A029907 -id:A029907 - cp's OEIS frontend

A268249 T(n,k)=Number of nXk 0..k arrays with every repeated value in every row greater than, and every column greater than or equal to, the previous repeated value.

2, 9, 4, 60, 81, 8, 570, 3600, 729, 15, 6960, 324900, 216000, 6084, 28, 104006, 48441600, 185193000, 12067722, 49284, 51, 1837752, 10817248036, 337153536000, 98956027822, 648144348, 386884, 92, 37478718, 3377332413504, 1125058699232216

Offset: 1

R. H. Hardin, Jan 29 2016

Table starts
...2........9............60...............570.................6960
...4.......81..........3600............324900.............48441600
...8......729........216000.........185193000.........337153536000
..15.....6084......12067722.......98956027822.....2213773305829075
..28....49284.....648144348....50740994294843.13958709698552890082
..51...386884...33600202086.25094770808231271
..92..2965284.1691726672094
.164.22268961
.290

			Some solutions for n=2 k=4
..2..3..1..1....3..1..4..3....0..1..0..3....4..0..0..4....0..2..4..3
..1..1..0..4....1..1..3..4....1..0..0..4....0..4..3..1....0..0..1..3

R. H. Hardin, Table of n, a(n) for n = 1..48

Column 1 is A029907(n+1).
Column 2 is A268014.
Column 3 is A267977.
Row 1 is A268205.
Row 2 is A268206.

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
k=2: [order 15]

A102702 Expansion of (2-x-2*x^2-x^3)/(1-x-x^2)^2.

Original entry on oeis.org

2, 3, 6, 10, 18, 31, 54, 93, 160, 274, 468, 797, 1354, 2295, 3882, 6554, 11046, 18587, 31230, 52401, 87812, 146978, 245736, 410425, 684818, 1141611, 1901454, 3164458, 5262330, 8744599, 14521158, 24097797, 39965224, 66241330, 109731132

Offset: 0

Creighton Dement, Feb 04 2005

A floretion-generated sequence which results from a certain transform of the Fibonacci numbers. Specifically, (a(n)) is the (type 1B) tesfor-transform of the Fibonacci numbers (A000045) with respect to the floretion + .5'i + .5i' Note, for example, that the sequence A001629, appearing in the formula given, has the name "Fibonacci numbers convolved with themselves" and that this sequence arises in FAMP (see program code) under the name: the lesfor-transform (type 1B) of the Fibonacci numbers (A000045) with respect to the floretion + .5'i + .5i' . The denominator of the generating function has roots at the golden ratio phi and -(1+phi).
Floretion Algebra Multiplication Program. FAMP Code: (a(n)) = 2tesforseq[ + .5'i + .5i' ], 2lesforseq = A001629, jesforseq = A029907, vesforseq = A000045, ForType: 1B.
a(n) is the total number of parts not greater than 2 among all compositions of n+3 in which only the last part may be equal to 1. - Andrew Yezhou Wang, Jul 14 2019

Thomas Koshy, Fibonacci and Lucas Numbers with Applications, Chapter 15, page 187, "Hosoya's Triangle".
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 183, Nr.(98).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. E. Hoggatt, Jr. and M. Bicknell-Johnson, Fibonacci convolution sequences, Fib. Quart., 15 (1977), 117-122.
Mengmeng Liu and Andrew Yezhou Wang, The number of designated parts in compositions with restricted parts, Journal of Integer Sequences, 23 (2020), Article 20.1.8.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A001629, A029907.

R:=PowerSeriesRing(Integers(), 35); Coefficients(R!( (2-x-2*x^2-x^3)/(1-x-x^2)^2)); // Marius A. Burtea, Dec 31 2019

CoefficientList[Series[(2-x-2x^2-x^3)/(x^4+2x^3-x^2-2x+1),{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2,-1},{2,3,6,10},40] (* Harvey P. Dale, Apr 21 2014 *)

G.f.: (2-x-2*x^2-x^3)/(1-x-x^2)^2.
a(n) = 2*F(n+1) + A001629(n+3) - 2*A029907(n+1);
F(n+1) = a(n+2) - a(n+1) - a(n).
a(0)=2, a(1)=3, a(2)=6, a(3)=10, a(n)=2*a(n-1)+a(n-2)-2*a(n-3)-a(n-4). - Harvey P. Dale, Apr 21 2014
a(n) = A010049(n+1) + A000045(n+2). - R. J. Mathar, May 21 2019
a(n) = ((2*n+10)*F(n+1)-(n-4)*F(n))/5. - Andrew Yezhou Wang, Jul 14 2019

