A180662 -id:A180662 - cp's OEIS frontend

A180845 a(n) = (16^n-3^n)/13.

0, 1, 19, 313, 5035, 80641, 1290499, 20648713, 330381595, 5286112081, 84577812979, 1353245066713, 21651921244555, 346430740444321, 5542891848703459, 88686269584038313, 1418980313358961915

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) appear in several triangle sums of Nicomachus' table A036561, i.e Gi1(4*n), Gi1(4*n+1)/2, Gi1(4*n+2)/4, Gi1(4*n+3)/8 and Gi4(n). See A180662 for information about these giraffe and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (19, -48).

Cf. A016153, A016140, A180844, A180845, A180846, A180847, A016185.

Table[(16^n-3^n)/13,{n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011*)

a(n) = (16^n-3^n)/13

G.f.: x/((16*x-1)*(3*x-1))

A185828 Half the number of n X 2 binary arrays with every element equal to exactly one or two of its horizontal and vertical neighbors.

Original entry on oeis.org

1, 3, 10, 23, 61, 162, 421, 1103, 2890, 7563, 19801, 51842, 135721, 355323, 930250, 2435423, 6376021, 16692642, 43701901, 114413063, 299537290, 784198803, 2053059121, 5374978562, 14071876561, 36840651123, 96450076810, 252509579303

Offset: 1

R. H. Hardin, Feb 05 2011

nonn

Column 2 of A185835.

			Some solutions for 4 X 2 with a(1,1)=0:
  0 0   0 1   0 0   0 0   0 1   0 0   0 0   0 0   0 0   0 0
  1 1   0 1   0 1   1 1   0 1   1 0   0 1   1 1   1 0   0 1
  0 1   0 0   0 1   0 1   1 0   1 0   1 1   1 1   1 1   0 1
  0 0   1 1   0 0   0 1   1 0   0 0   0 0   0 0   0 0   0 1
The logarithmic g.f. begins:
L(x) = x + 3*x^2/2 + 10*x^3/3 + 23*x^4/4 + 61*x^5/5 + 162*x^6/6 + ..., where
exp(L(x)) = 1 + x + 2*x^2 + 5*x^3 + 11*x^4 + 26*x^5 + 63*x^6 + ... + A051286(n)*x^n/n + ... - _Paul D. Hanna_, Mar 19 2011

R. H. Hardin, Table of n, a(n) for n = 1..200

Cf. A051286 (exp), A180662 (Fi1).

a := proc(n): n*add(binomial(2*n-2*k, 2*k)/(n-k), k=0..n-1) end: seq(a(n), n=1..28); # Johannes W. Meijer, Jun 18 2018

{a(n)=n*sum(k=0, n-1, binomial(2*n-2*k, 2*k)/(n-k))} /* Paul D. Hanna, Mar 19 2011 */

{a(n)=n*polcoeff(-log( (1+x+x^2)*(1-3*x+x^2) +x*O(x^n))/2, n)} /* Paul D. Hanna, Mar 19 2011 */

Empirical: a(n) = 2*a(n-1) + a(n-2) + 2*a(n-3) - a(n-4).
a(n) = n*Sum_{k=0..n-1} C(2n-2k, 2k)/(n-k). - Paul D. Hanna, Mar 19 2011
L.g.f.: Sum_{n>=1} a(n)*x^n/n = -log((1+x+x^2)*(1-3*x+x^2))/2. - Paul D. Hanna, Mar 19 2011
Logarithmic derivative of A051286, which is the Whitney number of level n of the lattice of the ideals of the fence of order 2n. - Paul D. Hanna, Mar 19 2011
Empirical g.f.: x*(1+x+3*x^2-2*x^3)/(1+x+x^2)/(1-3*x+x^2). - Colin Barker, Feb 22 2012
Empirical: a(n) = Sum_{k=0..floor(n/2)} A084534(n, 2*k). - Johannes W. Meijer, Jun 17 2018
Empirical: a(n) = A100886(2n). - Wojciech Florek, Jan 26 2020

A050187 a(n) = n * floor((n-1)/2).

