A086901 -id:A086901 - cp's OEIS frontend

A122542 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 2, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

1, 0, 1, 0, 2, 1, 0, 2, 4, 1, 0, 2, 8, 6, 1, 0, 2, 12, 18, 8, 1, 0, 2, 16, 38, 32, 10, 1, 0, 2, 20, 66, 88, 50, 12, 1, 0, 2, 24, 102, 192, 170, 72, 14, 1, 0, 2, 28, 146, 360, 450, 292, 98, 16, 1, 0, 2, 32, 198, 608, 1002, 912, 462, 128, 18, 1

Offset: 0

Philippe Deléham, Sep 19 2006, May 28 2007

Riordan array (1, x*(1+x)/(1-x)). Rising and falling diagonals are the tribonacci numbers A000213, A001590.

			Triangle begins:
  1;
  0, 1;
  0, 2,  1;
  0, 2,  4,   1;
  0, 2,  8,   6,   1;
  0, 2, 12,  18,   8,    1;
  0, 2, 16,  38,  32,   10,   1;
  0, 2, 20,  66,  88,   50,  12,   1;
  0, 2, 24, 102, 192,  170,  72,  14,   1;
  0, 2, 28, 146, 360,  450, 292,  98,  16,  1;
  0, 2, 32, 198, 608, 1002, 912, 462, 128, 18, 1;

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Sect. 2.3.
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 1.

Other versions: A035607, A113413, A119800, A266213.
Diagonals: A000012, A005843, A001105, A035597 - A035606.
Columns: A000007, A040000, A008575, A005899, A008412-A008416, A008418, A008420, A035706 - A035745.
Sums include: A000007, A001333 (row), A001590 (diagonal), A007483, A057077 (signed row), A078016 (signed diagonal), A086901, A091928, A104934, A122558, A122690.
Cf. A155161, A059283.

a122542 n k = a122542_tabl !! n !! k
a122542_row n = a122542_tabl !! n
a122542_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [0, 1])
-- Reinhard Zumkeller, Jul 20 2013, Apr 17 2013

function T(n, k) // T = A122542
  if k eq 0 then return 0^n;
  elif k eq n then return 1;
  else return T(n-1,k) + T(n-1,k-1) + T(n-2,k-1);
  end if;
end function;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 27 2024

CoefficientList[#, y]& /@ CoefficientList[(1-x)/(1 - (1+y)x - y x^2) + O[x]^11, x] // Flatten (* Jean-François Alcover, Sep 09 2018 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[k==n, 1, If[k==0, 0, T[n-1,k-1] +T[n-1,k] +T[n-2,k- 1] ]]; (* T = A122542 *)
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 27 2024 *)

def A122542_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)+2*sum(prec(n-i,k-1) for i in (2..n-k+1))
    return [prec(n, k) for k in (0..n)]
for n in (0..10): print(A122542_row(n)) # Peter Luschny, Mar 16 2016

Sum_{k=0..n} x^k*T(n,k) = A000007(n), A001333(n), A104934(n), A122558(n), A122690(n), A091928(n)  for x = 0, 1, 2, 3, 4, 5. - Philippe Deléham, Jan 25 2012
Sum_{k=0..n} 3^(n-k)*T(n,k) = A086901(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A007483(n-1), n >= 1. - Philippe Deléham, Oct 08 2006
T(2*n, n) = A123164(n).
T(n, k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1), n > 1. - Philippe Deléham, Jan 25 2012
G.f.: (1-x)/(1-(1+y)*x-y*x^2). - Philippe Deléham, Mar 02 2012
From G. C. Greubel, Oct 27 2024: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A057077(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A001590(n+1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A078016(n). (End)

A015530 Expansion of x/(1 - 4x - 3x^2).

Original entry on oeis.org

0, 1, 4, 19, 88, 409, 1900, 8827, 41008, 190513, 885076, 4111843, 19102600, 88745929, 412291516, 1915403851, 8898489952, 41340171361, 192056155300, 892245135283, 4145149007032, 19257331433977, 89464772757004

Offset: 0

Olivier Gérard

Let b(1)=1, b(k) = floor(b(k-1)) + 3/b(k-1); then for n>1, b(n) = a(n)/a(n-1). - Benoit Cloitre, Sep 09 2002
In general, x/(1 - a*x - b*x^2) has a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1,k)*b^k*a^(n-2k-1). - Paul Barry, Apr 23 2005
Pisano period lengths: 1, 2, 1, 4, 24, 2, 21, 4, 3, 24, 40, 4, 84, 42, 24, 8, 288, 6, 18, 24, ... . - R. J. Mathar, Aug 10 2012
This is the Lucas sequence U(4,-3). - Bruno Berselli, Jan 09 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lucyna Trojnar-Spelina and Iwona Włoch, On Generalized Pell and Pell-Lucas Numbers, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
Wikipedia, Lucas sequence.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Appears in A179596, A126473 and A179597. - Johannes W. Meijer, Aug 01 2010

Cf. A080042: Lucas sequence V(4,-3).

