A004070 -id:A004070 - cp's OEIS frontend

A115291 Expansion of (1+x)^3/(1-x).

1, 4, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 0

Paul Barry, Jan 19 2006

Partial sums are A086570. Partial sums of squares are A115295. Correlation triangle is A115292.
Let m=4. We observe that a(n) = Sum_{k=0..floor(n/2)} C(m,n-2*k). Then there is a link with A113311 and A040000: it is the same formula with respectively m=3 and m=2. We can generalize this result with the sequence whose G.f is given by (1+z)^(m-1)/(1-z). - Richard Choulet, Dec 08 2009
Also continued fraction expansion of (132-sqrt(17))/103. - Bruno Berselli, Sep 23 2011
Also decimal expansion of 1331/9000. - Vincenzo Librandi, Sep 23 2011

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

Cf. A040000, A113311, A171418, A171440, A171441, A171442, A171443.

CoefficientList[Series[(1+x)^3/(1-x),{x,0,100}],x] (* or *) PadRight[ {1,4,7},120,{8}] (* Harvey P. Dale, May 23 2016 *)

a(n) = 8 - C(2, n) - 2*C(1, n) - 4*C(0, n).
a(n) = Sum_{k=0..n} C(3, k).
a(n) = A004070(n, 3).
From Elmo R. Oliveira, Aug 09 2024: (Start)
E.g.f.: 8*exp(x) - 7 - 4*x - x^2/2.
a(n) = 8, n > 2. (End)

A054124 Left Fibonacci row-sum array, n >= 0, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 1, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 1, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 1, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2

Offset: 0

Clark Kimberling

Reflection of array in A054123 about vertical central line.
Starting with g(0) = {0}, generate g(n) for n > 0 inductively using these rules:
  (1)  if x is in g(n-1), then x+1 is in g(n); and
  (2)  if x is in g(n-1) and x < 2, then x/2 is in g(n).
Then g(1) = {1/1}, g(2) = {1/2,2/1}, g(3) = {1/4,3/2,3/1}, etc. The denominators in g(n) are 2^0, 2^1, ..., 2^(n-1), and T(n,k) is the number of occurrences of 2^k, for k = 0..n-1. - Clark Kimberling, Nov 09 2015
Variant of A004070 with an additional column of 1's on the left. - Jianing Song, May 30 2022

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Row sums: A000045. Central numbers: 1, 1, 2, 4, 8, ... (A000079).

First n numbers of n-th column for n >= 1 form the array in A008949.

a054124 n k = a054124_tabl !! n !! k
a054124_row n = a054124_tabl !! n
a054124_tabl = map reverse a054123_tabl
-- Reinhard Zumkeller, May 26 2015

t[, 0|1] = t[n, n_] = 1; t[n_, k_] := t[n, k] = t[n-1, k-1] + t[n-2, k-1]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 25 2013 *)

A052509(n,k) = sum(m=0, k, binomial(n-k, m));
T(n,k) = if(k==0, 1, A052509(n-1,n-k)) \\ Jianing Song, May 30 2022

T(n, 0) = T(n, n) = 1 for n >= 0; T(n, 1) = 1 for n >= 1; T(n, k) = T(n-1, k-1) + T(n-2, k-1) for k=2, 3, ..., n-1, n >= 3. [Corrected by Jianing Song, May 30 2022]

G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = (1-x^2*y) / ((1-x)*(1-x*y-x^2*y)). - Jianing Song, May 30 2022

A077846 Expansion of g.f. 1/(1 - 3x + 2x^3).

Original entry on oeis.org

1, 3, 9, 25, 69, 189, 517, 1413, 3861, 10549, 28821, 78741, 215125, 587733, 1605717, 4386901, 11985237, 32744277, 89459029, 244406613, 667731285, 1824275797, 4984014165, 13616579925, 37201188181, 101635536213, 277673448789, 758617970005, 2072582837589, 5662401615189

Offset: 0

N. J. A. Sloane, Nov 17 2002

Number of (s(0), s(1), ..., s(n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1..n+2, s(0) = 1, s(n+2) = 3. - Herbert Kociemba, Jun 17 2004

A Whitney transform of 2^n (see Benoit Cloitre formula and A004070). The Whitney transform maps the sequence with g.f. g(x) to that with g.f. (1/(1-x))g(x(1+x)). - Paul Barry, Feb 16 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Fan Chung and R. L. Graham, Primitive juggling sequences, Am. Math. Monthly 115 (3) (2008) 185-194.
William J. Keith, Partitions into parts simultaneously regular, distinct, and/or flat, Proceedings of CANT 2016; arXiv:1911.04755 [math.CO], 2019. Mentions this sequence.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 12.
Index entries for linear recurrences with constant coefficients, signature (3,0,-2).

First differences are in A002605.

Cf. A028859, A052945, A057960, A068911,

CoefficientList[Series[1 / (1 - 3 x + 2 x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 19 2013 *)
LinearRecurrence[{3,0,-2},{1,3,9},40] (* Harvey P. Dale, Apr 27 2014 *)

a(n)=sum(i=0,n,sum(j=0,n,2^j*binomial(j,i-j)))

Vec(1/(1-3*x+2*x^3) + O(x^100)) \\ Altug Alkan, Mar 24 2016

a(n) = 3*a(n-1) - 2*a(n-3) = 2*A057960(n) - 1 = round(2*A028859(n)/sqrt(3) - 1/3) = Sum_{i} b(n, i), where b(n, 0) = b(n, 6) = 0, b(0, 3) = 1, b(0, i) = 0 if i <> 3 and b(n+1, i) = b(n, i-1) + b(n, i) + b(n, i+1) if 0 < i < 6 (i.e., the number of three-choice paths along a corridor width 5, starting from the middle). - Henry Bottomley, Mar 06 2003
Binomial transform of A068911. a(n) = (1+sqrt(3))^n*(2+sqrt(3))/3 + (1-sqrt(3))^n*(2-sqrt(3))/3 - 1/3. - Paul Barry, Feb 17 2004
a(0)=1; for n >= 1, a(n) = ceiling((1+sqrt(3))*a(n-1)). - Benoit Cloitre, Jun 19 2004
a(n) = Sum_{i=0..n} Sum_{j=0..n} 2^j*binomial(j, i-j). - Benoit Cloitre, Oct 23 2004
a(n) = 2*(a(n-1) + a(n-2)) + 1, n > 1. - Gary Detlefs, Jun 20 2010
a(n) = (2*A052945(n+1) - 1)/3. - R. J. Mathar, Mar 31 2011
a(n) = floor((1+sqrt(3))^(n+2)/6). - Bruno Berselli, Feb 05 2013
a(n) = (-2 + (1-sqrt(3))^(n+2) + (1+sqrt(3))^(n+2))/6. - Alexander R. Povolotsky, Feb 13 2016
E.g.f.: exp(x)*(4*cosh(sqrt(3)*x) + 2*sqrt(3)*sinh(sqrt(3)*x) - 1)/3. - Stefano Spezia, Mar 02 2024

Name changed by Arkadiusz Wesolowski, Dec 06 2011

A164925 Array, binomial(j-i,j), read by rising antidiagonals.

Original entry on oeis.org

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

Offset: 0

Mark Dols, Aug 31 2009

Inverse of A052509, or A004070???