Original entry on oeis.org

0, 0, 0, 3, 4, 10, 12, 21, 24, 36, 40, 55, 60, 78, 84, 105, 112, 136, 144, 171, 180, 210, 220, 253, 264, 300, 312, 351, 364, 406, 420, 465, 480, 528, 544, 595, 612, 666, 684, 741, 760, 820, 840, 903, 924, 990, 1012, 1081, 1104, 1176

Offset: 0

Clark Kimberling, Dec 11 1999

T(n,2), array T as in A050186; a count of aperiodic binary words.
The Row2 triangle sums A159797 lead to the sequence given above for n >= 1 with a(1)=0. For the definitions of the Row2 and other triangle sums see A180662. - Johannes W. Meijer, May 20 2011
The number of chords joining n equally distributed points on a circle with a length less than the diameter. - Wesley Ivan Hurt, Nov 23 2013
a(n) is the maximum possible length of a circuit in the complete graph on n vertices. - Geoffrey Critzer, May 23 2014
For n > 0, a(n) is half the sum of the perimeters of the distinct rectangles that can be made with positive integer sides such that L + W = n, W < L. For example, a(14) = 84; the rectangles are 1 X 13, 2 X 12, 3 X 11, 4 X 10, 5 X 9, 6 X 8 (the 7 X 7 rectangle is not considered since we have W < L). The sum of the perimeters gives 28 + 28 + 28 + 28 + 28 + 28 = 168, half of which is 84. - Wesley Ivan Hurt, Nov 23 2017
Sum of the middle side lengths of all integer-sided triangles with perimeter 3n whose side lengths are in arithmetic progression (For example, when n=5 there are two triangles with perimeter 3(5) = 15 whose side lengths are in arithmetic progression: [3,5,7] and [4,5,6]; thus a(5) = 5+5 = 10). - Wesley Ivan Hurt, Nov 01 2020

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A050186, A159797, A180662.

[n*Floor((n-1)/2): n in [0..50]]; // Wesley Ivan Hurt, May 24 2014

A050187:=n->n*floor((n-1)/2); seq(A050187(n), n=0..100); # Wesley Ivan Hurt, Nov 23 2013

Table[n*Floor[(n-1)/2], {n,0,100}] (* Wesley Ivan Hurt, Nov 23 2013 *)

a(n) = n * floor((n-1)/2).
From R. J. Mathar, Aug 08 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: x^3*(3+x) / ((1+x)^2*(1-x)^3). (End)
a(n) = binomial(n,2) - (n/2) * ((n+1) mod 2). - Wesley Ivan Hurt, Nov 23 2013
E.g.f.: x*(x*cosh(x) + sinh(x)*(x - 1))/2. - Stefano Spezia, Nov 02 2020

Name change by Wesley Ivan Hurt, Nov 23 2013

A115730 a(n) = a(n-3) + A001654(n-1) with a(0)=0, a(1)=0 and a(2)=1.

Original entry on oeis.org

0, 0, 1, 2, 6, 16, 42, 110, 289, 756, 1980, 5184, 13572, 35532, 93025, 243542, 637602, 1669264, 4370190, 11441306, 29953729, 78419880, 205305912, 537497856, 1407187656, 3684065112, 9645007681, 25250957930, 66107866110

Offset: 0

Roger L. Bagula, Mar 13 2006

The a(n+1) represent the Ca2 and Ze4 sums of the Golden Triangle A180662. Furthermore the a(3*n) represent the Ze1 (terms doubled) and Ca3 sums of the Golden triangle. See A180662 for more information about these and other triangle sums.

			G.f. = x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 42*x^6 + 110*x^7 + 289*x^8 + ... - _Michael Somos_, Sep 05 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,0,-2,-2,1).

Cf. A000045, A001654, A064831, A079962, A180662, A180664, A180665, A180666, A182890.

function A115730(n)
  if n lt 3 then return Floor(n/2);
  else return A115730(n-3) + Fibonacci(n-1)*Fibonacci(n);
  end if; return A115730;
end function;
[A115730(n): n in [0..40]]; // G. C. Greubel, Jan 20 2022

nmax:=31: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: a(1):=0: a(2):=1: for n from 3 to nmax do a(n):=a(n-3) + A001654(n-1) od: seq(a(n),n=0..nmax);

LinearRecurrence[{2,2,0,-2,-2,1}, {0,0,1,2,6,16}, 40] (* modified by G. C. Greubel, Jan 20 2022 *)
a[ n_] := Floor[(2*Fibonacci[2*n+1] + Fibonacci[2*n+2] + 2)/20]; (* Michael Somos, Sep 05 2023 *)