I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 19 2012

LinearRecurrence[{4,3},{0,1},30] (* Vincenzo Librandi, Jun 19 2012 *)

x='x+O('x^30); concat([0], Vec(x/(1-4*x-3*x^2))) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,4,-3) for n in range(0, 23)]# Zerinvary Lajos, Apr 23 2009

a(n) = 4*a(n-1) + 3*a(n-2).
a(n) = (A086901(n+2) - A086901(n+1))/6. - Ralf Stephan, Feb 01 2004
a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1, k)*3^k*4^(n-2k-1). - Paul Barry, Apr 23 2005
a(n) = ((2+sqrt(7))^n - (2-sqrt(7))^n)/sqrt(28). Offset 1. a(3)=19. - Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit(a(n+k)/a(k), k=infinity) = A108851(n)+a(n)*sqrt(7).
Limit(A108851(n)/a(n), n=infinity) = sqrt(7). (End)
G.f.: x*G(0) where G(k)= 1 + (4*x+3*x^2)/(1 - (4*x+3*x^2)/(4*x + 3*x^2 + 1/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 28 2012
G.f.: G(0)*x/(2-4*x), where G(k)= 1 + 1/(1 - x*(7*k-4)/(x*(7*k+3) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013

A102900 a(n) = 3a(n-1) + 4a(n-2), a(0)=a(1)=1.

Original entry on oeis.org

1, 1, 7, 25, 103, 409, 1639, 6553, 26215, 104857, 419431, 1677721, 6710887, 26843545, 107374183, 429496729, 1717986919, 6871947673, 27487790695, 109951162777, 439804651111, 1759218604441, 7036874417767, 28147497671065

Offset: 0

Paul Barry, Jan 17 2005

Binomial transform of A102901.

Hankel transform is = 1,6,0,0,0,0,0,0,0,0,0,0,... - Philippe Deléham, Nov 02 2008

Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, A Meta-Algorithm for Creating Fast Algorithms for Counting ON Cells in Odd-Rule Cellular Automata, arXiv:1503.01796 [math.CO], 2015; see also the Accompanying Maple Package.
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, Odd-Rule Cellular Automata on the Square Grid, arXiv:1503.04249 [math.CO], 2015.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A001045, A004171, A046717, A086901, A102901, A247666 (which appears to be the run length transform of this sequence).

a102900 n = a102900_list !! n
a102900_list = 1 : 1 : zipWith (+)
               (map (* 4) a102900_list) (map (* 3) $ tail a102900_list)
-- Reinhard Zumkeller, Feb 13 2015

[n le 2 select 1 else 3*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 28 2015

a[n_]:=(MatrixPower[{{2,2},{3,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{3, 4}, {1, 1}, 30] (* Vincenzo Librandi, Dec 28 2015 *)

a(n)=([0,1; 4,3]^n*[1;1])[1,1] \\ Charles R Greathouse IV, Mar 28 2016

A102900=BinaryRecurrenceSequence(3,4,1,1)
[A102900(n) for n in range(51)] # G. C. Greubel, Dec 09 2022

G.f.: (1-2*x)/(1-3*x-4*x^2).
a(n) = (2*4^n + 3*(-1)^n)/5.
a(n) = ceiling(4^n/5) + floor(4^n/5) = (ceiling(4^n/5))^2 - (floor(4^n/5))^2.
a(n) + a(n+1) = 2^(2*n+1) = A004171(n).
a(n) = Sum_{k=0..n} binomial(2*n-k, 2*k)*2^k. - Paul Barry, Jan 20 2005
a(n) = upper left term in the 2 X 2 matrix [1,3; 2,2]^n. - Gary W. Adamson, Mar 14 2008
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(8*4^k-3*(-1)^k)/(x*(8*4^k-3*(-1)^k) + (2*4^k+3*(-1)^k)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = 2^(2*n-1) - a(n-1), a(1)=1. - Ben Paul Thurston, Dec 27 2015; corrected by Klaus Purath, Aug 02 2020
From Klaus Purath, Aug 02 2020: (Start)
a(n) = 4*a(n-1) + 3*(-1)^n.
a(n) = 6*4^(n-2) + a(n-2), n>=2. (End)

A234122 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with every 2X2 subblock having the absolute values of all six edge and diagonal differences no larger than 1.

