A076139 -id:A076139 - cp's OEIS frontend

A336626 Triangular numbers that are eight times another triangular number.

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)

A221075 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 12, 1, 24, 1, 192, 1, 360, 1, 2700, 1, 5040, 1, 37632, 1, 70224, 1, 524172, 1, 978120, 1, 7300800, 1, 13623480, 1, 101687052, 1, 189750624, 1, 1416317952, 1, 2642885280, 1, 19726764300, 1, 36810643320, 1

Offset: 0

Peter Bala, Jan 06 2013

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).

If we denote the present sequence by [2; 12, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below

			Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
= 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
product {n >= 0} {1 - 2*(26 - 15*sqrt(3))^(4*n+3)}/{1 - 2*(26 - 15*sqrt(3))^(4*n+1)} = 1.04000 05921 62729 43797 ... = 1 + 1/(24 + 1/(1 + 1/(2700 + 1/(1 + 1/(70224 + 1/(1 + 1/(7300800 + ...))))))).

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,14,0,-14,0,-1,0,1).

Cf. A001353, A007654, A045899, A076139, A076140, A098301, A123480, A174500, A217855, A221073 (m = 2), A221074 (m = 3), A221076 (m = 5).

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
a(4*n - 3) = 12*A098301(n) = 12*A001353(n)^2 = 4*A007654(n);
a(4*n - 1) = 24*A076139(n) = 12*A217855 = 8*A076140(n) = 6*A123480(n) = 3*A045899(n).
O.g.f.: 2 + x^2/(1 - x^2) + 12*x*(1 + x^2)^2/(1 - 15*x^4 + 15*x^8 - x^12) = 2 + 12*x + x^2 + 24*x^3 + x^4 + 192*x^5 + ....
O.g.f.: (x^10-2*x^8-14*x^6+28*x^4-12*x^3+x^2-12*x-2) / ((x-1)*(x+1)*(x^4-4*x^2+1)*(x^4+4*x^2+1)). - Colin Barker, Jan 10 2014

A014979 Numbers that are both triangular and pentagonal.

Original entry on oeis.org

0, 1, 210, 40755, 7906276, 1533776805, 297544793910, 57722156241751, 11197800766105800, 2172315626468283465, 421418033734080886426, 81752926228785223683195, 15859646270350599313653420

Offset: 1

Glenn Johnston (glennj(AT)sonic.net)

			G.f. = x^2 + 210*x^3 + 40755*x^4 + 7906276*x^5 + 1533776805*x^6 + ...
a(4) = 40755 which is 285*(285-1)/2 = 165*(3*165-1)/2.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 210, p. 61, Ellipses, Paris 2008.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 22.

C. Gill, solution to question no. 8, Mathematical Miscellany, 1 (1836), pp. 220-225, at p. 223.
J. C. Su, On some properties of two simultaneous polygonal sequences, JIS 10 (2007) 07.10.4, example 4.2.
Eric Weisstein's World of Mathematics, Pentagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Cf. A046174, A046175, A076139, A108281.

a[ n_] := ChebyshevU[ 2 n - 3, 7] / 14 + ChebyshevT[ 2 n - 3, 7] / 84 - 1/12; (* Michael Somos, Feb 24 2015 *)
LinearRecurrence[{195,-195,1},{0,1,210},20] (* Harvey P. Dale, May 19 2017 *)

{a(n) = polchebyshev( 2*n - 3, 2, 7) / 14 + polchebyshev( 2*n - 3, 1, 7) / 84 - 1 / 12}; /* Michael Somos, Jun 16 2011 */

a(n) = 194 * a(n-1) - a(n-2) + 16.
G.f.: x^2 * (1 + 15*x) / ((1 - x) * (1 - 194*x + x^2)).
a(n)=((((1+sqrt(3))^(4*n-1)-(1-sqrt(3))^(4*n-1))/(2^(2*n+1)*sqrt(3)))^2)/2-1/8. - John Sillcox (johnsillcox(AT)hotmail.com), Sep 01 2003
a(n+1) = 97*a(n)+8+7*(192*a(n)^2+32*a(n)+1)^(1/2) - Richard Choulet, Sep 19 2007
a(n) = A076139(2*n - 3) = A108281(2 - n). for all n in Z. - Michael Somos, Jun 16 2011

Corrected and extended by Warut Roonguthai

Edited by N. J. A. Sloane, Jul 24 2006

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

A068085 Numbers k such that k and 10*k are both triangular numbers.