			Array, A(n, k), begins as:
  1,  1,  1,   1,  1,   1,  1,  1,  1, ...
  1,  0,  0,   0,  0,   0,  0,  0,  0, ...
  1, -1,  0,   0,  0,   0,  0,  0,  0, ...
  1, -2,  1,   0,  0,   0,  0,  0,  0, ...
  1, -3,  3,  -1,  0,   0,  0,  0,  0, ...
  1, -4,  6,  -4,  1,   0,  0,  0,  0, ...
  1, -5, 10, -10,  5,  -1,  0,  0,  0, ...
  1, -6, 15, -20, 15,  -6,  1,  0,  0, ...
  1, -7, 21, -35, 35, -21,  7, -1,  0, ...
Antidiagonal triangle, T(n, k), begins as:
  1;
  1,  1;
  1,  0,  1;
  1, -1,  0,  1;
  1, -2,  0,  0,  1;
  1, -3,  1,  0,  0,  1;
  1, -4,  3,  0,  0,  0,  1;
  1, -5,  6, -1,  0,  0,  0,  1;
  1, -6, 10, -4,  0,  0,  0,  0,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A004070, A008346, A010892, A052509, A052992.

Cf. A109466, A130595, A164899, A164915, A164965.

A164925:= func< n,k | k eq 0 or k eq n select 1 else Binomial(2*k-n,k) >;
[A164925(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 10 2023

T[n_, k_]:= If[k==0 || k==n, 1, Binomial[2*k-n, k]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 10 2023 *)

{A(i, j) = if( i<0, 0, if(i==0 || j==0, 1, binomial(j-i, j)))}; /* Michael Somos, Jan 25 2012 */

def A164925(n,k): return 1 if (k==0 or k==n) else binomial(2*k-n, k)
flatten([[A164925(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 10 2023

Sum_{k=0..n} T(n, k) = A164965(n). - Mark Dols, Sep 02 2009
From G. C. Greubel, Feb 10 2023: (Start)
A(n, k) = binomial(k-n, k), with A(0, k) = A(n, 0) = 1 (array).
T(n, k) = binomial(2*k-n, k), with T(n, 0) = T(n, n) = 1 (antidiagonal triangle).
Sum_{k=0..n} (-1)^k*T(n, k) = A008346(n).
Sum_{k=0..n} (-2)^k*T(n, k) = (-1)^n*A052992(n). (End)

Edited by Michael Somos, Jan 26 2012

Offset changed by G. C. Greubel, Feb 10 2023

A178524 Triangle read by rows: T(n,k) is the number of leaves at level k in the Fibonacci tree of order n (n>=0, 0<=k<=n-1).

Original entry on oeis.org

1, 1, 0, 2, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 5, 2, 0, 0, 0, 4, 7, 2, 0, 0, 0, 1, 9, 9, 2, 0, 0, 0, 0, 5, 16, 11, 2, 0, 0, 0, 0, 1, 14, 25, 13, 2, 0, 0, 0, 0, 0, 6, 30, 36, 15, 2, 0, 0, 0, 0, 0, 1, 20, 55, 49, 17, 2, 0, 0, 0, 0, 0, 0, 7, 50, 91, 64, 19, 2, 0, 0, 0, 0, 0, 0, 1, 27, 105, 140, 81, 21, 2

Offset: 0

Emeric Deutsch, Jun 15 2010

A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.
Sum of entries in row n is the Fibonacci number F(n+1) (A000045(n+1)).
Sum(k*T(n,k),k>=0)=A067331(n-2).
T(n,k) is the number of vertices in the Fibonacci cube G_{n-1} that have eccentricity k (see Klavzar and Mollard reference). - Michel Mollard, Aug 20 2014

			Triangle starts:
1,
1,
0, 2,
0, 1, 2,
0, 0, 3, 2,
0, 0, 1, 5, 2,
0, 0, 0, 4, 7, 2,
0, 0, 0, 1, 9, 9, 2,
0, 0, 0, 0, 5, 16, 11, 2,
0, 0, 0, 0, 1, 14, 25, 13, 2,
0, 0, 0, 0, 0, 6, 30, 36, 15, 2,

D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

A. Castro and M. Mollard, The eccentricity sequences of Fibonacci and Lucas cubes, Discrete Math., 312 (2012), 1025-1037. See Table 1. [From _N. J. A. Sloane_, Mar 22 2012]
Y. Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, 20, No. 2, 1982, 168-178.
S. Klavzar, M. Mollard, Asymptotic Properties of Fibonacci Cubes and Lucas Cubes, Annals of Combinatorics, 18, 2014, 447-457.

Cf. A000045, A067331, A178522, A004070

G := (1+z-t*z)/(1-t*z-t*z^2): Gser := simplify(series(G, z = 0, 17)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 13 do seq(coeff(P[n], t, k), k = 0 .. n-1) end do; # yields sequence in triangular form

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

A104726 Triangle generated as the matrix product of A026729 and A000012 (triangular views), read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 3, 3, 1, 5, 5, 5, 4, 1, 8, 8, 8, 8, 5, 1, 13, 13, 13, 13, 12, 6, 1, 21, 21, 21, 21, 21, 17, 7, 1, 34, 34, 34, 34, 34, 33, 23, 8, 1, 55, 55, 55, 55, 55, 55, 50, 30, 9, 1, 89, 89, 89, 89, 89, 89, 88, 73, 38

Offset: 0

Gary W. Adamson, Mar 20 2005

If the triangular factors A026729 and A000012 are commuted in the product, A004070 results.

Riordan array (1/(1-x-x^2), x*(1+x)). - Philippe Deléham, Mar 06 2013

			First few rows of the triangle are
1;
1, 1;
2, 2, 1;
3, 3, 3, 1;
5, 5, 5, 4, 1;
8, 8, 8, 8, 5, 1;
13, 13, 13, 13, 12, 6, 1;
21, 21, 21, 21, 21, 17, 7, 1;
...
Production array begins
1, 1
1, 1, 1
-1, -1, 1, 1
2, 2, -1, 1, 1
-5, -5, 2, -1, 1, 1
14, 14, -5, 2, -1, 1, 1
-42, -42, 14, -5, 2, -1, 1, 1
132, 132, -42, 14, -5, 2, -1, 1, 1
-429, -429, 132, -42, 14, -5, 2, -1, 1, 1
... which is based on A000108 or A168491. - _Philippe Deléham_, Mar 06 2013

Cf. A001629 (row sums), A026729, A004070, A000071.

A104726 := proc(n,k)
        add( binomial(j,n-j),j=k..n) ;
end proc:
seq(seq(A104726(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2011

T(n,k) = sum_{j=k..n} binomial(j,n-j). - R. J. Mathar, Oct 30 2011
T(n,0) = T(n-1,0) + T(n-2,0), T(n,k) = T(n-1,k-1) + T(n-2,k-1) for k>0. - Philippe Deléham, Mar 06 2013
T(2*n,n) = A000045(2n+1) = A001519(n+1) = A122367(n). - Philippe Deléham, Mar 06 2013

A122438 Riordan array (1/(1-2x), x(1+2x)).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 8, 8, 6, 1, 16, 16, 16, 8, 1, 32, 32, 32, 28, 10, 1, 64, 64, 64, 64, 44, 12, 1, 128, 128, 128, 128, 120, 64, 14, 1, 256, 256, 256, 256, 256, 208, 88, 16, 1, 512, 512, 512, 512, 512, 496, 336, 116, 18, 1, 1024, 1024, 1024, 1024, 1024, 1024, 912, 512, 148

Offset: 0

Paul Barry, Sep 05 2006

Generalized Whitney triangle. Row sums are A045883(n+1). Diagonal sums are A122439.

			Number triangle begins
1,
2, 1,
4, 4, 1,
8, 8, 6, 1,
16, 16, 16, 8, 1,
32, 32, 32, 28, 10, 1,
64, 64, 64, 64, 44, 12, 1

Cf. A004070.