{a(n) = (2*fibonacci(2*n+1) + fibonacci(2*n+2) + 2)\20}; /* Michael Somos, Sep 05 2023 */

U=chebyshev_U
def A115730(n): return (1/60)*((-1)^n*(6 - 5*U(n, 1/2) + 10*U(n-1, 1/2)) - (10 - 9*U(n, 3/2) + 6*U(n-1, 3/2)))
[A115730(n) for n in (0..40)] # G. C. Greubel, Jan 20 2022

a(n) = -floor(g(Fibonacci(n+1))) where g(x) = (1-x^2)^2/(-4*x^2).
G.f.: x^2/( (1-x)*(1+x)*(1+x+x^2)*(1-3*x+x^2) ). - R. J. Mathar, Jun 20 2015
a(n) - a(n-2) = A182890(n-1). - R. J. Mathar, Jun 20 2015
a(n) = (1/60)*((-1)^n*(6 - 5*ChebyshevU(n, 1/2) + 10*ChebyshevU(n-1, 1/2)) - (10 - 9*ChebyshevU(n, 3/2) + 6*ChebyshevU(n-1, 3/2))). - G. C. Greubel, Jan 20 2022
a(n) = floor((2*Fibonacci(2*n+1) + Fibonacci(2*n+2) + 2)/20). - Michael Somos, Sep 05 2023

Corrected and information added by Johannes W. Meijer, Sep 22 2010

Edited by Editors-in-Chief. - N. J. A. Sloane, Jun 20 2015

A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 7, 11, 19, 31, 49, 79, 129, 209, 337, 545, 883, 1429, 2311, 3739, 6051, 9791, 15841, 25631, 41473, 67105, 108577, 175681, 284259, 459941, 744199, 1204139, 1948339, 3152479, 5100817, 8253295, 13354113, 21607409, 34961521

Offset: 0

Luis A Restrepo (luisiii(AT)mac.com), Mar 05 2007

nonn

This sequence has the same growth rate as the Fibonacci sequence, since x^4 - x^3 - x - 1 has the real roots phi and -1/phi.

The Ca1 sums, see A180662 for the definition of these sums, of triangle A035607 equal the terms of this sequence without the first term. - Johannes W. Meijer, Aug 05 2011

			G.f. = 1 + x + x^2 + x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 11*x^7 + 19*x^8 + 31*x^9 + ...

S. Wolfram, A New Kind of Science. Champaign, IL: Wolfram Media, pp. 82-92, 2002

Seiichi Manyama, Table of n, a(n) for n = 0..4786
K. T. Atanassov, D. R. Deford, A. G. Shannon, Pulsated Fibonacci recurrences, Fibonacci Quarterly, Vol. 52, No. 5, Dec. 2014, pp. 22-27.
Kelley L. Ross, The Golden Ratio and The Fibonacci Numbers
Eric Weisstein's World of Mathematics, Golden Ratio
Wikipedia, Golden Ratio
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. Fibonacci numbers A000045; Lucas numbers A000032; tribonacci numbers A000213; tetranacci numbers A000288; pentanacci numbers A000322; hexanacci numbers A000383; 7th-order Fibonacci numbers A060455; octanacci numbers A079262; 9th-order Fibonacci sequence A127193; 10th-order Fibonacci sequence A127194; 11th-order Fibonacci sequence A127624, A128429.

a:=[1,1,1,1];; for n in [5..50] do a[n]:=a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jul 15 2019

[n le 4 select 1 else Self(n-1) + Self(n-3) + Self(n-4): n in [1..50]]; // Vincenzo Librandi, Dec 25 2015

# From R. J. Mathar, Jul 22 2010: (Start)
A010684 := proc(n) 1+2*(n mod 2) ; end proc:
A000032 := proc(n) coeftayl((2-x)/(1-x-x^2),x=0,n) ; end proc:
A126116 := proc(n) ((-1)^floor(n/2)*A010684(n)+2*A000032(n))/5 ; end proc: seq(A126116(n),n=0..80) ; # (End)
with(combinat): A126116 := proc(n): fibonacci(n-1) + fibonacci(floor((n-4)/2)+1)* fibonacci(ceil((n-4)/2)+2) end: seq(A126116(n), n=0..38); # Johannes W. Meijer, Aug 05 2011

