A081585 -id:A081585 - cp's OEIS frontend

A158488 a(n) = 64*n^2 + 8.

72, 264, 584, 1032, 1608, 2312, 3144, 4104, 5192, 6408, 7752, 9224, 10824, 12552, 14408, 16392, 18504, 20744, 23112, 25608, 28232, 30984, 33864, 36872, 40008, 43272, 46664, 50184, 53832, 57608, 61512, 65544, 69704, 73992, 78408, 82952, 87624, 92424, 97352, 102408

Offset: 1

Vincenzo Librandi, Mar 20 2009

The identity (16*n^2 + 1)^2 - (64*n^2 + 8)*(2*n)^2 = 1 can be written as A108211(n)^2 - a(n)*A005843(n)^2 = 1. - rewritten by Bruno Berselli, Nov 16 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A081585, A108211.

I:=[72,264,584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 08 2012

A158488:=n->64*n^2+8: seq(A158488(n), n=1..50); # Wesley Ivan Hurt, Apr 08 2017

64Range[40]^2+8 (* or *) LinearRecurrence[{3,-3,1},{72,264,584},40] (* Harvey P. Dale, Nov 16 2011 *)

for(n=1, 40, print1(64*n^2 + 8", ")); \\ Vincenzo Librandi, Feb 08 2012

a(1)=72, a(2)=264, a(3)=584, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Nov 16 2011
G.f: x*(72 + 48*x + 8*x^2)/(1-x)^3. - Vincenzo Librandi, Feb 08 2012
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)) - 1)/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)))/16. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 8*(exp(x)*(8*x^2 + 8*x + 1) - 1).
a(n) = 8*A081585(n). (End)

A239331 Square array, read by antidiagonals: column k has g.f. (1+(k-1)*x)^2/(1-x)^3.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 6, 1, 1, 7, 13, 10, 1, 1, 9, 22, 25, 15, 1, 1, 11, 33, 46, 41, 21, 1, 1, 13, 46, 73, 79, 61, 28, 1, 1, 15, 61, 106, 129, 121, 85, 36, 1, 1, 17, 78, 145, 191, 201, 172, 113, 45, 1, 1, 19, 97, 190, 265, 301, 289, 232, 145, 55, 1, 1, 21

Offset: 0

Philippe Deléham, Mar 16 2014

			Square array begins:
n\k : 0......1......2......3......4......5......6......7......8......9
======================================================================
.0||  1......1......1......1......1......1......1......1......1......1
.1||  1......3......5......7......9.....11.....13.....15.....17.....19
.2||  1......6.....13.....22.....33.....46.....61.....78.....97....118
.3||  1.....10.....25.....46.....73....106....145....190....241....298
.4||  1.....15.....41.....79....129....191....265....351....449....559
.5||  1.....21.....61....121....201....301....421....561....721....901
.6||  1.....28.....85....172....289....436....613....820...1057...1324
.7||  1.....36....113....232....393....596....841...1128...1457...1828
.8||  1.....45....145....301....513....781...1105...1485...1921...2413
.9||  1.....55....181....379....649....991...1405...1891...2449...3079
10||  1.....66....221....466....801...1226...1741...2346...3041...3826
11||  1.....78....265....562....969...1486...2113...2850...3697...4654

Cf. Columns: A000012, A000217, A001844, A038764, A081585, A081587, A081589, A081591, A081593.

Cf. Rows: A000012, A005408, A028872, A100536, A239325, A069133.

T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k).
T(n,k) = 3*T(n,k-1) - 3*T(n,k-2) + T(n,k-3).
T(n,k) = (T(n,k-1) + T(n,k+1))/2 - A161680(n).
T(n,k) = (T(n-1,k) + T(n+1,k) - A000290(n))/2.

A336288 Numbers of squares formed by this procedure on n-th step: Step 1, draw a unit square. Step n, draw a unit square with center in every intersection of lines of the figure in step n-1.

Original entry on oeis.org

1, 10, 43, 116, 245, 446, 735, 1128, 1641, 2290, 3091, 4060, 5213, 6566, 8135, 9936, 11985, 14298, 16891, 19780, 22981, 26510, 30383, 34616, 39225, 44226, 49635, 55468, 61741, 68470, 75671, 83360, 91553, 100266, 109515, 119316, 129685, 140638, 152191, 164360, 177161

Offset: 1

Ilario Miriello, Jul 16 2020

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
Ilario Miriello, Step 1,2,3, Youtube video, Jul 16 2020.
Ilario Miriello, Illustration for a(2) and a(3)
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A081585.

[(8*n^3 - 12*n^2 + 7*n)/3 : n in [1..50]]; // Wesley Ivan Hurt, Jul 16 2020

Table[(8*n^3 - 12*n^2 + 7*n)/3, {n, 1, 50}] (* Amiram Eldar, Jul 16 2020 *)
LinearRecurrence[{4,-6,4,-1},{1,10,43,116},50] (* Harvey P. Dale, Sep 12 2021 *)

a(n) = (8*n^3 - 12*n^2 + 7*n)/3; \\ Michel Marcus, Jul 16 2020

Vec(x*(1 + 3*x)^2 / (1 - x)^4 + O(x^40)) \\ Colin Barker, Jul 17 2020

a(n) = (8*n^3 - 12*n^2 + 7*n)/3.
From Colin Barker, Jul 17 2020: (Start)
G.f.: x*(1 + 3*x)^2 / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4.
(End)
E.g.f.: exp(x)*x*(3 + 12*x + 8*x^2)/3. - Stefano Spezia, Jul 23 2020
a(n+1) - a(n) = 8*n^2 + 1 = A081585(n). - Charlie Marion, Mar 21 2022

More terms from Michel Marcus, Jul 16 2020

cp's OEIS Frontend

A158488 a(n) = 64*n^2 + 8.

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A239331 Square array, read by antidiagonals: column k has g.f. (1+(k-1)*x)^2/(1-x)^3.

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Crossrefs

Formula

A336288 Numbers of squares formed by this procedure on n-th step: Step 1, draw a unit square. Step n, draw a unit square with center in every intersection of lines of the figure in step n-1.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions