A026597 -id:A026597 - cp's OEIS frontend

A027285 a(n) = Sum_{k=0..2n-3} T(n,k) T(n,k+3), with T given by A026584.

12, 116, 682, 4908, 30272, 201648, 1273286, 8275894, 52783298, 340392020, 2180905198, 14035736838, 90149817980, 580197442656, 3732734480794, 24041345351898, 154874693823022, 998441294531516, 6439238635990250, 41552345665859196, 268252644944872486

Offset: 3

Clark Kimberling

nonn

G. C. Greubel, Table of n, a(n) for n = 3..1000

Cf. A026584, A026585, A026587, A026589, A026590, A026591, A026592, A026593, A026594, A026595, A026596, A026597, A026598, A026599, A027282, A027283, A027284, A027286.

T[n_, k_]:= T[n, k]= If[k==0 || k==2*n, 1, If[k==1 || k==2*n-1, Floor[n/2], If[EvenQ[n+k], T[n-1, k-2] + T[n-1, k], T[n-1, k-2] + T[n-1, k-1] + T[n-1, k] ]]]; (* T = A026584 *)
a[n_]:= a[n]= Sum[T[n, k]*T[n, k+3], {k, 0, 2*n-3}];
Table[a[n], {n, 3, 40}] (* G. C. Greubel, Dec 15 2021 *)

@CachedFunction
def T(n, k):  # T = A026584
    if (k==0 or k==2*n): return 1
    elif (k==1 or k==2*n-1): return (n//2)
    else: return T(n-1, k-2) + T(n-1, k) if ((n+k)%2==0) else T(n-1, k-2) + T(n-1, k-1) + T(n-1, k)
@CachedFunction
def A027285(n): return sum(T(n,j)*T(n, j+3) for j in (0..2*n-3))
[A027285(n) for n in (3..40)] # G. C. Greubel, Dec 15 2021

a(n) = Sum_{k=0..2*n-3} A026584(n,k) * A026584(n,k+3).

More terms from Sean A. Irvine, Oct 26 2019

A159612 INVERT transform of (1, 3, 1, 3, 1, ...).

Original entry on oeis.org

1, 4, 8, 24, 56, 152, 376, 984, 2488, 6424, 16376, 42072, 107576, 275864, 706168, 1809624, 4634296, 11872792, 30409976, 77901144, 199541048, 511145624, 1309309816, 3353892312, 8591131576, 22006700824, 56371227128, 144398030424, 369882938936, 947475060632, 2427006816376

Offset: 1

Gary W. Adamson, Apr 17 2009

The sequence 1,1,4,8,24,... is an eigensequence of the sequence triangle of 1,3,1,3,1,3,1,..., which is the Riordan array ((1+3x)/(1-x^2),x). - Paul Barry, Feb 10 2011
From Sean A. Irvine, Jun 07 2025: (Start)
Also, the number of walks of length n-1 starting at vertex 1 in the following graph:
  0   2
  |\ /|
  | 1 |
  |/ \|
  4   3. (End)

			a(4) = 24 = (1, 3, 1, 3) dot (8, 4, 1, 1) = (8 + 12, + 1 + 3).

Colin Barker, Table of n, a(n) for n = 1..1000
Sean A. Irvine, Walks on Graphs.
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (1,4).

Cf. A006131, A119473.

Cf. A026597 (vertices 0, 2, 3, 4), A384604 (missing edge {0,4}).

LinearRecurrence[{1, 4}, {1, 4}, 50] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *)

Vec(x*(1+3*x)/(1-x-4*x^2) + O(x^40)) \\ Colin Barker, Dec 22 2016

G.f.: x*(1+3*x)/(1-x-4*x^2). - Philippe Deléham, Mar 01 2012
a(n) = a(n-1) + 4*a(n-2), a(1)=1, a(2)=4. - Vincenzo Librandi, Mar 11 2011
a(n+1) = Sum_{k=0..n} A119473(n,k)*3^k. - Philippe Deléham, Oct 05 2012
a(n) = 2^(-3-n)*((1-sqrt(17))^n*(-5+3*sqrt(17)) + (1+sqrt(17))^n*(5+3*sqrt(17))) / sqrt(17) for n > 0. - Colin Barker, Dec 22 2016
a(n) = A006131(n)+3*A006131(n-1). - R. J. Mathar, Jun 07 2025
E.g.f.: (exp(x/2)*(51*cosh(sqrt(17)*x/2) + 5*sqrt(17)*sinh(sqrt(17)*x/2)) - 51)/68. - Stefano Spezia, Jun 07 2025

A103631 Triangle read by rows: T(n,k) = abs(qStirling2(n,k,q)) for q = -1, with 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 2, 1, 0, 1, 1, 4, 3, 3, 1, 0, 1, 1, 5, 4, 6, 3, 1, 0, 1, 1, 6, 5, 10, 6, 4, 1, 0, 1, 1, 7, 6, 15, 10, 10, 4, 1, 0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1

Offset: 0

Paul Barry, Feb 11 2005

Previous name: An invertible triangle whose row sums are F(n+1).
Triangle inverse has general term (-1)^(n-k)*binomial(floor(n/2),n-k). Diagonal sums are A103632.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 08 2005
Row sums are Fibonacci numbers (A000045).
Another version of triangle in A065941. - Philippe Deléham, Jan 01 2009
From Johannes W. Meijer, Aug 11 2011: (Start)
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(r) coefficients are all equal to 1; see the second Maple program.
The T(n,k) triangle is related to a linear (n+1)-th order differential equation with coefficients a(n,k), see triangle A194005.
Parks's triangle appears to be an appropriate name for the triangle given above. (End)

			From _Paul Barry_, Oct 02 2009: (Start)
Triangle begins:
  1,
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 2, 1,
  0, 1, 1, 3, 2,  1,
  0, 1, 1, 4, 3,  3,  1,
  0, 1, 1, 5, 4,  6,  3,  1,
  0, 1, 1, 6, 5, 10,  6,  4, 1,
  0, 1, 1, 7, 6, 15, 10, 10, 4, 1
Production matrix is:
  0, 1,
  0, 1, 1,
  0, 0, 0, 1,
  0, 0, 0, 1, 1,
  0, 0, 0, 0, 0, 1,
  0, 0, 0, 0, 0, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 1 (End)

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.
P. C. Parks, A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) pp. 694-702.

Cf. A103633 (signed version).

a103631 n k = a103631_tabl !! n !! k
a103631_row n = a103631_tabl !! n
a103631_tabl = [1] : [0,1] : f [1] [0,1] where
   f xs ys = zs : f ys zs where
     zs = zipWith (+)  ([0,0] ++ xs)  (ys ++ [0])
-- Reinhard Zumkeller, May 07 2012

/* As triangle: */ [[Binomial(Floor((2*n-k-1)/2), n-k): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Aug 28 2016

From Johannes W. Meijer, Aug 11 2011: (Start)
A103631 := proc(n,k): binomial(floor((2*n-k-1)/2),n-k) end: seq(seq(A103631(n,k), k=0..n), n=0..12);
nmax:=12: for n from 0 to nmax+1 do b(n):=1 od: A103631 := proc(n,k) option remember: local j: if k=0 and n=0 then b(1) elif k=0 and n>=1 then 0 elif k=1 then b(n+1) elif k=2 then b(1)*b(n+1) elif k>=3 then expand(b(n+1)*add(procname(j,k-2), j=k-2..n-2)) fi: end: for n from 0 to nmax do seq(A103631(n,k), k=0..n) od: seq(seq(A103631(n,k),k=0..n), n=0..nmax); # (End)

p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = x; p[x, 2] = x + x^2; p[x_, n_] := p[x, n] = p[x, n - 1] + x^2*p[x, n - 2]; (* with *) Table[ExpandAll[p[x, n]], {n, 0, 10}]; (* or *) a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 27 2008 *)
Table[Binomial[Floor[(2*n - k - 1)/2], n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 27 2016 *)
qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k - 1) qStirling2[n - 1, k - 1, q] + Sum[q^j, {j, 0, k - 1}] qStirling2[n - 1, k, q];
qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
qStirling2[0, k_, _] := KroneckerDelta[0, k];
qStirling2[, , _] = 0;
Table[Abs[qStirling2[n, k, -1]], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *)

from sage.combinat.q_analogues import q_stirling_number2
for n in (0..9):
    print([abs(q_stirling_number2(n,k).substitute(q=-1)) for k in [0..n]])
# Peter Luschny, Mar 09 2020

T(n,k) = binomial(floor((2*n-k-1)/2), n-k).
A polynomial recursion which produces this triangle: p(x, n) = p(x, n - 1) + x^2*p(x, n - 2). - Roger L. Bagula, Apr 27 2008
Sum_{k=0..n} T(n,k)*x^k = A152163(n), A000007(n), A000045(n+1), A026597(n), A122994(n+1), A158608(n), A122995(n+1), A158797(n), A122996(n+1), A158798(n), A158609(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Jun 12 2009
G.f.: (1+(y-1)*x)/(1-x-y^2*x^2). - Philippe Deléham, Mar 09 2012
T(n,k) = T(n-1,k) + T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 09 2012

New name from Peter Luschny, Mar 09 2020

A026581 Expansion of (1 + 2x) / (1 - x - 4x^2).