T[n_, k_] := Sum[ Binomial[k, n - k - j]*2^(n - k), {j, 0, n - k}]; Table[ T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Sep 14 2006 *)

Number triangle T(n,k)=sum{j=0..n-k, C(k,n-k-j)}*2^(n-k).

More terms from Robert G. Wilson v, Sep 14 2006

A180975 Array of the "egg-drop" numbers, read by downwards antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 3, 6, 4, 1, 3, 7, 10, 5, 1, 3, 7, 14, 15, 6, 1, 3, 7, 15, 25, 21, 7, 1, 3, 7, 15, 30, 41, 28, 8, 1, 3, 7, 15, 31, 56, 63, 36, 9, 1, 3, 7, 15, 31, 62, 98, 92, 45, 10

Offset: 1

Francis Carr (fcarr(AT)alum.mit.edu), Sep 30 2010

The egg-drop problem is as follows. We wish to determine the highest floor in a building such that an egg dropped from that floor will not shatter. We have a set of k identical eggs for dropping; note that an egg that does not shatter can be reused to test higher floors. Then T(n,k) is the largest number of floors that can be tested using at most n drops and k eggs.
Suppose one is given k eggs and a building of f floors. Then the worst-case number of drops WC(k,f) to test all the floors satisfies the dynamic programming equation
. WC(k,f) = 1 + max_{g in 1..f} { WC(k-1, g-1), WC(k, f-g) }
with boundary conditions
. WC(1,f) = f and WC(k,0) = 0
where g is the floor chosen for the next drop. One can substitute f -> T(n,k) and g -> T(n-1,k-1)+1 = f-T(k-1,j). Then by an inductive argument, it is verified that WC(j,T(n,j)) = n as expected.
T(n,2) = n(n+1)/2. This leads to the commonly-used heuristic solution for 2 eggs of dropping from the sqrt(f), 2*sqrt(f), 3*sqrt(f) etc. floors until the first egg breaks. T(n,j) is a j-th order polynomial in n, and so this heuristic may be generalized: drop from multiples of f^((j-1)/j) until the j-th egg breaks.

			T(n,1) = n. With only a single egg, one must drop the egg from the first floor, then the second, and so on until it finally breaks. At most n floors may be tested this way.
If one is allowed fewer drops than eggs, a simple binary search is optimal. Hence T(n,k) = 2^n-1 for n <= k. Note that one cannot test 2^n floors in this case. For example, suppose one had 2 drops and a 4-story building; dropping from either the second or third floor could leave another two floors to test, but only one drop remaining.
The square array starts in row n=1 with columns k >= 1:
  1:  1  1  1  1  1  1  1  1  1
  2:  2  3  3  3  3  3  3  3  3
  3:  3  6  7  7  7  7  7  7  7
  4:  4 10 14 15 15 15 15 15 15
  5:  5 15 25 30 31 31 31 31 31
  6:  6 21 41 56 62 63 63 63 63

J. Kleinberg and E. Tardos, "Algorithm Design", Addison-Wesley Longman Publishing Co. Inc., 2005. Problem 2.8.
D. Velleman, "Which Way Did The Bicycle Go? And Other Intriguing Mathematical Mysteries (Dolciani Mathematical Expositions)", The Mathematical Association of America, 1996, "166. An Egg-Drop Experiment" pages 53 and 204-205.

G. C. Greubel, Antidiagonals n = 1..100 of array, flattened
Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372.

Cf. A004070, A117670.
Cf. A131251 (transpose).
Cf. A001924 (antidiagonal sums).

nmax:=11;; T:=List([1..nmax],n->List([1..nmax],k->Sum([1..k],j->Binomial(n,j))));;
b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Oct 09 2018

[[(&+[Binomial(k,j): j in [1..n-k]]): k in [1..n-1]]: n in [2..15]]; // G. C. Greubel, Oct 09 2018

T:= proc(n,k) option remember;
if n = 0 or k = 0 then 0
else T(n-1,k-1) + 1 + T(n-1,k)
fi
end proc:
seq(seq(T(i,m-i),i=1..m-1),m=1..20); # Robert Israel, Jan 20 2015

