A015445 -id:A015445 - cp's OEIS frontend

A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

1, 1, 2, 1, 3, 3, 1, 4, 5, 5, 1, 5, 7, 11, 8, 1, 6, 9, 19, 21, 13, 1, 7, 11, 29, 40, 43, 21, 1, 8, 13, 41, 65, 97, 85, 34, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1, 10, 17, 71, 133, 301, 441, 508, 341, 89, 1, 11, 19, 89, 176, 463, 781, 1165, 1159, 683, 144, 1, 12, 21, 109, 225, 673

Offset: 1

R. H. Hardin, Jul 24 2011

Transposed variant of A083856. - R. J. Mathar, Aug 23 2011

As to the sequences by columns beginning (1, N, ...), let m = (N-1). The g.f. for the sequence (1, N, ...) is 1/(1 - x - m*x^2). Alternatively, the corresponding matrix generator is [[1,1], [m,0]]. Another equivalency is simply: The sequence beginning (1, N, ...) is the INVERT transform of (1, m, 0, 0, 0, ...). Convergents to the sequences a(n)/a(n-1) are (1 + sqrt(4*m+1))/2. - Gary W. Adamson, Feb 25 2014

			Array T(n,k) (with rows n >= 1 and column k >= 1) begins as follows:
  ..1...1....1....1.....1.....1.....1......1......1......1......1......1...
  ..2...3....4....5.....6.....7.....8......9.....10.....11.....12.....13...
  ..3...5....7....9....11....13....15.....17.....19.....21.....23.....25...
  ..5..11...19...29....41....55....71.....89....109....131....155....181...
  ..8..21...40...65....96...133...176....225....280....341....408....481...
  .13..43...97..181...301...463...673....937...1261...1651...2113...2653...
  .21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425...
  .34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261...
  .55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361...
  .89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493...
  ...
Some solutions for n = 5 and k = 3 with colors = 1, 2, 3 and empty = 0:
..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1

R. H. Hardin, Table of n, a(n) for n = 1..9999
Ron H. Hardin, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.
Robert Israel, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.

Column 1 is A000045(n+1), column 2 is A001045(n+1), column 3 is A006130, column 4 is A006131, column 5 is A015440, column 6 is A015441(n+1), column 7 is A015442(n+1), column 8 is A015443, column 9 is A015445, column 10 is A015446, column 11 is A015447, and column 12 is A053404,
Row 2 is A000027(n+1), row 3 is A004273(n+1), row 4 is A028387, row 5 is A000567(n+1), and row 6 is A106734(n+2).
Diagonal is A171180, superdiagonal 1 is A083859(n+1), and superdiagonal 2 is A083860(n+1).

T:= proc(n,k) option remember; `if`(n<0, 0,
      `if`(n<2 or k=0, 1, k*T(n-2, k) +T(n-1, k)))
    end;
seq(seq(T(n, d+1-n), n=1..d), d=1..12); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2 || k == 0, 1, k*T[n-2, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. Thus, T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n = 0, 1, ..., z-1.
The solution is T(n,k) = Sum_r r^(-n-1)/(1 + z*k*r^(z-1)), where the sum is over the roots r of the polynomial k*x^z + x - 1.
For z = 2, T(n,k) = ((2*k / (sqrt(1 + 4*k) - 1))^(n+1) - (-2*k/(sqrt(1 + 4*k) + 1))^(n+1)) / sqrt(1 + 4*k).
T(n,k) = Sum_{s=0..[n/2]} binomial(n-s,s) * k^s.
For z X 1 tiles, T(n,k,z) = Sum_{s = 0..[n/z]} binomial(n-(z-1)*s, s) * k^s. - R. H. Hardin, Jul 31 2011

Formula and proof from Robert Israel in the Sequence Fans mailing list.

A122994 a(n) = a(n-1)+9*a(n-2) initialized with a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 12, 39, 147, 498, 1821, 6303, 22692, 79419, 283647, 998418, 3551241, 12537003, 44498172, 157331199, 557814747, 1973795538, 6994128261, 24758288103, 87705442452, 310530035379, 1099879017447, 3894649335858, 13793560492881, 48845404515603, 172987448951532

Offset: 0

Roger L. Bagula, Sep 22 2006

The two roots of the denominator of the g.f. (for Binet's formula) are -0.393486... and 0.2823756...

Pisano period lengths: 1, 3, 1, 6, 6, 3, 6, 12, 1, 6, 10, 6, 84, 6, 6, 24,144, 3, 72, 6,... - R. J. Mathar, Aug 10 2012

Index entries for linear recurrences with constant coefficients, signature (1,9).

Cf. A026597.

CoefficientList[Series[(-2 z - 1)/(9 z^2 + z - 1), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)

G.f.: -(1+2*x)/(-1+x+9*x^2). a(n) = A015445(n)+2*A015445(n-1). [R. J. Mathar, Aug 12 2009]
a(n) = (1/2+5*sqrt(37)/74) *(1/2+sqrt(37)/2)^(n-1) +(1/2-5*sqrt(37)/74) *(1/2-sqrt(37)/2)^(n-1). [Antonio Alberto Olivares, Jun 07 2011]
a(n) = Sum_{k, 0<=k<=n} A103631(n,k)*3^k. - Philippe Deléham, Dec 17 2011
a(n) = A015445(n) + 2*A015445(n-1), n>0. - Ralf Stephan, Jul 21 2013

Definition replaced with the Deleham recurrence of Mar 2009 by the Assoc. Editors of the OEIS, Mar 12 2010

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 44, 312, 2224, 15840, 112832, 803712, 5724928, 40779264, 290475008, 2069084160, 14738305024, 104982503424, 747801460736, 5326668791808, 37942424436736, 270267896954880, 1925146777223168, 13713023838978048, 97679317251653632, 695780094221746176

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,8).

Sequences of the form a(n) = c*a(n-1) + d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A000045,A001045,A006130,A006131,A015440,A015441,A015442,A015443,A015445,A015446
.A000129,A002605,A015518,A063727,A002532,A083099,A015519,A003683,A002534,A083102
.A006190,A007482,A030195,A015521,A015523,A083858,A015524,A015525,A099012,A015528
.A001076,A090017,A015530,A057087,A015531,A085939,A015532,A180222,A015533,A180226
.A052918,A015535,A015536,A015537,A057088,A015540,A015541,A015544,A015545,A180250
.A005668,A135030,A090018,A135032,A015551,A057089,A015552,A189800,A189801,A190005
.A054413,A015555,A015559,A015561,A015562,A015564,A057090,A015565,A015566,A015568
.A041025,A190331,A015574,A190510,A015575,A190560,A015576,A057091,A015577,A190953
.A099371,A015579,A181353,A015580,A015581,A153191,A015583,A015584,A057092,A015585
A041041,A191014,A015588,A190954,A190955,A190956,A015589,A190957,A015591,A057093

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

LinearRecurrence[{6, 8}, {0, 1}, 50]
CoefficientList[Series[-(x/(-1+6 x+8 x^2)),{x,0,50}],x] (* Harvey P. Dale, Jul 26 2011 *)

a(n)=([0,1; 8,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

G.f.: x/(1 - 2*x*(3+4*x)). - Harvey P. Dale, Jul 26 2011

A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 5, 5, 4, 1, 1, 0, 1, 8, 11, 7, 5, 1, 1, 0, 1, 13, 21, 19, 9, 6, 1, 1, 0, 1, 21, 43, 40, 29, 11, 7, 1, 1, 0, 1, 34, 85, 97, 65, 41, 13, 8, 1, 1, 0, 1, 55, 171, 217, 181, 96, 55, 15, 9, 1, 1, 0, 1, 89, 341, 508, 441, 301, 133, 71, 17, 10, 1, 1, 0

Offset: 0

Henry Bottomley, May 10 2001

			Square array begins as:
  0, 1, 1, 1,  1,  1,  1, ...
  0, 1, 1, 2,  3,  5,  8, ...
  0, 1, 1, 3,  5, 11, 21, ...
  0, 1, 1, 4,  7, 19, 40, ...
  0, 1, 1, 5,  9, 29, 65, ...
  0, 1, 1, 6, 11, 41, 96, ...

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows include A057427, A000045, A001045, A006130, A006131, A015440, A015441, A015442, A015443, A015445, A015446, A015447, A053404.

Columns include A000004, A000012 (twice), A000027, A000567 A005408, A028387.

Flat(List([0..12], n-> List([0..n], k-> (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k))/Sqrt(1+4*k) ))); # G. C. Greubel, Jan 15 2020

[Round( (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k) )/Sqrt(1+4*k) ): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020

seq(seq( round((((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020

T[n_, k_]:= If[n==k==0, 0, Round[(((1+Sqrt[1+4n])/2)^k - ((1-Sqrt[1+4n])/2)^k)/Sqrt[1+4n]]]; Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 15 2020 *)

T(n,k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n);
for(n=0,12, for(k=0,n, print1( round(T(k,n-k)), ", "))) \\ G. C. Greubel, Jan 15 2020

[[ round( (((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020

T(n, k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n).

A097041 Expansion of (1+x)/(1-x^2-9*x^3).

Original entry on oeis.org

1, 1, 1, 10, 10, 19, 100, 109, 271, 1009, 1252, 3448, 10333, 14716, 41365, 107713, 173809, 479998, 1143226, 2044279, 5463208, 12333313, 23861719, 61502185, 134861536, 276257656, 688381201, 1490011480, 3174700105, 7685442289

Offset: 0

Paul Barry, Jul 20 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,9).

Cf. A015445.

CoefficientList[Series[(1+x)/(1-x^2-9x^3),{x,0,30}],x] (* or *) LinearRecurrence[{0,1,9},{1,1,1},30] (* Harvey P. Dale, Mar 12 2015 *)

x='x+O('x^50); Vec((1+x)/(1-x^2-9*x^3)) \\ G. C. Greubel, Apr 30 2017

a(n) = a(n-2) + 9*a(n-3).

a(n) = Sum_{k=0..floor(n/2)} binomial(floor((n-k)/2), k)*9^k.

A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

Original entry on oeis.org

1, 2, 1, 3, 4, 4, 1, 10, 5, 6, 20, 6, 1, 21, 35, 7, 8, 56, 56, 8, 1, 36, 126, 84, 9, 10, 120, 252, 120, 10, 1, 55, 330, 462, 165, 11, 12, 220, 792, 792, 220, 12, 1, 78, 715, 1716, 1287, 286, 13, 14, 364, 2002, 3432, 2002, 364, 14, 1, 105, 1365, 5005, 6435, 3003, 455, 15

Offset: 1

Philippe Deléham, Aug 27 2005

The same as A119900 without 0's. A reflected version of A034867 or A202064. - Alois P. Heinz, Feb 07 2014
From Vladimir Shevelev, Feb 07 2014: (Start)
Also table of coefficients of polynomials  P_1(x)=1, P_2(x)=2, for n>=2, P_(n+1)(x) = 2*P_n(x)+(x-1)* P_(n-1)(x).  The polynomials P_n(x)/2^(n-1) are connected with sequences A000045 (x=5), A001045 (x=9), A006130 (x=13), A006131 (x=17), A015440 (x=21), A015441 (x=25), A015442 (x=29), A015443 (x=33), A015445 (x=37), A015446 (x=41), A015447 (x=45), A053404 (x=49); also the polynomials P_n(x) are connected with sequences A000129, A002605, A015518, A063727, A085449, A002532, A083099, A015519, A003683, A002534, A083102, A015520. (End)

			Starred terms in Pascal's triangle (A007318), read by rows:
1;
1*, 1;
1, 2*, 1;
1*, 3, 3*, 1;
1, 4*, 6, 4*, 1;
1*, 5, 10*, 10, 5*, 1;
1, 6*, 15, 20*, 15, 6*, 1;
1*, 7, 21*, 35, 35*, 21, 7*, 1;
1, 8*, 28, 56*, 70, 56*, 28, 8*, 1;
1*, 9, 36*, 84, 126*, 126, 84*, 36, 9*, 1;
Triangle T(n,k) begins:
1;
2;
1,    3;
4,    4;
1,   10,  5;
6,   20,  6;
1,   21,  35,   7;
8,   56,  56,   8;
1,   36, 126,  84,  9;
10, 120, 252, 120, 10;

Alois P. Heinz, Rows n = 1..200, flattened

Cf. A109446.