LinearRecurrence[{1,0,1,1},{1,1,1,1},50] (* Harvey P. Dale, Nov 08 2011 *)

Vec((x-1)*(1+x+x^2)/((x^2+x-1)*(x^2+1)) + O(x^50)) \\ Altug Alkan, Dec 25 2015

((1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2))).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Jul 15 2019

From R. J. Mathar, Jul 22 2010: (Start)
G.f.: (1-x)*(1+x+x^2)/((1-x-x^2)*(1+x^2)).
a(n) = ( (-1)^floor(n/2) * A010684(n) + 2*A000032(n))/5.
a(2*n) = A061646(n). (End)
From Johannes W. Meijer, Aug 05 2011: (Start)
a(n) = F(n-1) + A070550(n-4) with F(n) = A000045(n).
a(n) = F(n-1) + F(floor((n-4)/2) + 1)*F(ceiling((n-4)/2) + 2). (End)
a(n) = (1/5)*((sqrt(5)-1)*(1/2*(1+sqrt(5)))^n - (1+sqrt(5))*(1/2*(1-sqrt(5)))^n + sin((Pi*n)/2) - 3*cos((Pi*n)/2)). - Harvey P. Dale, Nov 08 2011
(-1)^n * a(-n) = a(n) = F(n) - A070550(n - 6). - Michael Somos, Feb 05 2012
a(n)^2 + 3*a(n-2)^2 + 6*a(n-5)^2 + 3*a(n-7)^2 = a(n-8)^2 + 3*a(n-6)^2 + 6*a(n-3)^2 + 3*a(n-1)^2. - Greg Dresden, Jul 07 2021
a(n) = A293411(n)-A293411(n-1). - R. J. Mathar, Jul 20 2025

Edited by Don Reble, Mar 09 2007

A166300 Number of Dyck paths of semilength n, having no ascents and no descents of length 1, and having no UUDD's starting at level 0.

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 5, 10, 22, 50, 113, 260, 605, 1418, 3350, 7967, 19055, 45810, 110637, 268301, 653066, 1594980, 3907395, 9599326, 23643751, 58374972, 144442170, 358136905, 889671937, 2214015802, 5518884019, 13778312440, 34448765740

Offset: 0

Emeric Deutsch, Nov 07 2009

nonn

a(n) = A166299(n,0).
a(n) is the number of peakless Motzkin paths of length n with no (1,0)-steps at level 0. Example: a(5)=2 because, denoting  U=(1,1), H=(1,0), and D=(1,-1), we have UHHHD and UUHDD. - Emeric Deutsch, May 03 2011
From Paul Barry, Mar 31 2011: (Start)
The Hankel transform of a(n+3) is A188444(n+1).
a(n+3) gives the diagonal sums of the triangle A100754.
a(n+3) has g.f. 1/(1-x-x^2/(1-2x+3x^2/(1+2x+x^2/(1-2x-(1/3)x^2/(1-x-(2/3)x^2/(1-2x+(5/2)x^2/(1+2x+(3/2)x^2/(1-...)))))))) (continued fraction) where the coefficients of x^2 have denominators A188442 and numerators A188443. (End)
The Ca1 triangle sums of triangle A175136 lead to this sequence (n>=3). For the definitions of the Ca1 and other triangle sums see A180662. - Johannes W. Meijer, May 06 2011
a(n) is the number of closed Deutsch paths of n steps with all peaks at even height.  A Deutsch path is a lattice path of up-steps (1,1) and down-steps (1,-j), j>=1, starting at the origin that never goes below the x-axis, and it is closed if it ends on the x-axis. For example a(5) = 2 counts UUUU4, UU1U2, where  U denotes an up-step and a down-step is denoted by its length, and a(6) = 5 counts UUUU13, UUUU22, UUUU31, UU1U11, UU2UU2. - David Callan, Dec 08 2021

			a(5)=2 because we have UUUDDUUDDD and UUUUUDDDDD.
G.f. = 1 + x^3 + x^4 + 2*x^5 + 5*x^6 + 10*x^7 + 22*x^8 + 50*x^9 + 113*x^10 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jean Luc Baril, Rigoberto Flórez, and José L. Ramirez, Generalized Narayana arrays, restricted Dyck paths, and related bijections, Univ. Bourgogne (France, 2025). See p. 22.

