A079291 -id:A079291 - cp's OEIS frontend

A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

1, 2, 4, 10, 25, 60, 144, 348, 841, 2030, 4900, 11830, 28561, 68952, 166464, 401880, 970225, 2342330, 5654884, 13652098, 32959081, 79570260, 192099600, 463769460, 1119638521, 2703046502, 6525731524, 15754509550, 38034750625, 91824010800

Offset: 0

Paul Barry, Nov 15 2003

a(n) is the number of tilings of an n-board (a board of size n X 1) using white squares, black squares, and white (1,1)-fences. A (1,1)-fence is a tile composed of two squares separated by a gap of width 1. - Michael A. Allen, Mar 12 2021

a(n) is the number of tilings of an n-board using white squares, black squares, white trominoes, black trominoes, and white tetrominoes. - Michael A. Allen, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, J. Int. Seq. 24 (2021) Article 21.3.8.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,0,2,1).

Cf. A000129, A001333, A006498, A057077, A079291, A114620.

[(Evaluate(DicksonFirst(n+2,-1), 2) + 2*(-1)^Binomial(n,2))/8: n in [0..40]]; // G. C. Greubel, Aug 18 2022

CoefficientList[Series[1/(1-2x-2x^3-x^4),{x,0,30}],x] (* Michael A. Allen, Mar 12 2021 *)
LinearRecurrence[{2,0,2,1}, {1,2,4,10}, 41] (* G. C. Greubel, Aug 18 2022 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+2b+a}; NestList[nxt,{1,2,4,10},30][[;;,1]] (* Harvey P. Dale, Jul 18 2024 *)

[(lucas_number2(n+2,2,-1) +2*(-1)^binomial(n,2))/8 for n in (0..40)] # G. C. Greubel, Aug 18 2022

a(n) = ( (1+sqrt(2))^(n+2) + (1-sqrt(2))^(n+2) + 2*(-1)^floor(n/2) )/8.
a(n) = (-i)^n*Sum_{k=0..floor(n/2)} U(n-2*k, i) with i^2 = -1.
a(n) + a(n+2) = A000129(n+3). - Alex Ratushnyak, Aug 06 2012
G.f.: 1/ ( (1+2*x)*(1-2*x-x^2) ). - R. J. Mathar, Apr 26 2013
4*a(n) = A057077(n) + A001333(n+2). - R. J. Mathar, Apr 26 2013
a(2*n) = (A000129(n+1))^2 = A079291(n+1). - Michael A. Allen, Mar 12 2021
a(2*n+1) = A000129(n+1)*A000129(n+2) = A114620(n+1). - Michael A. Allen, Mar 12 2021

Formula corrected by Max Alekseyev, Aug 22 2013

A276914 Subsequence of triangular numbers obtained by adding a square and two smaller triangles, a(n) = n^2 + 2*A000217(A052928(n)).

Original entry on oeis.org

0, 1, 10, 15, 36, 45, 78, 91, 136, 153, 210, 231, 300, 325, 406, 435, 528, 561, 666, 703, 820, 861, 990, 1035, 1176, 1225, 1378, 1431, 1596, 1653, 1830, 1891, 2080, 2145, 2346, 2415, 2628, 2701, 2926, 3003, 3240, 3321, 3570, 3655, 3916, 4005, 4278, 4371, 4656

Offset: 0

Daniel Poveda Parrilla, Sep 22 2016

All terms of this sequence are triangular numbers. Graphically, for each term of the sequence, one corner of the square will be part of the corresponding triangle's hypotenuse if the term is an odd number. Otherwise, it will not be part of it.
a(A276915(n)) is a triangular pentagonal number.
a(A079291(n)) is a triangular square number, as A275496 is a subsequence of this.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A004526, A016813, A042948, A052928, A079291, A168277, A275496, A276915.

[n*(2*n+(-1)^n): n in [0..40]]; // G. C. Greubel, Aug 19 2022

Table[n (2 n + (-1)^n), {n, 0, 48}] (* Michael De Vlieger, Sep 23 2016 *)

concat(0, Vec(x*(1+9*x+3*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^50))) \\ Colin Barker, Sep 23 2016

[n*(2*n+(-1)^n) for n in (0..40)] # G. C. Greubel, Aug 19 2022

a(n) = n^2 + 2*A000217(A052928(n)).
a(n) = A000217(A042948(n)).
a(n) = n*(2*n + (-1)^n).
a(n) = n*A168277(n + 1).
a(n) = n*A016813(A004526(n)).
From Colin Barker, Sep 23 2016: (Start)
G.f.: x*(1 + 9*x + 3*x^2 + 3*x^3) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = n*(2*n+1) for n even.
a(n) = n*(2*n-1) for n odd. (End)
E.g.f.: x*( 2*(1+x)*exp(x) - exp(-x) ). - G. C. Greubel, Aug 19 2022
Sum_{n>=1} 1/a(n) = 2 - log(2). - Amiram Eldar, Aug 21 2022

A098212 Expansion of (5-x^2)/((1+x)(1-6x+x^2)).

Original entry on oeis.org

5, 25, 149, 865, 5045, 29401, 171365, 998785, 5821349, 33929305, 197754485, 1152597601, 6717831125, 39154389145, 228208503749, 1330096633345, 7752371296325, 45184131144601, 263352415571285, 1534930362283105, 8946229758127349, 52142448186480985

Offset: 0

Creighton Dement, Oct 25 2004

Old name was: Relates the squares of Pell numbers with the squares of the numerators of continued fraction convergents to sqrt(2).

Floretion Algebra Multiplication Program, FAMP Code: 1vesseq[(j' + k' + 'ii')*('j + 'k + 'ii')] - Creighton Dement, Aug 16 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001333, A001541, A079291, A075848, A090390.

I:=[5,25,149]; [n le 3 select I[n] else 5*Self(n-1)+5*Self(n-2)-Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 26 2015

a[0]= 5; a[1]= 25; a[2]= 149; a[n_]:= a[n]= 5 a[n-1] + 5 a[n-2] - a[n-3]; Table[ a[n], {n,0,40}] (* Robert G. Wilson v, Nov 05 2004 *)
CoefficientList[Series[(5-x^2)/((1+x)(1-6x+x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{5,5,-1},{5,25,149},40] (* Harvey P. Dale, Jun 09 2011 *)

Vec((5-x^2)/((1+x)*(1-6*x+x^2))+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

def Pell(n): return lucas_number1(n,2,-1)
[4*Pell(n+1)^2 +(Pell(n+1) +Pell(n))^2  for n in (0..40)] # G. C. Greubel, Aug 20 2022

G.f.: (5-x^2)/((1+x)*(1-6*x+x^2)).
a(n) = 4*A079291(n+1) + A090390(n+1) = 4(A000129(n+1))^2 + (A001333(n+1))^2.
a(n) + a(n+1) = A075848(n+2) - A075848(n+1).
a(n) = A001541(n+1) + 2*A079291(n+1). - Creighton Dement, Oct 26 2004
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3), a(0) = 5, a(1) = 25, a(2) = 149. - Robert G. Wilson v, Nov 05 2004
2*a(n) = (-1)^n + 3*A001541(n+1). - R. J. Mathar, Sep 11 2019

A105058 Expansion of g.f. (1+8x-x^2)/((1+x)(1-6*x+x^2)).

Original entry on oeis.org

1, 13, 69, 409, 2377, 13861, 80781, 470833, 2744209, 15994429, 93222357, 543339721, 3166815961, 18457556053, 107578520349, 627013566049, 3654502875937, 21300003689581, 124145519261541, 723573111879673

Offset: 0

Creighton Dement, Apr 04 2005

A floretion-generated sequence relating the squares of the numerators of continued fraction convergents to sqrt(2) to the squares of the denominators of continued fraction convergents to sqrt(2) (Pell numbers).
Floretion Algebra Multiplication Program, FAMP Code:
1dia[J]tesseq[ - .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e ]. Identity used: dia[I]tes + dia[J]tes + dia[K]tes = jes + fam + 3tes.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1)

Cf. A000129, A001109, A001333, A046729, A077444.

Cf. A077445, A079291, A082639, A084158, A090390.

[Evaluate(DicksonSecond(2*n+1, -1), 2) -(-1)^n: n in [0..30]]; // G. C. Greubel, Aug 21 2022

CoefficientList[ Series[(1+8x-x^2)/((1+x)(1-6x+x^2)), {x,0,30}], x] (* Robert G. Wilson v, Apr 06 2005 *)
LinearRecurrence[{5,5,-1}, {1,13,69}, 30] (* Harvey P. Dale, Jun 03 2017 *)

[lucas_number1(2*n+2,2,-1) -(-1)^n for n in (0..30)] # G. C. Greubel, Aug 21 2022

a(n) = 2 * A001109(n+1) - (-1)^n.
G.f.: G(0)/(1-3*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
From G. C. Greubel, Aug 21 2022: (Start)
a(n) = A000129(2*n+2) - (-1)^n.
E.g.f.: exp(3*x)*( 2*cosh(2*sqrt(2)*x) + (3/sqrt(2))*sinh(2*sqrt(2)*x)) - exp(-x). (End)

A244352 a(n) = Pell(n)^3 - Pell(n)^2, where Pell(n) is the n-th Pell number (A000129).