Corrected by T. D. Noe, Nov 02 2006

A211228 Shallow diagonal sums of A211226.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 5, 8, 8, 15, 13, 28, 21, 51, 34, 92, 55, 164, 89, 290, 144, 509, 233, 888, 377, 1541, 610, 2662, 987, 4580, 1597, 7852, 2584, 13419, 4181, 22868, 6765, 38871, 10946, 65920, 17711

Offset: 0

Peter Bala, Apr 05 2012

The even-indexed terms a(2*n) count the compositions of n+2 into odd parts while the odd-indexed terms a(2*n+3) count the total number of parts in the compositions of n+2 into odd parts.

			The compositions of 5 into odd parts are 1+1+1+1+1, 1+1+3, 1+3+1, 3+1+1 and 5. Hence a(6) = 5 and a(9) = 15.

Cf. A000045, A029907, A211226.

Let f(n) := (floor(n/2))! and define c(n,k) = f(n)/(f(k)*f(n-k)) = A211226(n,k). Then a(n) = sum {k = 0..floor(n/2)} c(n-k,k).
a(2*n) = A000045(n+2); a(2*n-1) = A029907(n).
O.g.f.: (1+x-2*x^4-x^5-x^6)/(1-x^2-x^4)^2 = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + ....

A240847 a(n) = 2a(n-1) + a(n-2) - 2a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, -2, -5, -12, -25, -50, -96, -180, -331, -600, -1075, -1908, -3360, -5878, -10225, -17700, -30509, -52390, -89664, -153000, -260375, -442032, -748775, -1265832, -2136000, -3598250, -6052061, -10164540

Offset: 0

Paul Curtz, Apr 13 2014

F1(m, n) is the difference table of a(n):
    0,   0,   1,   0,   1,   0,   0,  -2, ...
    0,   1,  -1,   1,  -1,   0,  -2,  -3, ...
    1,  -2,   2,  -2,   1,  -2,  -1,  -4, ...
   -3,   4,  -4,   3,  -3,   1,  -3,  -2, ...
    7,  -8,   7,  -6,   4,  -4,   1,  -4, ...
  -15,  15, -13,  10,  -8,   5,  -5,   1, ...
   30, -28,  23, -18,  13, -10,   6,  -6, ...
The recurrence holds for every row and every signed column.
Main diagonal: F1(n, n) = A001477(n).
First upper diagonal: F1(n, n+1) = -A001477(n).
F1(m, n) = F1(m, n-1) + F1(m+1, n-1).
Inverse binomial transform: 0, 0, 1, -3, 7, -15, 30, ... = 0, 0, followed by (-1)^n*A023610(n). Without signs: F2(0, n) = 0, 0, 1, 3, 7, 15, 30, ... = b(n) has the same recurrence.
F1(0, n) + F2(0, n) = 0, followed by A099920(n).
a(n) and b(n) are reciprocal by their inverse binomial transform.
0, followed by A001629(n) is an autosequence.
F1(m, 1) = (-1)^n*A029907(n).
F1(1, n) = 0, 1, -1, 1, -1, followed by -A226432(n+3).
F1(m, 2) = (-1)^n*A208354(n).

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

Cf. A000032, A000045, A001629 (main sequence for the recurrence), A067331.

List([0..40], n-> (6*Fibonacci(n-3) - (n-3)*Lucas(1,-1,n-3)[2])/5 ); # G. C. Greubel, Feb 06 2020

[(6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5: n in [0..40]]; // G. C. Greubel, Feb 06 2020

with(combinat): seq( ((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5, n=0..40); # G. C. Greubel, Feb 06 2020

a[n_]:= a[n]= 2*a[n-1] +a[n-2] -2*a[n-3] -a[n-4]; a[0]= a[1]= a[3]= 0; a[2]= 1; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Apr 17 2014 *)
CoefficientList[Series[x^2*(1-2*x)/(1-x-x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+c-2b-a}; NestList[nxt,{0,0,1,0},40][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)