T:= (n, k)-> binomial(n, 2*k+1-irem(n, 2)):
seq(seq(T(n, k), k=0..ceil((n-2)/2)), n=1..20);  # Alois P. Heinz, Feb 07 2014

Flatten[ Table[ If[ OddQ[n - k], Binomial[n, k], {}], {n, 0, 15}, {k, 0, n}]] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Aug 30 2005

Corrected offset by Alois P. Heinz, Feb 07 2014

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 5, 1, 8, 10, 3, 13, 20, 9, 1, 21, 38, 22, 4, 34, 71, 51, 14, 1, 55, 130, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Emeric Deutsch, Feb 18 2007

Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 24 2012
Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012
Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014
Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - Emeric Deutsch, Aug 12 2014
Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - Clark Kimberling, Oct 22 2014

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)

C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

G.f.: 1/(1-z-(1+t)z^2).
Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - Philippe Deléham, Jan 24 2012
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - Emeric Deutsch, Aug 12 2014

A317051 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 9 * T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, 9, 1, 18, 1, 27, 81, 1, 36, 243, 1, 45, 486, 729, 1, 54, 810, 2916, 1, 63, 1215, 7290, 6561, 1, 72, 1701, 14580, 32805, 1, 81, 2268, 25515, 98415, 59049, 1, 90, 2916, 40824, 229635, 354294, 1, 99, 3645, 61236, 459270, 1240029, 531441, 1, 108, 4455, 87480, 826686, 3306744, 3720087

Offset: 0

Zagros Lalo, Jul 20 2018

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013616 ((1+9*x)^n) and along skew diagonals pointing top-left in center-justified triangle given in A038291 ((9+x)^n).
The coefficients in the expansion of 1/(1-x-9*x^2) are given by the sequence generated by the row sums.
The row sums are Generalized Fibonacci numbers (see A015445).
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.5413812651491... ((1+sqrt(37))/2), when n approaches infinity.

			Triangle begins:
  1;
  1;
  1, 9;
  1, 18;
  1, 27, 81;
  1, 36, 243;
  1, 45, 486, 729;
  1, 54, 810, 2916;
  1, 63, 1215, 7290, 6561;
  1, 72, 1701, 14580, 32805;
  1, 81, 2268, 25515, 98415, 59049;
  1, 90, 2916, 40824, 229635, 354294;
  1, 99, 3645, 61236, 459270, 1240029, 531441;
  1, 108, 4455, 87480, 826686, 3306744, 3720087;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 100

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 9x)^n
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (9 + x)^n

Row sums give A015445.

Cf. A013616, A038291.

Flat(List([0..13],n->List([0..Int(n/2)],k->9^k*Binomial(n-k,k)))); # Muniru A Asiru, Jul 20 2018

/* As triangle */ [[9^k*Binomial(n-k,k): k in [0..Floor(n/2)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 9 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten
Table[9^k Binomial[n - k, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten

T(n, k) = 9^k*binomial(n-k,k);
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018

T(n,k) = 9^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

A374749 Number of zygodrome numbers in base-10 (A033023) with n or fewer digits.

Original entry on oeis.org

0, 9, 18, 108, 279, 1260, 3780, 15129, 49158, 185328, 627759, 2295720, 7945560, 28607049, 100117098, 357580548, 1258634439, 4476859380, 15804569340, 56096303769, 198337427838, 703204161768, 2488241012319, 8817078468240, 31211247579120, 110564953793289

Offset: 1

Ruud H.G. van Tol, Sep 06 2024

Maarten Looijen, Over getallen gesproken - Talking about numbers, Van Haren Publishing, 2018, page 513.

Giovanni Resta, Zygodrome.
Index entries for linear recurrences with constant coefficients, signature (2,8,-9).

Cf. A015445, A033023.

LinearRecurrence[{1, 9}, {1, 10}, 26] - 1 (* Amiram Eldar, Sep 06 2024 *)

PARI
```
a(n) = ([0, 1; 9, 1]^n)[2, 2] - 1;
```

a(n) = A015445(n) - 1.
From Stefano Spezia, Sep 07 2024: (Start)
G.f.: 9*x^2/((1 - x)*(1 - x - 9*x^2)).
E.g.f.: exp(x/2)*(cosh(sqrt(37)*x/2) + sinh(sqrt(37)*x/2)/sqrt(37)) - exp(x). (End)

cp's OEIS Frontend

A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

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A122994 a(n) = a(n-1)+9*a(n-2) initialized with a(0)=1, a(1)=3.

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A189800 a(n) = 6*a(n-1) + 8*a(n-2), with a(0)=0, a(1)=1.

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A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

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A097041 Expansion of (1+x)/(1-x^2-9*x^3).

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A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

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A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

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A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.