A005565 -id:A005565 - cp's OEIS frontend

A093768 Positive first differences of the rows of triangle A088459, which enumerates symmetric Dyck paths.

1, 1, 1, 1, 2, 3, 1, 3, 8, 6, 1, 4, 15, 20, 20, 1, 5, 24, 45, 75, 50, 1, 6, 35, 84, 189, 210, 175, 1, 7, 48, 140, 392, 588, 784, 490, 1, 8, 63, 216, 720, 1344, 2352, 2352, 1764, 1, 9, 80, 315, 1215, 2700, 5760, 7560, 8820, 5292, 1, 10, 99, 440, 1925, 4950, 12375, 19800

Offset: 0

Paul D. Hanna, Apr 16 2004

Suggested by Bozydar Dubalski (slawb(AT)atr.bydgoszcz.pl). Related to walks on a square lattice: main diagonal forms A005558, secondary diagonals form A005559, A005560, A005561, A005562, A005563.

Apparently row-reversed version of A052174. - R. J. Mathar, Feb 03 2025

			1;
1, 1;
1, 2, 3;
1, 3, 8, 6;
1, 4, 15, 20, 20;
1, 5, 24, 45, 75, 50;
1, 6, 35, 84, 189, 210, 175;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A088459, A005558-A005562, A005563 (column 3), A005564 (column 4), A005565 (column 5), A005566 (row sums).

A093768 := proc(n,k)
    if k = 0 then
        A088459(n,k);
    else
        A088459(n,k)-A088459(n,k-1);
    end if;
end proc:
seq(seq(A093768(n,k),k=0..n-1),n=1..10) ; # R. J. Mathar, Apr 02 2017

T[n_, k_] := Binomial[n + 1, Ceiling[k/2]]*Binomial[n, Floor[k/2]] - Binomial[n + 1, Ceiling[(k - 1)/2]]*Binomial[n, Floor[(k - 1)/2]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 25 2017 *)

{T(n,k) =binomial(n+1,ceil(k/2))*binomial(n,floor(k/2)) -binomial(n+1,ceil((k-1)/2))*binomial(n,floor((k-1)/2))}

T(n, k) = C(n+1, ceiling(k/2))*C(n, floor(k/2)) - C(n+1, ceiling((k-1)/2))*C(n, floor((k-1)/2)) for n>=k>=0.

A227168 a(n) = gcd(2n, n(n+1)/2)^2.

Original entry on oeis.org

1, 1, 36, 4, 25, 9, 196, 16, 81, 25, 484, 36, 169, 49, 900, 64, 289, 81, 1444, 100, 441, 121, 2116, 144, 625, 169, 2916, 196, 841, 225, 3844, 256, 1089, 289, 4900, 324, 1369, 361, 6084, 400

Offset: 1

Paul Curtz, Jul 03 2013

a(n) is defined as A062828(n)^2 for n >= 1. If we extend the sequence to n=0 and negative n by use of the recurrence that relates a(n) to a(n+12), a(n+8) and a(n+4), we obtain a(0)=0, a(-1)=4 and a(-n) = A176743(n-2)^2 for n >= 2.
Define c(n) = a(n+2) - a(n-2) for c >= 0. Because a(n) is a shuffle of three interleaved 2nd-order polynomials, c(n) is a shuffle of three interleaved 1st-order polynomials: c(n) = 4* A062828(n)*(periodically repeated 1, 8, 1, 1).
The sequence a(n) is case p=0 of the family A062828(n)*A062828(n+p):
0, 1, 1, 36,  4, 25,  9, 196, ... = a(n).
0, 1, 6, 12, 10, 15, 42,  56, ... = A130658(n)*A000217(n) = A177002(n-1)*A064038(n+1).
0, 6, 2, 30,  6, 70, 12, 126, ... = 2*A198148(n)
0, 2, 5, 18, 28, 20, 27,  70, ... = A177002(n+2)*A160050(n+1) = A014695(n+2)*A000096(n).

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A062828, A016814, A017138, A005565, A006752, A061037, A061038.