Cf. A166299, A180662, A188442, A188443, A188444.

m:=40; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!(2/(1+x+x^2+Sqrt((1+x+x^2)*(1-3*x+x^2))))); // G. C. Greubel, Sep 22 2018

G := 2/(1+z+z^2+sqrt((1+z+z^2)*(1-3*z+z^2))): Gser := series(G, z = 0, 35): seq(coeff(Gser, z, n), n = 0 .. 32);

CoefficientList[Series[2/(1+x+x^2+Sqrt[(1+x+x^2)*(1-3*x+x^2)]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

a(n):=sum((j+1)*sum((binomial(j+2*i+1,i)*sum(binomial(k,n-k-j-2*i)*binomial(k+j+2*i,k)*(-1)^(n-k),k,0,n-j-2*i))/(j+2*i+1),i,0,n-j),j,0,n); /*  Vladimir Kruchinin, Mar 07 2016 */

{a(n) = local(A); A = 1 + O(x); for( k=1, ceil(n / 5), A = 1 / (1 - x^3 / (1 - x / (1 - x * A)))); polcoeff( A, n)}; /* Michael Somos, May 12 2012 */

x='x+O('x^40); Vec(2/(1+x+x^2+((1+x+x^2)*(1-3*x+x^2))^(1/2))) \\ Altug Alkan, Sep 23 2018

G.f. = G(z)=2/(1 + z + z^2 + sqrt((1 + z + z^2)*(1 - 3*z + z^2))).
G.f.: 1 / (1 - x^3 / (1 - x / (1 - x / (1 - x^3 / (1 - x / (1 - x / ...)))))). - Michael Somos, May 12 2012
G.f. A(x) satisfies A(x) = 1 / (1 - x^3 / (1 - x / (1 - x *A(x)))). - Michael Somos, May 12 2012
Conjecture: (n+1)*a(n) +2*(-n+1)*a(n-1) +(-n+1)*a(n-2) +2*(-n+1)*a(n-3) +(n-3)*a(n-4)=0. - R. J. Mathar, Nov 24 2012
a(n) ~ (3+sqrt(5))^(n+2) * sqrt(7*sqrt(5)-15) / (2 * sqrt(Pi) * n^(3/2) * 2^(n+9/2)). - Vaclav Kotesovec, Feb 12 2014. Equivalently, a(n) ~  5^(1/4) * phi^(2*n + 2) / (8 * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
a(n) = Sum_{j=0..n}((j+1)*Sum_{i=0..n-j}((binomial(j+2*i+1,i)*Sum_{k=0..n-j-2*i}(binomial(k,n-k-j-2*i)*binomial(k+j+2*i,k)*(-1)^(n-k)))/(j+2*i+1))). - Vladimir Kruchinin, Mar 07 2016

A180665 Golden Triangle sums: a(n)=a(n-2)+A001654(n) with a(0)=0 and a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 17, 47, 121, 320, 835, 2190, 5730, 15006, 39282, 102847, 269252, 704917, 1845491, 4831565, 12649195, 33116030, 86698885, 226980636, 594243012, 1555748412, 4073002212, 10663258237, 27916772486, 73087059235

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n) are the Kn21, Kn22, Kn23, Fi2, and Ze2 sums of the Golden Triangle A180662. Furthermore the a(2*n) are the Kn3, Fi1 (terms doubled) and Ze3 (terms tripled) sums. See A180662 for information about these and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (2, 3, -3, -2, 1).

Cf. A064831, A180664, A180665, A115730, A180666.

nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: for n from 2 to nmax do a(n) := a(n-2) + A001654(n) od: seq(a(n),n=0..nmax);

a(n) = a(n-2)+A001654(n) with a(0)=0 and a(1)=1.
GF(x) = (-x)/((x-1)*(x+1)^2*(x^2-3*x+1)).
a(n) = ((-1)^(-n-1)*(15+10*n)-25+(16-4*A)*A^(-n-1)+(16-4*B)*B^(-n-1))/100 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.

A180666 Golden Triangle sums: a(n)=a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.

Original entry on oeis.org

0, 1, 2, 6, 15, 41, 106, 279, 729, 1911, 5001, 13095, 34281, 89752, 234971, 615165, 1610520, 4216400, 11038675, 28899630, 75660210, 198081006, 518582802, 1357667406, 3554419410, 9305590831, 24362353076, 63781468404

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n) are the Gi2 sums of the Golden Triangle A180662. See A180662 for information about these giraffe and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (2,2,-1,1,-2,-2,1).