Original entry on oeis.org

1, 3, 7, 19, 47, 123, 311, 803, 2047, 5259, 13447, 34483, 88271, 226203, 579287, 1484099, 3801247, 9737643, 24942631, 63893203, 163663727, 419236539, 1073891447, 2750837603, 7046403391, 18049753803, 46235367367, 118434382579, 303375852047, 777113382363

Offset: 0

Clark Kimberling

T(n,0) + T(n,1) + ... + T(n,2n), T given by A026568.
Row sums of Riordan array ((1+2x)/(1+x),x(1+2x)/(1+x)). Binomial transform is A055099. - Paul Barry, Jun 26 2008
Equals row sums of triangle A153341. - Gary W. Adamson, Dec 24 2008
Also, the number of walks of length n starting at vertex 0 in the graph with 4 vertices and edges {{0,1}, {0,2}, {0,3}, {1,2}, {2,3}}. - Sean A. Irvine, Jun 02 2025

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4).

Cf. A006131, A026568, A026583, A026597, A026599, A052923, A055099.

Cf. A153341. - Gary W. Adamson, Dec 24 2008

a:=[1,3];; for n in [3..30] do a[n]:=a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Aug 03 2019

I:=[1,3]; [n le 2 select I[n] else Self(n-1) +4*Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 03 2019

CoefficientList[Series[(1+2x)/(1-x-4x^2),{x,0,30}],x] (* or *) LinearRecurrence[{1,4},{1,3},30] (* Harvey P. Dale, Aug 04 2015 *)

Vec((1+2*x)/(1-x-4*x^2) + O(x^30)) \\ Colin Barker, Dec 22 2016

((1+2*x)/(1-x-4*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

G.f.: (1 + 2*x) / (1 - x - 4*x^2).
a(n) = a(n-1) + 4*a(n-2), n>1.
a(n) = 2*A006131(n-1) + A006131(n), n>0.
a(n) = (2^(-1-n)*((1-sqrt(17))^n*(-5+sqrt(17)) + (1+sqrt(17))^n*(5+sqrt(17))))/sqrt(17). - Colin Barker, Dec 22 2016

Edited by Ralf Stephan, Jul 20 2013

A285397 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 3; a(n) is the number of cells after n iterations.

Original entry on oeis.org

1, 26, 646, 15818, 385822, 9401330, 229023958, 5578844858, 135894050926, 3310204057250, 80632220390758, 1964094376340522, 47842741143064894, 1165385872796078546, 28387257791866411894, 691476036231391881242, 16843441238514542846350, 410283940250387099210114

Offset: 0

Peter Karpov, Apr 23 2017

Cell configuration converges to a fractal with dimension 2.906...

Colin Barker, Table of n, a(n) for n = 0..700
Peter Karpov, InvMem, Item 26
Index entries for linear recurrences with constant coefficients, signature (32,-195,216).

Cf. A007482, A026597, A285391, A285392, A285393, A285394, A285395, A285396, A285398, A285399, A285400.

I:=[1, 26, 646]; [n le 3 select I[n] else 32*Self(n-1) - 195*Self(n-2) + 216*Self(n-3) : n in [1..41]]; // G. C. Greubel, Dec 09 2021

LinearRecurrence[{32, -195, 216}, {1, 26, 646}, 18]

Vec((1 - 3*x)^2 / (1 - 32*x + 195*x^2 - 216*x^3) + O(x^20)) \\ Colin Barker, Apr 23 2017

def A285397_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-6*x+9*x^2)/(1-32*x+195*x^2-216*x^3) ).list()
A285397_list(40) # G. C. Greubel, Dec 09 2021

a(0) = 1, a(1) = 26, a(2) = 646, a(n) = 28*a(n-1) - 195*a(n-2) + 216*a(n-3).

G.f.: (1-6*x+9*x^2)/(1-32*x+195*x^2-216*x^3).

A285398 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 0; a(n) is the number of cells after n iterations.

Original entry on oeis.org

1, 19, 452, 10948, 266300, 6484372, 157936172, 3847025764, 93707895260, 2282596837492, 55601016789068, 1354367059315396, 32990588541122684, 803607076375862356, 19574804963320797548, 476816346057854861860, 11614615234500986326556, 282916657894827156657460

Offset: 0

Peter Karpov, Apr 23 2017

Cell configuration converges to a fractal with dimension 2.906...