[GCD(2*n, n*(n+1)/2)^2: n in [1..50]]; // G. C. Greubel, Sep 20 2018

A227168 := proc(n)
    A062828(n)^2 ;
end proc: # R. J. Mathar, Jul 25 2013

a[n_] := GCD[2*n, n*(n + 1)/2]^2; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Jul 03 2013 *)

a(n)=if(n%2, n*if(n%4>2, 2, 1), n/2)^2 \\ Charles R Greathouse IV, Jul 07 2013

a(n) = A062828(n)^2.
a(4n) = (4*n+1)^2; a(2n+1) = (n+1)^2; a(4n+2) = 4*(4*n+3)^2.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) * (period 4: repeat 4, 1, 1, 4) = A061038(n).
A005565(n-3) = a(n+1) * A061037(n). - Corrected by R. J. Mathar, Jul 25 2013
a(n) = A130658(n-1)^2 * A181318(n). - Corrected by R. J. Mathar, Aug 01 2013
G.f.: -x*(1 + x + 36*x^2 + 4*x^3 + 22*x^4 + 6*x^5 + 88*x^6 + 4*x^7 + 9*x^8 + x^9 + 4*x^10) / ( (x-1)^3*(1+x)^3*(x^2+1)^3 ). - R. J. Mathar, Jul 20 2013
Sum_{n>=1} 1/a(n) = 47*Pi^2/192 + 3*G/8, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022

A378062 Array read by ascending antidiagonals: A(n, k) = (n + 1)binomial(2k + n - 1, k - 1)^2 / (2*k + n - 1) for k > 0, and A(n, 0) = 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 8, 20, 0, 1, 15, 75, 175, 0, 1, 24, 189, 784, 1764, 0, 1, 35, 392, 2352, 8820, 19404, 0, 1, 48, 720, 5760, 29700, 104544, 226512, 0, 1, 63, 1215, 12375, 81675, 382239, 1288287, 2760615, 0, 1, 80, 1925, 24200, 196625, 1145144, 5010005, 16359200, 34763300

Offset: 0

Peter Luschny, Dec 07 2024

			Array A(n, k) starts:
  [0] 0, 1,  3,   20,   175,    1764,    19404, ... A000891
  [1] 0, 1,  8,   75,   784,    8820,   104544, ... A145600
  [2] 0, 1, 15,  189,  2352,   29700,   382239, ... A145601
  [3] 0, 1, 24,  392,  5760,   81675,  1145144, ... A145602
  [4] 0, 1, 35,  720, 12375,  196625,  3006003, ... A145603
  [5] 0, 1, 48, 1215, 24200,  429429,  7154784, ...
  [6] 0, 1, 63, 1925, 44044,  869505, 15767024, ...
  [7] 0, 1, 80, 2904, 75712, 1656200, 32626944, ...
.
Seen as a triangle, T(n, k) = A(n-k, k). Compare the descending antidiagonals of A378061.
  [0] 0;
  [1] 0, 1;
  [2] 0, 1,  3;
  [3] 0, 1,  8,  20;
  [4] 0, 1, 15,  75,  175;
  [5] 0, 1, 24, 189,  784,  1764;
  [6] 0, 1, 35, 392, 2352,  8820,  19404;
  [7] 0, 1, 48, 720, 5760, 29700, 104544, 226512;

Rows: A000891, A145600, A145601, A145602, A145603.

Columns: A005563, A005565, A378061.

A := (n, k) -> ifelse(k = 0, 0, (n + 1)*binomial(2*k + n - 1, k - 1)^2/(2*k + n - 1)):
for n from 0 to 7 do seq(A(n, k), k = 0..7);

A[n_, k_] := If[k==0, 0, (n + 1)*Binomial[2*k + n - 1, k - 1]^2 / (2*k + n - 1)]; Table[A[n-k,k],{n,0,9},{k,0,n}]//Flatten (* Stefano Spezia, Dec 08 2024 *)

cp's OEIS Frontend

A093768 Positive first differences of the rows of triangle A088459, which enumerates symmetric Dyck paths.

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A227168 a(n) = gcd(2n, n(n+1)/2)^2.

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A378062 Array read by ascending antidiagonals: A(n, k) = (n + 1)binomial(2k + n - 1, k - 1)^2 / (2*k + n - 1) for k > 0, and A(n, 0) = 0.

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cp's OEIS Frontend

A093768 Positive first differences of the rows of triangle A088459, which enumerates symmetric Dyck paths.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A227168 a(n) = gcd(2*n, n*(n+1)/2)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A378062 Array read by ascending antidiagonals: A(n, k) = (n + 1)*binomial(2*k + n - 1, k - 1)^2 / (2*k + n - 1) for k > 0, and A(n, 0) = 0.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

A227168 a(n) = gcd(2n, n(n+1)/2)^2.

A378062 Array read by ascending antidiagonals: A(n, k) = (n + 1)binomial(2k + n - 1, k - 1)^2 / (2*k + n - 1) for k > 0, and A(n, 0) = 0.