380, 115444312483502, 1607931846049650, 22395601532211600, 311930489604912752, 4344631252936566930, 60512907051507024270, 842836067468161772852

Offset: 0

Arnaldo Vicentini, Oct 12 2012

If S(h) is the sum of the squares of the integers from 1 to h, then S(h)/h = (h+1)*(2*h+1)/6. If S(h)/h is a square, then h+1 is even and 2*h+1 is a multiple of 3, so that (h+1)/2 and (2*h+1)/3 must be odd squares. The numbers h that satisfy this condition are in A084231. Let h = 1+24*k+72*k^2; for any natural k we have (h+1)/2 = (6*k+1)^2, but only if 16*k*(3*k+1)+1 is a square also (2*h+1)/3 is a square. Searching what integer numbers k satisfy this condition leads to this sequence.

			a(0) = ((2+sqrt(3))^1+(2-sqrt(3))^1-4)/24 = 0,
a(1) = ((2+sqrt(3))^3+(2-sqrt(3))^3-4)/24 = 2,
a(2) = ((2+sqrt(3))^5+(2-sqrt(3))^5-4)/24 = 30;
a(2) = 14*a(1)-a(0)+2 = 30,
a(3) = 14*a(2)-a(1)+2 = 420,
a(4) = 14*a(3)-a(2)+2 = 5852;
a(3) = 15*a(2)-15*a(1)+a(0) = 420,
a(4) = 15*a(3)-15*a(2)+a(1) = 5852.
Regarding to the comment, a(3) = 420 and so 72*a(3)^2+24*a(3)+1 = A084231(4) = 12710881, therefore Sum_{i=1..12710881} i^2/12710881 = 7338631^2 = A084232(3)^2. - _Bruno Berselli_, Oct 17 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Forum Rudi Mathematici, Un po' di calcoli...un po' di logica, post #923 (in Italian)
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A028230, A076139, A084231, A084232. A045899, A076140, A123480.

m:=19; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2/((1-x)*(1-14*x+x^2)))); // Bruno Berselli, Oct 16 2012

LinearRecurrence[{15, -15, 1}, {0, 2, 30}, 20] (* Bruno Berselli, Oct 16 2012 *)

makelist(expand(((2+sqrt(3))^(2*n+1)+(2-sqrt(3))^(2*n+1)-4)/24), n, 0, 19); /* Bruno Berselli, Oct 16 2012 */

a=vector(20); a[1]=0; a[2]=2; a[3]=30; for(i=4, #a, a[i]=15*a[i-1]-15*a[i-2]+a[i-3]); a \\ Bruno Berselli, Oct 16 2012

a(n) = ((2 + sqrt(3))^(2*n + 1) + (2 - sqrt(3))^(2*n + 1) - 4)/24.
a(n) = (cosh((2*n + 1)*log(2 + sqrt(3))) - 2)/12.
a(n) = a(-n-1) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = 14*a(n-1) - a(n-2) + 2.
a(n) = 2*A076139(n).
72*a(n)^2 + 24*a(n)+1 = A084231(n+1).
16*a(n)*(3*a(n) + 1) + 1 = A028230(n+1)^2.
G.f.: 2*x/((1 - x)*(1 - 14*x + x^2)). - Bruno Berselli, Oct 16 2012
a(n) = 2*A076139(n) = 1/2*A123480(n) = 1/4*A045899(n) = 2/3*A076140(n). - Peter Bala, Dec 31 2012

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)

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A001570 Numbers k such that k^2 is centered hexagonal.

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A212336 Expansion of 1/(1 - 23*x + 23*x^2 - x^3).

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A076140 Triangular numbers T(k) that are three times another triangular number: T(k) such that T(k) = 3*T(m) for some m.

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A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

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A123480 Coefficients of the series giving the best rational approximations to sqrt(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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A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

A217855 Numbers m such that 16m(3*m+1)+1 is a square.