Colin Barker, Table of n, a(n) for n = 0..700
Peter Karpov, InvMem, Item 26
Peter Karpov, Illustration of initial terms (n = 1..4)
Index entries for linear recurrences with constant coefficients, signature (32,-195,216).

Cf. A007482, A026597, A285391, A285392, A285393, A285394, A285395, A285396, A285397, A285399, A285400.

I:=[19, 452, 10948]; [1] cat [n le 3 select I[n] else 32*Self(n-1) - 195*Self(n-2) + 216*Self(n-3) : n in [1..41]]; // G. C. Greubel, Dec 09 2021

{1}~Join~LinearRecurrence[{32, -195, 216}, {19, 452, 10948}, 17]

Vec((1 - x)*(1 - 3*x)*(1 - 9*x) / (1 - 32*x + 195*x^2 - 216*x^3) + O(x^20)) \\ Colin Barker, Apr 23 2017

def A285398_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-13*x+39*x^2-27*x^3)/(1-32*x+195*x^2-216*x^3) ).list()
A285398_list(40) # G. C. Greubel, Dec 09 2021

a(0) = 1, a(1) = 19, a(2) = 452, a(3) = 10948, a(n) = 28*a(n-1) - 195*a(n-2) + 216*a(n-3).

G.f.: (1-13*x+39*x^2-27*x^3)/(1-32*x+195*x^2-216*x^3).

A285399 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 0 or 2; a(n) is the number of cells after n iterations.

Original entry on oeis.org

1, 13, 182, 2548, 35672, 499408, 6991712, 97883968, 1370375552, 19185257728, 268593608192, 3760310514688, 52644347205632, 737020860878848, 10318292052303872, 144456088732254208, 2022385242251558912, 28313393391521824768, 396387507481305546752

Offset: 0

Peter Karpov, Apr 23 2017

Cell configuration converges to a fractal with dimension 2.402...

Colin Barker, Table of n, a(n) for n = 0..850
Peter Karpov, InvMem, Item 26
Peter Karpov, Illustration of initial terms (n = 1..4)
Index entries for linear recurrences with constant coefficients, signature (14).

Cf. A007482, A026597, A285391, A285392, A285393, A285394, A285395, A285396, A285397, A285398, A285400.

[1] cat [13*14^(n-1): n in [1..40]]; // G. C. Greubel, Dec 09 2021

A285399:=n->13*14^(n-1): 1,seq(A285399(n), n=1..30); # Wesley Ivan Hurt, Apr 23 2017

{1}~Join~LinearRecurrence[{14}, {13}, 18]

Vec((1-x) / (1-14*x) + O(x^20)) \\ Colin Barker, Apr 23 2017

[1]+[13*14^(n-1) for n in (1..40)] # G. C. Greubel, Dec 09 2021

a(0) = 1, a(1) = 13, a(n) = 14*a(n-1).
G.f.: (1-x)/(1-14*x).
a(n) = 13 * 14^(n-1) for n>0. - Colin Barker, Apr 23 2017
E.g.f.: (1 + 13*exp(14*x))/14. - G. C. Greubel, Dec 09 2021

A285400 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 0 or 3; a(n) is the number of cells after n iterations.

Original entry on oeis.org

1, 18, 378, 7938, 166698, 3500658, 73513818, 1543790178, 32419593738, 680811468498, 14297040838458, 300237857607618, 6304995009759978, 132404895204959538, 2780502799304150298, 58390558785387156258, 1226201734493130281418, 25750236424355735909778

Offset: 0

Peter Karpov, Apr 23 2017

Cell configuration converges to a fractal with dimension 2.771...

Colin Barker, Table of n, a(n) for n = 0..750
Peter Karpov, InvMem, Item 26
Peter Karpov, Illustration of initial terms (n = 1..4)
Index entries for linear recurrences with constant coefficients, signature (21).

Cf. A007482, A026597, A285391, A285392, A285393, A285394, A285395, A285396, A285397, A285398, A285399.

[1] cat [18*21^(n-1): n in [1..40]]; // G. C. Greubel, Dec 09 2021

{1}~Join~LinearRecurrence[{21}, {18}, 17]

Vec((1-3*x) / (1-21*x) + O(x^20)) \\ Colin Barker, Apr 23 2017

[1]+[18*21^(n-1) for n in (1..40)] # G. C. Greubel, Dec 09 2021

a(0) = 1, a(1) = 18, a(n) = 21*a(n-1).
G.f.: (1-3*x)/(1-21*x).
a(n) = 2 * 3^(n+1) * 7^(n-1) for n>0. - Colin Barker, Apr 23 2017
E.g.f.: (1 + 6*exp(21*x))/7. - G. C. Greubel, Dec 09 2021

A122995 Expansion of x(1+4x)/(1-x-25*x^2).

Original entry on oeis.org

1, 5, 30, 155, 905, 4780, 27405, 146905, 832030, 4504655, 25305405, 137921780, 770556905, 4218601405, 23482524030, 128947559155, 716010659905, 3939699638780, 21839966136405, 120332457105905, 666331610516030, 3674643038163655, 20332933301064405, 112199009255155780

Offset: 1

Roger L. Bagula, Sep 22 2006

The sequence can also be generated by adding of the top-row elements of the (n-1)st power of the matrix [[0,1],[1,1/5]] and multiplying with 5^(n-1).

Nathaniel Johnston, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,25)

Cf. A026597.

M := {{0, 1}, {1, 1/5}}; v[1] = {1, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[v[n][[1]]*5^(n - 1), {n, 1, 30}]
LinearRecurrence[{1,25},{1,5},30] (* Harvey P. Dale, Mar 11 2017 *)

Vec(x*(1+4*x)/(1-x-25*x^2) + O(x^30)) \\ Michel Marcus, Jan 28 2015

a(n) = a(n-1)+25*a(n-2). [Philippe Deléham, Mar 26 2009]
a(n) = A122999(n-1) + 4*A122999(n-2). [R. J. Mathar, Aug 12 2009]
a(n) = (1/2+9*sqrt(101)/202)*(1/2+sqrt(101)/2)^(n-1) + (1/2-9*sqrt(101)/202)*(1/2-sqrt(101)/2)^(n-1). [Antonio Alberto Olivares, Jun 07 2011]

Replaced definition by a specific one - The Assoc. Eds. of the OEIS, Jun 07 2010

More terms from Michel Marcus, Jan 28 2015

A180168 a(n) = 2a(n-1) + 5a(n-2), a(0) = 1, a(1) = 3.

Original entry on oeis.org

1, 3, 11, 37, 129, 443, 1531, 5277, 18209, 62803, 216651, 747317, 2577889, 8892363, 30674171, 105810157, 364991169, 1259033123, 4343022091, 14981209797, 51677530049, 178261109083, 614909868411, 2121125282237, 7316799906529, 25239226224243, 87062451981131

Offset: 0

Gary W. Adamson, Aug 14 2010

			a(5) = 443 = 2*a(4) + 5*a(3) = 2*129 + 5*37.
Using the INVERT operation, a(4) = 129 = (38, 14, 6, 2, 1) dot (1, 1, 3, 11, 37)
= (38 + 14 + 18 + 22 + 37); where A026597 = (1, 2, 6, 14, 38, 94,...).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,5).

Cf. A026597.

LinearRecurrence[{2, 5}, {1, 3}, 50] (* G. C. Greubel, Feb 18 2017 *)

x='x+O('x^25); Vec((1 + x)/(1 - 2*x - 5*x^2)) \\ G. C. Greubel, Feb 18 2017

G.f.: (1 + x)/(1 - 2*x - 5*x^2).
Equals INVERT transform of A026597: (1, 2, 6, 14, 38, 94,...).
a(n) = (1/6)*( -(1-sqrt(6))^n*sqrt(6) + sqrt(6)*(1+sqrt(6))^n + 3*(1-sqrt(6))^n + 3*(1 +sqrt(6))^n ). - Alexander R. Povolotsky, Aug 15 2010
a(n) = A176812(n)/3 = A002532(n) + A002532(n+1). - R. J. Mathar, Oct 11 2011

cp's OEIS Frontend

A027285 a(n) = Sum_{k=0..2*n-3} T(n,k) * T(n,k+3), with T given by A026584.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Sage

Formula

Extensions

A159612 INVERT transform of (1, 3, 1, 3, 1, ...).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A103631 Triangle read by rows: T(n,k) = abs(qStirling2(n,k,q)) for q = -1, with 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Sage

Formula

Extensions

A026581 Expansion of (1 + 2*x) / (1 - x - 4*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

Extensions

A285397 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 3; a(n) is the number of cells after n iterations.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A285398 Start with a single cell at coordinates (0, 0, 0), then iteratively subdivide the grid into 3 X 3 X 3 cells and remove the cells whose sum of modulo 2 coordinates is 0; a(n) is the number of cells after n iterations.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

A027285 a(n) = Sum_{k=0..2n-3} T(n,k) T(n,k+3), with T given by A026584.

A026581 Expansion of (1 + 2x) / (1 - x - 4x^2).

A122995 Expansion of x(1+4x)/(1-x-25*x^2).

A180168 a(n) = 2a(n-1) + 5a(n-2), a(0) = 1, a(1) = 3.