Original entry on oeis.org

0, 1, 21, 78, 1540, 30381, 112575, 2220778, 43809480, 162333171, 3202360435, 63173239878, 234084320106, 4617801526591, 91095768094695, 337549427259780, 6658866598983886, 131360034419310411, 486746040024282753, 9602081017933237120, 189421078536877518066, 701887452165588470145

Offset: 1

Amarnath Murthy, Feb 18 2002

Let y=sqrt(8*k+1) and x=sqrt(80*k+1), which must be integers if k and 10*k are triangular. These quantities satisfy the Pell-like equation x^2 - 10*y^2 = -9. All solutions x+y*sqrt(10) are obtained from 1+sqrt(10), 9+3*sqrt(10) and 41+13*sqrt(10) by multiplying by powers of the fundamental unit 19+6*sqrt(10).
Conjecture: satisfies a linear recurrence having signature (1, 0, 1442, -1442, 0, -1, 1). - Harvey P. Dale, Sep 03 2020
This conjecture is true because of the connection between (generalized) Pell equations and continued fractions of quadratic irrationals. - Georg Fischer, Feb 23 2021
From Vladimir Pletser, Feb 26 2021: (Start)
The triangular numbers T(t) that are one-tenth of other triangular numbers T(u) : T(t)=T(u)/10. The t's are in A341893, and the u's are in A341895.
Can be defined for negative n by setting a(n) = a(1-n) for all n in Z. (End)

			21 and 210 are both triangular numbers.

Georg Fischer, 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 (1,0,1442,-1442,0,-1,1).

Cf. for k and m*k both triangular: A075528 (m=2), A076139 (m=3), 0 (m=4), A077260 (m=5), A077289 (m=6), A077399 (m=7), A336624 (m=8), 0 (m=9), this sequence (m=10).

f := gfun:-rectoproc({a(-3) = 21, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 21, a(n) = 1442*a(n-3)-a(n-6)+99}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ; # Vladimir Pletser, Feb 26 2021

a[0]=0; a[1]=1; a[2]=21; a[n_] := a[n]=(99+1442a[n-3]+57Sqrt[(1+8a[n-3])(1+80a[n-3])])/2

a(n) = (99 + 1442*a(n-3) + 57*sqrt((1 + 8*a(n-3))*(1 + 88*a(n-3))))/2.
G.f.: -x^2*(x^4+20*x^3+57*x^2+20*x+1) / ((x-1)*(x^6-1442*x^3+1)). - Colin Barker, Jun 24 2014
From _Vladimir Pletser, Feb 26 2021: (Start)
a(n) = 1442 *a(n-3) - a(n-6) + 99, for n > 3, with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.
a(n) = a(n - 1) + 1442 ( a(n - 3) - a(n - 4) ) - ( a(n - 6) - a(n - 7) ) for n >= 4 with a(-2) = 21, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 21.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A341893(n). (End)

Edited by Dean Hickerson, Feb 20 2002

More terms from Georg Fischer, Feb 23 2021

A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

Original entry on oeis.org

1, -3, 13, -48, 181, -675, 2521, -9408, 35113, -131043, 489061, -1825200, 6811741, -25421763, 94875313, -354079488, 1321442641, -4931691075, 18405321661, -68689595568, 256353060613, -956722646883, 3570537526921, -13325427460800, 49731172316281

Offset: 0

Creighton Dement, Jul 21 2005

In reference to program code, 2baseiseq[X](n) = ((-1)^n)*A001353(n) (a(n)^2 + 1 is a perfect square.) 1tesseq[X](n) = (-1^(n+1))*A097948(n).

Floretion Algebra Multiplication Program, FAMP Code: 1ibaseiseq[X] with X = .5'i + .5i' + 'ii' - .5'jj' + 1.5'kk' - 1 (* Corrected by Creighton Dement, Dec 11 2009 *)

Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,4,1).

Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).

Cf. A001353, A097948.