T[n,k] = Sum[Binomial[n, j], {j, 1, k}]; Table[T[k, n-k], {n, 2, 15}, {k, 1, n-1}]//Flatten (* modified by G. C. Greubel, Oct 09 2018 *)

for(n=2, 15, for(k=1,n-1, print1(sum(j=1, n-k, binomial(k,j)), ", "))) \\ G. C. Greubel, Oct 09 2018

from functools import lru_cache
@lru_cache(maxsize=None)
def T(n, k): return 0 if n*k == 0 else T(n-1, k-1) + 1 + T(n-1, k)
print([T(k, n-k) for n in range(1, 12) for k in range(1, n)]) # Michael S. Branicky, Apr 04 2021

Recursive formula: T(n,k) = T(n-1,k-1) + 1 + T(n-1,k) with boundary conditions T(0,k) = T(n,0) = 0.
T(n,k) = Sum_{j=1..k} binomial(n,j).
From the journal article by Boardman, the generating function for a fixed value of n is g_n(x) = ((1+x)^n - 1) / (1-x).
G.f.: x*y/((1-y)*(1-x)*(1-x-x*y)). - Vladimir Kruchinin, Oct 09 2018

A210561 Triangle of coefficients of polynomials u(n,x) jointly generated with A210562; see the Formula section.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 3, 8, 8, 1, 3, 9, 20, 16, 1, 3, 9, 26, 48, 32, 1, 3, 9, 27, 72, 112, 64, 1, 3, 9, 27, 80, 192, 256, 128, 1, 3, 9, 27, 81, 232, 496, 576, 256, 1, 3, 9, 27, 81, 242, 656, 1248, 1280, 512, 1, 3, 9, 27, 81, 243, 716, 1808, 3072, 2816, 1024, 1, 3, 9

Offset: 1

Clark Kimberling, Mar 22 2012

Last term in row n:  2^(n-1)
Limiting row:  3^(k-1)
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...2
1...3...4
1...3...8...8
1...3...9...20...16
First three polynomials u(n,x): 1, 1 + 2x, 1 + 3x + 4x^2.

P. Bala, A note on the diagonals of a proper Riordan Array

Cf. A210562, A208510, A004070, A048739.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A210559 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210560 *)

u(n,x)=x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x)=(x+1)*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Peter Bala, Mar 06 2017: (Start)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1).
E.g.f. for n-th subdiagonal: exp(2*x)*(1 + x + x^2/2! + x^3/3! + ... + x^n/n!). Cf. A004070.
Riordan array (1/(1 - x), x*(2 + x)).
Row sums A048739.
(End)

A103819 Whitney transform of Jacobsthal numbers.

Original entry on oeis.org

0, 1, 3, 8, 23, 63, 172, 471, 1287, 3516, 9607, 26247, 71708, 195911, 535239, 1462300, 3995079, 10914759, 29819676, 81468871, 222577095, 608091932, 1661338055, 4538859975, 12400396060, 33878512071, 92557816263, 252872656668

Offset: 0

Paul Barry, Feb 16 2005

The Whitney transform maps the sequence with g.f. g(x) to that with g.f. (1/(1-x))*g(x(1+x)).

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

Cf. A004070, A001045, A002605, A099837.

LinearRecurrence[{2,2,1,-2,-2},{0,1,3,8,23},30] (* Harvey P. Dale, Nov 02 2024 *)

G.f.: x(1+x)/((1-x)(1+x+x^2)(1-2x-2x^2)).
a(n) = 2a(n-1)+2a(n-2)+a(n-3)-2a(n-4)-2a(n-5).
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(k, i-k)*A001045(k).
9*a(n) = -2 +2*(A002605(n)+2*A002605(n+1))-A099837(n+3). - R. J. Mathar, Oct 23 2011

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