A049016 -id:A049016 - cp's OEIS frontend

A306915 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. 1/((1-x)^k-x^k).

1, 1, 2, 1, 2, 4, 1, 3, 4, 8, 1, 4, 6, 8, 16, 1, 5, 10, 11, 16, 32, 1, 6, 15, 20, 21, 32, 64, 1, 7, 21, 35, 36, 42, 64, 128, 1, 8, 28, 56, 70, 64, 85, 128, 256, 1, 9, 36, 84, 126, 127, 120, 171, 256, 512, 1, 10, 45, 120, 210, 252, 220, 240, 342, 512, 1024

Offset: 0

Seiichi Manyama, Mar 16 2019

			Square array begins:
     1,   1,   1,   1,   1,    1,    1,    1, ...
     2,   2,   3,   4,   5,    6,    7,    8, ...
     4,   4,   6,  10,  15,   21,   28,   36, ...
     8,   8,  11,  20,  35,   56,   84,  120, ...
    16,  16,  21,  36,  70,  126,  210,  330, ...
    32,  32,  42,  64, 127,  252,  462,  792, ...
    64,  64,  85, 120, 220,  463,  924, 1716, ...
   128, 128, 171, 240, 385,  804, 1717, 3432, ...
   256, 256, 342, 496, 715, 1365, 3017, 6436, ...

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

Columns (1+2),3-9 give A000079, A024495(n+2), A000749(n+3), A049016, A192080, A049017, A290995(n+7), A306939.

Cf. A039912, A101508, A306846, A306913, A306914, A307047, A307078, A307393.

A[n_, k_] := Sum[Binomial[n + k - 1, k*j + k - 1], {j, 0, Floor[n/k]}]; Table[A[n - k, k], {n, 0, 11}, {k, n, 1, -1}] // Flatten (* Amiram Eldar, May 25 2021 *)

A(n,k) = Sum_{j=0..floor(n/k)} binomial(n+k-1,k*j+k-1).

A(n,2*k) = Sum_{i=0..n} Sum_{j=0..n-i} binomial(i+k-1,k*j+k-1) * binomial(n-i+k-1,k*j+k-1). - Seiichi Manyama, Apr 07 2019

A078789 Expansion of (1-4x+2x^2)/(1-7x+13x^2-4*x^3).

Original entry on oeis.org

1, 3, 10, 35, 127, 474, 1807, 6995, 27370, 107883, 427351, 1698458, 6765175, 26985675, 107746282, 430470899, 1720537327, 6878624730, 27505271455, 109996928003, 439924466026, 1759532283963, 7037695641415, 28149647662490, 112595619434887, 450374698997499

Offset: 0

Michael Somos, Dec 03 2002

Number of walks of length 2*n+1 between two adjacent vertices in the cycle graph C_10. - Herbert Kociemba, Jul 02 2004

Also, with offset 1, the cogrowth sequence of the 20-element group D10 = . - Sean A. Irvine, Nov 20 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
Index entries for linear recurrences with constant coefficients, signature (7,-13,4).

Cf. A000032, A001622, A049016.

I:=[1,3,10]; [n le 3 select I[n] else 7*Self(n-1)-13*Self(n-2)+4*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 18 2018

LinearRecurrence[{7, -13, 4}, {1, 3, 10}, 25] (* Vincenzo Librandi, Dec 18 2018 *)
CoefficientList[Series[(1-4x+2x^2)/(1-7x+13x^2-4x^3),{x,0,40}],x] (* Harvey P. Dale, Feb 17 2024 *)

{a(n) = polcoeff( (1 - 4*x + 2*x^2) / (1 - 7*x + 13*x^2 - 4*x^3) + x * O(x^n), n)}

{a(n) = sum( k=-n\5, n\5, binomial(2*n + 1, n+1 + 5*k))}

G.f.: (1 - 4*x + 2*x^2) / (1 - 7*x + 13*x^2 - 4*x^3).
a(5*n + 3) = A049016(10*n + 3), a(5*n + 4) = A049016(10*n + 5).
From Herbert Kociemba, Jul 02 2004: (Start)
a(n) = [4^(n+1) + (sqrt(5)+3)*phi^(2n) - (sqrt(5)-3)*phi^(-2n)]/10, where phi is the golden ratio (1+sqrt(5))/2.
a(n) = 7a(n-1)-13a(n-2)+4a(n-3). (End)
a(n) = Sum_{k= -floor(n/5)..floor(n/5)} binomial(2*n,n+5*k)/2 = ((3+sqrt(5))^n + (3-sqrt(5))^n + 2^(3*n-1))/(5*2^n). - Mircea Merca, Jan 28 2012
a(n) = (Lucas(2*n+2) + 2^(2*n+1))/5. - Greg Dresden, Jan 26 2023

A140344 Catalan triangle A009766 prepended by n zeros in its n-th row.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 0, 1, 3, 5, 5, 0, 0, 0, 0, 1, 4, 9, 14, 14, 0, 0, 0, 0, 0, 1, 5, 14, 28, 42, 42, 0, 0, 0, 0, 0, 0, 1, 6, 20, 48, 90, 132, 132, 0, 0, 0, 0, 0, 0, 0, 1, 7, 27, 75, 165, 297, 429, 429, 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 35, 110, 275, 572, 1001, 1430, 1430

Offset: 0

Paul Curtz, May 29 2008

The triangle's n-th row is also related to recurrences for sequences f(n) which p-th differences, p=n+2: The denominator of the generating function contains a factor 1-2x in these cases.
This factor may be "lifted" either by looking at auxiliary sequences f(n+1)-2f(n) or by considering the corresponding "degenerate" shorter recurrences right away. In the case p=4, the recurrence is f(n)=4f(n-1)-6f(n-2)+4f(n-3) from the 4th row in A135356, the denominator in the g.f. is 1-4x+6x^2-4x^3=(1-2x)(1-2x+2x^2), which yields the degenerate recurrence f(n)=2f(n-1)-2f(n-2) from the 2nd factor and leaves the first three coefficients of 1/(1-2x+2x^2)=1+2x+2x^2+.. in row 2.
A000749 is an example which follows the recurrence but not the degenerate recurrence, but still A000749(n+1)-2A000749(n) = 0, 0, 1, 2, 2,.. starts with the 3 coefficients. A009545 follows both recurrences and starts with the three nonzero terms because there is only a power of x in the numerator of the g.f.
In the case p=5, the recurrence is f(n)=5f(n-1)-10f(n-2)+10f(n-3)-5f(n-4)+2f(n-5), the denominator in the g.f. is 1-5x+10x^2-10x^3+5x^4-2x^5= (1-2x)(1-3x+4x^2-2x^3+x^4), where 1/(1-3x+4x^2-2x^3+x^4) = 1+3x+5x^2+5x^3+... and the 4 coefficients populate row 3.
A049016 obeys the main recurrence but not the degenerate recurrence f(n)=3f(n-1)-4f(n-2)+2f(n-3)-f(n-4), yet A049016(n+1)-2A049016(n)=1, 3, 5, 5,.. starts with the 4 coefficients. A138112 obeys both recurrences and is constructed to start with the 4 coefficients themselves.
In the nomenclature of Foata and Han, this is the doubloon polynomial triangle d_{n,m}(0), up to index shifts. - R. J. Mathar, Jan 27 2011

			Triangle starts
1;
0,1,1;
0,0,1,2,2;
0,0,0,1,3,5,5;
0,0,0,0,1,4,9,14,14;

D. Foata, G.-N. Han, The doubloon polynomial triangle, Ramanujan J. 23 (2010), 107-126

Cf. A135356, A130020, A139687, A140343 (p=6), A140342 (p=7).

Table[Join[Array[0&, n], Table[Binomial[n+k, n]*(n-k+1)/(n+1), {k, 0, n}]], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 16 2014 *)

Edited by R. J. Mathar, Jul 10 2008

A085476 Periodic Pascal array, read by upward antidiagonals.

Original entry on oeis.org

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

Offset: 0

Paul Barry, Jul 02 2003

G.f. of binomial transform of n-th row is given by 1/((1-x)^(n+1)-x^(n+1)).

			Rows begin:
n\k | 0 1 2 3 4 5
----+------------
  0 | 1 1 1 1 1 1 ...
  1 | 1 1 1 1 1 1 ...
  2 | 1 2 1 1 2 1 ...
  3 | 1 3 3 1 1 3 ...
  4 | 1 4 6 4 1 1 ...

John Tyler Rascoe, Antidiagonals n = 0..139, flattened

Cf. A007318, A000749, A049016, A006090, A049017.

from math import comb
def A085476(n,k): return(comb(n,k%(n+1))) # John Tyler Rascoe, Dec 01 2023

Square array T(n, k) = C(n, k mod (n+1)).

cp's OEIS Frontend

A306915 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. 1/((1-x)^k-x^k).

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A078789 Expansion of (1-4*x+2*x^2)/(1-7*x+13*x^2-4*x^3).

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A140344 Catalan triangle A009766 prepended by n zeros in its n-th row.

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A085476 Periodic Pascal array, read by upward antidiagonals.

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A078789 Expansion of (1-4x+2x^2)/(1-7x+13x^2-4*x^3).