Original entry on oeis.org

31, 145, 145, 673, 1361, 673, 3127, 12593, 12593, 3127, 14527, 116801, 231713, 116801, 14527, 67489, 1082977, 4279065, 4279065, 1082977, 67489, 313537, 10041953, 79003521, 157630963, 79003521, 10041953, 313537, 1456615, 93113761

Offset: 1

R. H. Hardin, Dec 19 2013

Table starts
.......31.........145.............673...............3127.................14527
......145........1361...........12593.............116801...............1082977
......673.......12593..........231713............4279065..............79003521
.....3127......116801.........4279065..........157630963............5807422543
....14527.....1082977........79003521.........5807422543..........427196005695
....67489....10041953......1458813409.......214027901025........31446640848897
...313537....93113761.....26937444801......7888454356625......2315408571668225
..1456615...863396401....497411686793....290756314787875....170502665692732079
..6767071..8005833073...9184935953377..10716964158533127..12556134956123911615
.31438129.74233997105.169604155276817.395017615132720993.924677153131389366689

			Some solutions for n=2 k=4
..0..1..0..0..1....0..1..2..2..2....1..2..1..2..2....0..1..0..1..2
..1..0..1..0..1....1..1..1..1..2....1..2..2..2..2....1..0..0..1..2
..1..1..0..0..1....1..2..2..1..2....2..1..1..1..2....0..1..0..1..1

R. H. Hardin, Table of n, a(n) for n = 1..364

Column 1 is A086901(n+3)

Empirical for column k:
k=1: a(n) = 4*a(n-1) +3*a(n-2)
k=2: a(n) = 10*a(n-1) -4*a(n-2) -26*a(n-3) +5*a(n-4)
k=3: a(n) = 20*a(n-1) -10*a(n-2) -324*a(n-3) -277*a(n-4) +144*a(n-5)
k=4: [order 11]
k=5: [order 17]
k=6: [order 35]
k=7: [order 62]

A154246 a(n) = ( (5 + sqrt(7))^n - (5 - sqrt(7))^n )/(2*sqrt(7)).

Original entry on oeis.org

1, 10, 82, 640, 4924, 37720, 288568, 2206720, 16872976, 129008800, 986374432, 7541585920, 57661119424, 440862647680, 3370726327168, 25771735613440, 197044282245376, 1506551581411840, 11518718733701632, 88069258871603200

Offset: 1

Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009

Second binomial transform of A086901 without initial term 1.

Lim_{n -> infinity} a(n)/a(n-1) = 5 + sqrt(7) = 7.6457513110....

G. C. Greubel, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (10,-18).

Cf. A010465 (decimal expansion of square root of 7), A086901.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-7); S:=[((5+r)^n-(5-r)^n)/(2*r): n in [1..25]]; [Integers()!S[j]: j in [1..#S]]; // Klaus Brockhaus, Jan 07 2009

I:=[1,10]; [n le 2 select I[n] else 10*Self(n-1)-18*Self(n-2): n in [1..25]]; // Vincenzo Librandi, Sep 08 2016

Table[Simplify[((5+Sqrt[7])^n -(5-Sqrt[7])^n)/(2*Sqrt[7])], {n,1,25}] (* or *) LinearRecurrence[{10, -18}, {1, 10}, 25] (* G. C. Greubel, Sep 07 2016 *)

my(x='x+O('x^25)); Vec(x/(1-10*x+18*x^2)) \\ G. C. Greubel, May 31 2019

[lucas_number1(n,10,18) for n in range(1, 25)] # Zerinvary Lajos, Apr 26 2009

From Philippe Deléham, Jan 06 2009: (Start)
a(n) = 10*a(n-1) - 18*a(n-2) for n > 1, with a(0)=0, a(1)=1.
G.f.: x/(1 - 10*x + 18*x^2). (End)
E.g.f.: (1/sqrt(7))*exp(5*x)*sinh(sqrt(7)*x). - G. C. Greubel, Sep 07 2016

Extended beyond a(7) by Klaus Brockhaus, Jan 07 2009

Edited by Klaus Brockhaus, Oct 06 2009

cp's OEIS Frontend

A122542 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 2, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

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A015530 Expansion of x/(1 - 4*x - 3*x^2).

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A102900 a(n) = 3*a(n-1) + 4*a(n-2), a(0)=a(1)=1.

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A234122 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with every 2X2 subblock having the absolute values of all six edge and diagonal differences no larger than 1.

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A154246 a(n) = ( (5 + sqrt(7))^n - (5 - sqrt(7))^n )/(2*sqrt(7)).

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A015530 Expansion of x/(1 - 4x - 3x^2).

A102900 a(n) = 3a(n-1) + 4a(n-2), a(0)=a(1)=1.