Vec(x^2*(1-2*x)/(1-x-x^2)^2 + O(x^100)) \\ Colin Barker, Apr 13 2014

vector(41, n, my(m=n-1); ((m+3)*fibonacci(m-3) - 2*(m-3)*fibonacci(m-2) )/5 ) \\ G. C. Greubel, Feb 06 2020

[((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5 for n in (0..40)] # G. C. Greubel, Feb 06 2020

a(n) = 0, 0, 1, 0, 1, 0, 0, followed by -A067331.
G.f.: x^2*(1-2*x)/(1-x-x^2)^2. - Colin Barker, Apr 13 2014
a(n) = ( (10*n + (3-5*n)*t)*(1+t)^n + (10*n-(3-5*n)*t)*(1-t)^n )/(25*2^n), where t=sqrt(5). - Bruno Berselli, Apr 17 2014
a(n) = (6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5 = ((n+3)*Fibonacci(n-3) - 2*(n-3)*Fibonacci(n-2))/5. - G. C. Greubel, Feb 06 2020

A091594 Triangle read by rows: T(n,m) := Sum_{k=0..floor((n-m)/2)} binomial(n-2k,m) * binomial(n-m-k,k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 5, 8, 7, 4, 1, 8, 15, 16, 11, 5, 1, 13, 28, 34, 28, 16, 6, 1, 21, 51, 70, 66, 45, 22, 7, 1, 34, 92, 140, 148, 116, 68, 29, 8, 1, 55, 164, 274, 320, 281, 190, 98, 37, 9, 1, 89, 290, 527, 672, 651, 494, 295, 136, 46, 10, 1, 144, 509, 999, 1379, 1456, 1219, 819, 439, 183, 56, 11, 1

Offset: 0

Paul Barry, Jan 23 2004

A Fibonacci related number triangle.

			Rows begin:
   1,
   1,  1,
   2,  2,  1,
   3,  4,  3,  1,
   5,  8,  7,  4,  1,
   8, 15, 16, 11,  5,  1,
  13, 28, 34, 28, 16,  6, 1,
  21, 51, 70, 66, 45, 22, 7, 1,
  ...

Seiichi Manyama, Rows n = 0..139, flattened

Columns include A000045, A029907, A054455. Row sums are A006054.

k-th column has g.f. 1/(1-x-x^2) * ( x*(1-x^2)/(1-x-x^2) )^k.

A134508 Row sums of triangle A134507.

Original entry on oeis.org

1, 2, 5, 11, 22, 42, 77, 138, 243, 423, 730, 1252, 2137, 3634, 6161, 10419, 17582, 29614, 49797, 83610, 140191, 234767, 392690, 656136, 1095217, 1826402, 3043037, 5065883, 8426758, 14006898, 23265725, 38618922, 64062987, 106206519, 175972426

Offset: 1

Gary W. Adamson, Oct 28 2007

			a(4) = 11 = sum of row 4 terms of triangle A134507 = (4 + 5 + 1 + 1).
a(4) = 11 = A000071(6) + A029907(4) - 4 = (7 + 8 - 4).

Index entries for linear recurrences with constant coefficients, signature (4,-4,-2,4,0,-1).

Cf. A134507, A000071, A029907.

a(n)=fibonacci(n+2) + ((n+4)*fibonacci(n)+2*n*fibonacci(n-1))/5 - n - 1 \\ Charles R Greathouse IV, Feb 13 2018

a(n) = A000071(n+2) + A029907(n) - n. [Corrected by Charles R Greathouse IV, Feb 13 2018]

More terms from Charles R Greathouse IV, Feb 13 2018

A182725 Bisection of A005291.

Original entry on oeis.org

1, 1, 2, 4, 8, 15, 28, 51, 92, 163, 285, 490, 833, 1396, 2313, 3789

Offset: 1

Omar E. Pol, Dec 02 2010

This sequence contains the string "1, 2, 4, 8, 15, 28, 51, 92" the same as A005682 and A029907. The next terms is similar. I think that A005291 is a sequence formed from two similar mechanisms but not from only one.

Cf. A005291, A182712, A182722, A182726.

A276129 a(n) is the number of ordered ways to tile a strip of length n+2 with white tiles of odd lengths summing to length n and two red squares.