Original entry on oeis.org

0, 0, 4, 100, 1584, 23548, 338100, 4798248, 67750848, 954701400, 13441659268, 189185124940, 2662308356400, 37463104912660, 527155118240244, 7417689205890000, 104375121328998144, 1468671237346368048, 20665783224031936900, 290789699203441908148

Offset: 0

Colin Barker, Jun 26 2014

			a(3) = Pell(3)^3 - Pell(3)^2 = 5^3 - 5^2 = 100.

Colin Barker, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (17,-25,-223,-79,95,-7,-1).

Cf. A000129, A079291, A110272.

Pell:= func< n | n eq 0 select 0 else Evaluate(DicksonSecond(n-1,-1),2) >;
[Pell(n)^3 - Pell(n)^2: n in [0..40]]; // G. C. Greubel, Aug 20 2022

CoefficientList[Series[4*x^2*(3*x^3-4*x^2+8*x+1) / ((x+1)*(x^2-6*x+1)*(x^2-2*x-1)*(x^2+14*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Jun 26 2014 *)

pell(n) = round(((1+sqrt(2))^n-(1-sqrt(2))^n)/(2*sqrt(2)))
vector(50, n, pell(n-1)^3-pell(n-1)^2)

def Pell(n): return lucas_number1(n,2,-1)
[Pell(n)^3 -Pell(n)^2 for n in (0..40)] # G. C. Greubel, Aug 20 2022

a(n) = A110272(n) - A079291(n).
G.f.: 4*x^2*(1+8*x-4*x^2+3*x^3) / ((1+x)*(1-6*x+x^2)*(1+2*x-x^2)*(1-14*x-x^2)).
a(n) = A045991(A000129(n)). - Michel Marcus, Jun 26 2014

A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

Original entry on oeis.org

1, 5, 41, 61, 145, 181, 313, 365, 545, 613, 841, 925, 1201, 1301, 1625, 1741, 2113, 2245, 2665, 2813, 3281, 3445, 3961, 4141, 4705, 4901, 5513, 5725, 6385, 6613, 7321, 7565, 8321, 8581, 9385, 9661, 10513, 10805, 11705, 12013, 12961, 13285, 14281, 14621, 15665

Offset: 0

Daniel Poveda Parrilla, Sep 27 2016

All terms of this sequence are centered square numbers. Graphically, each term of the sequence is made of four squares, eight triangles and a central element.

a(A220185(n+1)) = A008844(2n) = A079291(4n+1), which is a square of a Pell number.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008844, A079291, A220185, A276914.

[4*n*(2*n+(-1)^n)+1 : n in [0..60]]; // Wesley Ivan Hurt, Sep 27 2016

A276916:=n->4*n*(2*n+(-1)^n)+1: seq(A276916(n), n=0..60); # Wesley Ivan Hurt, Sep 27 2016

Table[4 n (2 n + (-1)^n) + 1, {n, 0, 44}] (* or *)
CoefficientList[Series[(1 +4x +34x^2 +12x^3 +13x^4)/((1-x)^3*(1+x)^2), {x, 0, 44}], x] (* Michael De Vlieger, Sep 28 2016 *)

Vec((1+4*x+34*x^2+12*x^3+13*x^4)/((1-x)^3*(1+x)^2) + O(x^50)) \\ Colin Barker, Sep 27 2016

[4*n*(2*n+(-1)^n) +1 for n in (0..60)] # G. C. Greubel, Aug 19 2022

a(n) = 4*n*(2*n + (-1)^n) + 1.
a(n) = 4*n*(2*n + 1) + 1 for n even.
a(n) = 4*n*(2*n - 1) + 1 for n odd.
a(n) is sum of two squares; a(n) = k^2 + (k+1)^2 where k = 2n-(n mod 2). - David A. Corneth, Sep 27 2016
From Colin Barker, Sep 27 2016: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
G.f.: (1+4*x+34*x^2+12*x^3+13*x^4) / ((1-x)^3*(1+x)^2). (End)
E.g.f.: (1+8*x+8*x^2)*exp(x) - 4*x*exp(-x). - G. C. Greubel, Aug 19 2022

cp's OEIS Frontend

A089928 a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

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A276914 Subsequence of triangular numbers obtained by adding a square and two smaller triangles, a(n) = n^2 + 2*A000217(A052928(n)).

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A098212 Expansion of (5-x^2)/((1+x)*(1-6*x+x^2)).

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A105058 Expansion of g.f. (1+8*x-x^2)/((1+x)*(1-6*x+x^2)).

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A244352 a(n) = Pell(n)^3 - Pell(n)^2, where Pell(n) is the n-th Pell number (A000129).

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A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

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A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

A098212 Expansion of (5-x^2)/((1+x)(1-6x+x^2)).

A105058 Expansion of g.f. (1+8x-x^2)/((1+x)(1-6*x+x^2)).