Cf. A064831, A180664, A180665, A115730, A180666.

nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: a(2):=2: a(3):=6: for n from 4 to nmax do a(n):=a(n-4)+A001654(n) od: seq(a(n),n=0..nmax);
A180666 := proc(n)
    option remember;
    if n <=3 then
        op(n+1,[0,1,2,6]) ;
    else
        procname(n-4)+A001654(n) ;
    end if;
end proc:
seq(A180666(n),n=0..100 ) ; # R. J. Mathar, Aug 18 2016

Take[Total@{#, PadLeft[Drop[#, -4], Length@ #]}, Length@ # - 4] &@ Table[Times @@ Fibonacci@ {n, n + 1}, {n, 0, 31}] (* or *)
CoefficientList[Series[(-x)/((x^2 - 3 x + 1) (x - 1) (x + 1)^2 (x^2 + 1)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 18 2016 *)

a(n) = a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.
G.f.: (-x)/((x^2-3*x+1)*(x-1)*(x+1)^2*(x^2+1)).
a(n) = Sum_{k=0..floor(n/4)} A180662(n-3*k,n-4*k).
120*a(n) = 8*A001519(n) -10*A087960(n) -9*(-1)^n -15 -6*(n+1)*(-1)^n. - R. J. Mathar, Aug 18 2016

A180844 a(n) = (27^n - 2^n)/25.

Original entry on oeis.org

0, 1, 29, 787, 21257, 573955, 15496817, 418414123, 11297181449, 305023899379, 8235645283745, 222362422662139, 6003785411879801, 162102206120758723, 4376759565260493713, 118172508262033346635, 3190657723074900391913

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) appear in several triangle sums of Nicomachus's table A036561, i.e., Ca2(3*n), Ca2(3*n+1)/3, Ca2(3*n+2)/9 and Ca3(n). See A180662 for information about these camel sums and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (29, -54).

Cf. A016153, A016140, A180844, A180845, A180846, A180847, A016185.

(#[[1]]-#[[2]])/25&/@Partition[Riffle[27^Range[0,20],2^Range[0,20]],2]  (* Harvey P. Dale, Jan 22 2011 *)

a(n) = (27^n - 2^n)/25 \\ Iain Fox, Dec 12 2017

first(n) = Vec(x/((27*x-1)*(2*x-1)) + O(x^n), -n) \\ Iain Fox, Dec 12 2017

a(n) = (27^n - 2^n)/25.

G.f.: x/((27*x-1)*(2*x-1)).

A180846 a(n) = (81^n - 2^n)/79.

Original entry on oeis.org

0, 1, 83, 6727, 544895, 44136511, 3575057423, 289579651327, 23455951757615, 1899932092367071, 153894499481733263, 12465454458020395327, 1009701811099652023535, 81785846699071813910431, 6624653582624816926753103

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) appear in several triangle sums of Nicomachus's table A036561, i.e., Gi2(4*n), Gi2(4*n+1)/2, Gi2(4*n+2)/4, Gi2(4*n+3)/8 and Gi3(n). See A180662 for information about these giraffe and other chess sums.

Nathaniel Johnston, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (83,-162).

Cf. A016153, A016140, A180844, A180845, A180846, A180847, A016185.

[(81^n-2^n)/79: n in [0..50]]; // Vincenzo Librandi, Apr 15 2011

Table[(81^n-2^n)/79,{n,0,15}] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2011 *)

a(n) = (81^n - 2^n)/79.

G.f.: x/((81*x-1)*(2*x-1)).

cp's OEIS Frontend

A180845 a(n) = (16^n-3^n)/13.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A185828 Half the number of n X 2 binary arrays with every element equal to exactly one or two of its horizontal and vertical neighbors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

PARI

Formula

A050187 a(n) = n * floor((n-1)/2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

Extensions

A115730 a(n) = a(n-3) + A001654(n-1) with a(0)=0, a(1)=0 and a(2)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A126116 a(n) = a(n-1) + a(n-3) + a(n-4), with a(0)=a(1)=a(2)=a(3)=1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A166300 Number of Dyck paths of semilength n, having no ascents and no descents of length 1, and having no UUDD's starting at level 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs