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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. 5
0, 15, 1485, 145530, 14260470, 1397380545, 136929032955, 13417647849060, 1314792560174940, 128836253249295075, 12624638025870742425, 1237085690282083462590, 121221773009618308591410, 11878496669252312158495605, 1163971451813716973223977895 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

For n > 1, a(n) =  98*a(n-1) - a(n-2) + 15.  In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n).  Then b(0) = 0, b(1) = A014105(m) and for n > 1,  b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m).  Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

LINKS

Colin Barker, Table of n, a(n) for n = 0..500

Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

FORMULA

From Colin Barker, Mar 02 2016: (Start)

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n>2.

G.f.: 15*x / ((1-x)*(1-98*x+x^2)). (End)

a(n) = (-10+(5-2*sqrt(6))*(49+20*sqrt(6))^(-n)+(5+2*sqrt(6))*(49+20*sqrt(6))^n)/64. - Colin Barker, Mar 03 2016

EXAMPLE

2*0 = 3*0.

2*15 = 3*10.

2*1485 = 3*990.

2*145530 = 3*97020.

MATHEMATICA

LinearRecurrence[{99, -99, 1}, {0, 15, 1485}, 20] (* T. D. Noe, Feb 15 2012 *)

PROG

(PARI) concat(0, Vec(15*x/((1-x)*(1-98*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

(MAGMA) m:=25; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(15*x/((1-x)*(1-98*x+x^2)))); // G. C. Greubel, Jul 15 2018

CROSSREFS

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

Sequence in context: A206376 A331465 A223058 * A338634 A209680 A122469

Adjacent sequences:  A200991 A200992 A200993 * A200995 A200996 A200997

KEYWORD

nonn,easy

AUTHOR

Charlie Marion, Feb 15 2012

STATUS

approved

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Last modified September 20 19:02 EDT 2021. Contains 347588 sequences. (Running on oeis4.)