/* By definition: */
m:=15; R:=PowerSeriesRing(Integers(), m);
A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
&cat[[A001570[i],-A007654[i]]: i in [1..m-2]]; // Bruno Berselli, Feb 05 2013

seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,4,1},{1,-3,13,-48},30] (* Harvey P. Dale, Jun 15 2018 *)

G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
Floor(((2 + sqrt(3))^n + (2 - sqrt(3))^n)/4) produces this sequence with a different offset and without signs. - James R. Buddenhagen, May 20 2010
Define c(n) = a(n) - 4*a(n+1) - a(n+2) and d(n) = -a(n) - 4*a(n+1) - a(n+2); Conjectures: I: c(2n) = 24*A076139(n); (Triangular numbers that are one-third of another triangular number) II: c(2n+1) = -A011943(n+1); (Numbers n such that any group of n consecutive integers has integral standard deviation) III: d(2n) = -2; IV: d(2n+1) = -1

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

A108281 Numbers that are both triangular and pentagonal of the second kind.

Original entry on oeis.org

0, 15, 2926, 567645, 110120220, 21362755051, 4144264359690, 803965923024825, 155965244802456376, 30256453525753512135, 5869596018751378897830, 1138671371184241752666901, 220896376413724148638480980

Offset: 1

Michael Somos, May 30 2005

nonn

			15*x^2 + 2926*x^3 + 567645*x^4 + 110120220*x^5 + 21362755051*x^6 + ...
a(4) = 567645 which is 1065*(1065-1)/2 = 615*(3*615+1)/2.

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 22.

Index entries for linear recurrences with constant coefficients, signature (195, -195, 1).

Cf. A076139, A014979.

{a(n) = polchebyshev( 2*n - 2, 2, 7) / 14 + polchebyshev( 2*n - 2, 1, 7) / 84 - 1 / 12} /* Michael Somos, Jun 16 2011 */

a(n) = 194 * a(n-1) - a(n-2) + 16.
G.f.: x^2 *(15 + x) / ((1 - x) * (1 - 194*x + x^2)).
a(n) = A076139(2*n - 2) = A014979(2 - n).

A077829 Expansion of 1/(1-3x-3x^2-2*x^3).

Original entry on oeis.org

1, 3, 12, 47, 183, 714, 2785, 10863, 42372, 165275, 644667, 2514570, 9808261, 38257827, 149227404, 582072215, 2270414511, 8855914986, 34543132921, 134737972743, 525555146964, 2049965624963, 7996038261267, 31189121952618, 121655411891581, 474525678055131

Offset: 0

N. J. A. Sloane, Nov 17 2002

Index entries for linear recurrences with constant coefficients, signature (3,3,2).

Partial sums of S(n, x), for x=1...14, A021823, A000217, A027941, A061278, A089817, A053142, A092521, A076765, A092420, A097784, A097826-A097828, A076139.

CoefficientList[Series[1/(1 - 3*x - 3*x^2 - 2*x^3), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jan 20 2024 *)
LinearRecurrence[{3,3,2},{1,3,12},30] (* Harvey P. Dale, Dec 20 2024 *)

Vec(1/(1-3*x-3*x^2-2*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

G.f.: 1/(1-3*x-3*x^2-2*x^3).

a(n) = 3*a(n-1) + 3*a(n-2) + 2*a(n-3). - Wesley Ivan Hurt, Jan 20 2024

A077831 Expansion of 1/(1-3x-2x^2-2*x^3).

Original entry on oeis.org

1, 3, 11, 41, 151, 557, 2055, 7581, 27967, 103173, 380615, 1404125, 5179951, 19109333, 70496151, 260067021, 959412031, 3539362437, 13057045415, 48168685181, 177698871247, 655548074933, 2418379337655, 8921631905325, 32912750541151, 121418274109413

Offset: 0

N. J. A. Sloane, Nov 17 2002

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,2,2).

Partial sums of S(n, x), for x=1...16, A021823, A000217, A027941, A061278, A089817, A053142, A092521, A076765, A092420, A097784, A097826-A097828, A076139, A097829, A097830.

CoefficientList[Series[1/(1-3x-2x^2-2x^3),{x,0,30}],x] (* or *) LinearRecurrence[{3,2,2},{1,3,11},30] (* Harvey P. Dale, Feb 28 2025 *)

Vec(1/(1-3*x-2*x^2-2*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

cp's OEIS Frontend

A336626 Triangular numbers that are eight times another triangular number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Maple

Mathematica

Formula

A221075 Simple continued fraction expansion of an infinite product.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A014979 Numbers that are both triangular and pentagonal.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A068085 Numbers k such that k and 10*k are both triangular numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

Views

Author

Keywords

Comments

Examples

Links

A077829 Expansion of 1/(1-3x-3x^2-2*x^3).

A077831 Expansion of 1/(1-3x-2x^2-2*x^3).