Original entry on oeis.org

1, 3, 6, 13, 27, 54, 106, 204, 387, 725, 1344, 2469, 4500, 8145, 14652, 26213, 46665, 82704, 145982, 256722, 449937, 786109, 1369494, 2379447, 4123944, 7130895, 12303714, 21186013, 36411399, 62466906, 106987282, 182946888, 312367887, 532587461, 906840060

Offset: 0

Gregory L. Simay, Aug 21 2016

nonn

a(n) is a specific case of b(r,n), the number of ordered ways to rearrange a tiling of length n + r, with odd(1,3,5...) white tiles summing to n and r red squares.
Define the following summation: b(0,r,n) = b(r,n); b(1,r,n) = b(r, n-2) + b(r, n-4) + b(r, n-6) + ..; b(s, r, n) = b(s-1, r, n-2) + b(s-1, r, n-4) + b(r-1, s, n-6) + ...
The number of compositions of n with exactly r even numbers is b(r, r, n-2r).
Except for the initial 1, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = (1 - S)^3. See A291219. - Clark Kimberling, Sep 04 2017

			Let 1,3 be the lengths of the odd tiles summing to 3 and let r,r be the two odd squares. Then the resulting number of compositions is a(3) = 13. The 6 compositions are 3,r,r; r,3,r; r,r,3; 1,1,1,r,r; 1,1,r,r,1; 1,r,r,1,1; r,r,1,1,1; 1,1,r,1,r; 1,r,1,r,1; r,1,r,1,1; r,1,1,r,1; 1,r,1,1,r; r,1,1,1,r.

Cf. A000045, A029907, A239242.

b:= proc(n, m) option remember;
      `if`(n+m=0, 1, `if`(m>0, b(n, m-1), 0)+
      add(`if`(j::odd, b(n-j, m), 0), j=1..n))
    end:
a:= n-> b(n, 2):
seq(a(n), n=0..50);  # Alois P. Heinz, Aug 29 2016

a[0] = 1; a[n_] := Sum[Binomial[n - 2*k + 2, 2]*Binomial[n - k - 1, k], {k, 0, (n - 1)/2}];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Dec 27 2017, after Andrew Howroyd *)

a(n)={if(n<=0, n==0, sum(k=0, (n-1)\2, binomial(n-2*k+2, 2)*binomial(n-k-1, k)))} \\ Andrew Howroyd, Dec 26 2017

a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-2*k+2, 2)*binomial(n-k-1, k) for n > 0. - Andrew Howroyd, Dec 26 2017
b(0,0)=1. For n>=1, b(0,n) = b(0,0,n) = the n-th Fibonacci number, A000045(n).
b(1,0)=1. For n>=1, b(1,n) = A239342(n+1).
b(2,n) = a(n) = a(n-1) + a(n-2) + A239342(n+1) + A239342(n-1).
G.f. for b(2,n): ((1-x^2)/(1-x-x^2))^3.
G.f. for b(r,n): ((1-x^2)/(1-x-x^2))^(r+1).
b(1,1,n) = A029907(n+1).
b(r,n) = b(r, n-1) + b(r, n-2) + b(r-1, n) - b(r-1, n-2).
b(r,r,n) = b(r-1, r-1, n) + b(r, r, n-1) + b(r, r, n-2).
G.f. for b(r,r,n): (1-x^2)/((1-x-x^2)^(r+1)).

More terms from Alois P. Heinz, Aug 29 2016

A100898 Triangle read by rows: T(n,k) is the number of k-matchings of the fan graph on n+1 vertices (i.e., the join of the path graph on n vertices with one extra vertex).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 5, 2, 1, 7, 7, 1, 9, 15, 3, 1, 11, 26, 13, 1, 13, 40, 34, 4, 1, 15, 57, 70, 21, 1, 17, 77, 125, 65, 5, 1, 19, 100, 203, 155, 31, 1, 21, 126, 308, 315, 111, 6, 1, 23, 155, 444, 574, 301, 43, 1, 25, 187, 615, 966, 686, 175, 7, 1, 27, 222, 825, 1530, 1386, 532, 57

Offset: 0

Emeric Deutsch, Jan 10 2005

Row n contains 1 + ceiling(n/2) terms. The row sums yield A029907.

			T(3,2)=2 because in the graph with vertex set {O,A,B,C} and edge set {AB,BC,OA,OB,OC} the 2-matchings are: {OA,BC} and {OC,AB}.
The triangle starts:
  1;
  1,  1;
  1,  3;
  1,  5,  2;
  1,  7,  7;
  1,  9, 15,  3;
  1, 11, 26, 13;

Cf. A029907.

