A084159 -id:A084159 - cp's OEIS frontend

A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321

Offset: 0

Peter Bala, Apr 30 2012

The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1) + r) with initial conditions a(0) = 1, a(1) = 1 + r has the solution a(n) = 1/2 + (1/2)*Sum_{k = 0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613 (r = -2), A146983 (r = -3) and A084159(r = -4).
Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - Colin Barker, Jan 01 2015

Seiichi Manyama, Table of n, a(n) for n = 0..1116 (first 201 terms from Vincenzo Librandi)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A011900, A054318, A070997, A101265, A109613, A133872, A146983, A211955.

I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // Vincenzo Librandi, May 18 2012

RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* Harvey P. Dale, May 14 2012 *)

Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n) = 1/2 + (1/2)*Sum_{k = 0..n} 6^k*binomial(n+k,2*k).
a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.
a(n) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the n-th Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2) - 3 with a(0) = 1 and a(1) = 4.
O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....
Sum_{n >= 0} 1/a(n) = sqrt(5/3); 5 - 3*(Sum_{k = 0..2*n} 1/a(k))^2 = 2/A070997(n)^2.
a(0) = 1, a(1) = 4, a(2) = 28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - Harvey P. Dale, May 14 2012

A110048 Expansion of 1/((1+2x)(1-4x-4x^2)).

Original entry on oeis.org

1, 2, 16, 64, 336, 1568, 7680, 36864, 178432, 860672, 4157440, 20070400, 96915456, 467935232, 2259419136, 10909384704, 52675280896, 254338531328, 1228055511040, 5929575645184, 28630525673472, 138240403177472

Offset: 0

Creighton Dement, Jul 10 2005

Floretion Algebra Multiplication Program, FAMP Code:
-kbasejseq[A*B] with A = + 'i - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki' and B = - .5'i + .5'j + 'k - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'
See also comment for A110047.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (2,12,8).

Cf. A079291, A084159, A086346, A086348, A110046, A110047, A110049, A110050.

[2^(n-2)*(Evaluate(DicksonFirst(n+1,-1), 2) +2*(-1)^n): n in [0..40]]; // G. C. Greubel, Aug 18 2022

seriestolist(series(1/((1+2*x)*(1-4*x-4*x^2)), x=0,40));

CoefficientList[Series[1/((1+2x)(1-4x-4x^2)), {x,0,40}], x] (* or *) LinearRecurrence[{2,12,8}, {1,2,16}, 41] (* Harvey P. Dale, Nov 02 2011 *)

[2^(n-2)*(lucas_number2(n+1,2,-1) +2*(-1)^n) for n in (0..40)] # G. C. Greubel, Aug 18 2022

Superseeker finds: a(n+1) = 2*A086348(n+1) (A086348's offset is 1: On a 3 X 3 board, number of n-move routes of chess king ending at central cell); binomial transform matches A084159 (Pell oblongs); j-th coefficient of g.f.*(1+x)^j matches A079291 (Squares of Pell numbers); a(n) + a(n+1) = A086346(n+2) (A086346's offset is 1: On a 3 X 3 board, the number of n-move paths for a chess king ending in a given corner cell.)
From Maksym Voznyy (voznyy(AT)mail.ru), Jul 24 2008: (Start)
a(n) = 2*a(n-1) + 12*a(n-2) + 8*a(n-3), where a(1)=1, a(2)=2, a(3)=16.
a(n) = 2^(n-3)*( 4*(-1)^(1-n) + (sqrt(2)-1)^(-n) + (-sqrt(2)-1)^(-n)) . (End)
a(n) = 2^n*A097076(n+1). - R. J. Mathar, Mar 08 2021

A046727 Related to Pythagorean triples: alternate terms of A001652 and A046090.