G:=(1-z)*(1+t*z)/(1-z-t*z^2)^2:Gser:=simplify(series(G,z=0,18)):P[0]:=1: for n from 1 to 16 do P[n]:=sort(coeff(Gser,z^n)) od:for n from 0 to 15 do seq(coeff(t*P[n],t^k),k=1..1+ceil(n/2)) od; # yields sequence in triangular form

G.f.: (1-z)(1+t*z)/(1 - z - t*z^2)^2.

A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 4, 4, 6, 7, 8, 8, 8, 7, 6, 4, 4, 5, 5, 8, 10, 13, 15, 16, 16, 15, 13, 10, 8, 5, 5, 6, 6, 10, 13, 18, 23, 28, 31, 32, 31, 28, 23, 18, 13, 10, 6, 6, 7, 7, 12, 16, 23, 31, 41, 51, 59, 63, 63, 59, 51, 41, 31, 23, 16, 12, 7, 7

Offset: 1

Lechoslaw Ratajczak, Jul 04 2020

The number of terms in row n is 3*n-1 = A016789(n-1).
The sum of row n is equal to 2*A094002(n-1) = 2*A188589(n).
Fibonacci(n) = T(n+k,n) - T(n+k-1,n) for n >= 1, k = 1,2,3,...
The elements b(k) of the main diagonal, superdiagonal 1 and all subdiagonals have the recursive formula: b(k) = 2*b(k-1) + b(k-2) - 2*b(k-3) - b(k-4) for k > 4.

			Triangle begins:
n\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20...
1   1  1
2   2  2  2  2  2
3   3  3  4  4  4  4  3  3
4   4  4  6  7  8  8  8  7  6  4  4
5   5  5  8 10 13 15 16 16 15 13 10  8  5  5
6   6  6 10 13 18 23 28 31 32 31 28 23 18 13 10  6  6
7   7  7 12 16 23 31 41 51 59 63 63 59 51 41 31 23 16 12  7  7
...

Superdiagonal 1 is A029907 for n >= 1.
The main diagonal is A208354 for n >= 1.
Subdiagonal 1 is A102702(n-1) for n >= 1.
Subdiagonal 2 is A206268(n+2) for n >= 1 (conjectured).
Subdiagonal 3 is A191830(n+3) for n >= 1.
Cf. A007318, A051597, A228196.

T(n,k) = T(n,3*k-n) for 1 <= k <= 3*n-1.
T(n,k) = Sum_{u=2*(n-k)+3..2*n-k+1} ceiling(u/2)*A065941(k-2,u-2*(n-k)-3) for n >= 3, 3 <= k <= n.
T(n,k) = Sum_{m1=1..k-n} A208354(m1)*binomial(n-m1-1, k-n-m1) + Sum_{m2=1..2*n-k} A208354(m2)*binomial(n-m2-1, 2*n-k-m2) for n >= 2, n+1 <= k <= 2*n-1.
T(n,k) = Sum_{u=2*(k-2*n)+3..k-n+1} ceiling(u/2)*A065941(3*n-k-2,u-2*(k-2*n)-3) for n>= 3, 2*n <= k <= 3*(n-1).
T(n,k) = A208354(k) + (n-k)*Fibonacci(k) for n >= 3, 3 <= k <= n.
T(n,k) = A029907(k-1) + (n-k+1)*Fibonacci(k) for n >= 2, 3 <= k <= n+1.

cp's OEIS Frontend

A268249 T(n,k)=Number of nXk 0..k arrays with every repeated value in every row greater than, and every column greater than or equal to, the previous repeated value.

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A102702 Expansion of (2-x-2*x^2-x^3)/(1-x-x^2)^2.

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A211228 Shallow diagonal sums of A211226.

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A240847 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

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A091594 Triangle read by rows: T(n,m) := Sum_{k=0..floor((n-m)/2)} binomial(n-2k,m) * binomial(n-m-k,k).

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A134508 Row sums of triangle A134507.

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PARI

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A182725 Bisection of A005291.

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A276129 a(n) is the number of ordered ways to tile a strip of length n+2 with white tiles of odd lengths summing to length n and two red squares.

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A240847 a(n) = 2a(n-1) + a(n-2) - 2a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).