Original entry on oeis.org

0, 3, 21, 119, 697, 4059, 23661, 137903, 803761, 4684659, 27304197, 159140519, 927538921, 5406093003, 31509019101, 183648021599, 1070379110497, 6238626641379, 36361380737781, 211929657785303, 1235216565974041, 7199369738058939, 41961001862379597, 244566641436218639

Offset: 0

N. J. A. Sloane

For a triple (a,b,c) there exist k,m such that (a,b,c) = (k^2 - m^2, 2*k*m, k^2 + m^2). Here k = A001333(n) and m = A001333(n+1), so this sequence is identical to the Pell oblongs A084159 for n > 0. - Lambert Klasen (Lambert.Klasen(AT)gmx.de), Nov 10 2004

a(n), for n >= 1, gives the odd length (in some unit) catheti (legs) of the (primitive) Pythagorean triples which have absolute length difference of the catheti equal to one. See a W. Lang comment on A001653 on how to generate all such Pythagorean triples. - Wolfdieter Lang, Mar 08 2012

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. Bogomolny, The Trinary Tree(s) underlying Primitive Pythagorean Triples.
Dan Romik, The dynamics of Pythagorean Triples, Trans. Amer. Math. Soc. 360 (2008), 6045-6064.
P. E. Trier, "Almost Isosceles" Right-Angled Triangles, Eureka, No. 4, May 1940, pp. 9 - 11.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A001109, A001333, A001652, A005319, A046090, A046729.

Essentially the same as A084159.

a046727 n = a046727_list !! n
a046727_list = 0 : f (tail a001652_list) (tail a046090_list) where
   f (x::xs) (:y:ys) = x : y : f xs ys
-- Reinhard Zumkeller, Jan 10 2012

I:=[0,3,21,119]; [n le 4 select I[n] else 5*Self(n-1)+5*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Nov 04 2016

RecurrenceTable[{a[n+2]==6a[n+1] -a[n] -4*(-1)^n, a[0]==3, a[1]==21}, a, {n, 30}] (* Ron Knott, Jul 01 2013 *)
LinearRecurrence[{5,5,-1}, {0,3,21,119}, 30] (* Vincenzo Librandi, Nov 04 2016 *)

concat(0, Vec(x*(3+6*x-x^2)/((1+x)*(1-6*x+x^2)) + O(x^30))) \\ Colin Barker, Nov 03 2016

[(lucas_number2(2*n+1,2,-1) +2*(-1)^n)/4 -int(n==0) for n in range(41)] # G. C. Greubel, Feb 11 2023

Values of x obtained by repeatedly multiplying the triple (x, y, z) = (3, 4, 5) by the matrix A = ([1 2 2], [2 1 2], [2 2 3]), the Across matrix of "The Trinary Tree(s) underlying Primitive Pythagorean Triples" generating matrices. - Vim Wenders, Jan 14 2004
For n > 0, a(n) = A001333(n)*A001333(n+1). - Lambert Klasen (Lambert.Klasen(AT)gmx.de), Nov 10 2004
G.f.: x*(3+6*x-x^2)/((1+x)*(1-6*x+x^2)). - R. J. Mathar, Jul 08 2009
a(n) + a(n+1) = A005319(n+1), n > 0. - R. J. Mathar, Jul 13 2009
a(n) = 6*a(n-1) - a(n-2) - 4*(-1)^n. - Ron Knott, Jul 01 2013
From Colin Barker, Nov 03 2016: (Start)
a(n) = (2*(-1)^n + (1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1))/4 for n > 0.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3) for n > 3. (End)
From G. C. Greubel, Feb 11 2023: (Start)
a(n) = (1/2)*(A001109(n+1) + A001109(n) + (-1)^n) - [n=0].
a(n) = (A001333(2*n+1) + (-1)^n)/2 - [n=0]. (End)
E.g.f.: exp(-x)*(1 + exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2 - 1. - Stefano Spezia, Aug 03 2024

More terms from Sascha Kurz, Jan 23 2003

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A182432 Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1) + 3) with a(0) = 1, a(1) = 4.

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A110048 Expansion of 1/((1+2*x)*(1-4*x-4*x^2)).

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A046727 Related to Pythagorean triples: alternate terms of A001652 and A046090.

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A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

A110048 Expansion of 1/((1+2x)(1-4x-4x^2)).