A131941 -id:A131941 - cp's OEIS frontend

A171218 a(n) = Sum_{k=0..n} A109613(k)*A005843(n-k).

0, 2, 6, 16, 32, 58, 94, 144, 208, 290, 390, 512, 656, 826, 1022, 1248, 1504, 1794, 2118, 2480, 2880, 3322, 3806, 4336, 4912, 5538, 6214, 6944, 7728, 8570, 9470, 10432, 11456, 12546, 13702, 14928, 16224, 17594, 19038, 20560, 22160, 23842, 25606

Offset: 0

Reinhard Zumkeller, Dec 05 2009

a(n) is the number of triples (w,x,y) with all terms in {0,...,n} and 2|w-x|Clark Kimberling, Jun 11 2012]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

[&+[(2*k+(-1)^k+1)*(n-k): k in [0..n]]: n in [0..42]]; // Bruno Berselli, Nov 16 2011

CoefficientList[Series[2x (1+x^2)/((1+x)(1-x)^4),{x,0,50}],x] (* or *) LinearRecurrence[ {3,-2,-2,3,-1},{0,2,6,16,32},50] (* Harvey P. Dale, Jan 22 2023 *)

a(n+1) - a(n) = A137928(n+1).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: 2*x*(1+x^2)/((1+x)*(1-x)^4).
a(n) = 2*A131941(n) = (2*n*(2*n^2+3*n+4)-3*(-1)^n+3)/12.
a(n) = -a(-n-1) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5). (End)

A172482 a(n) = (1+n)(9 + 11n + 4*n^2)/3.

Original entry on oeis.org

3, 16, 47, 104, 195, 328, 511, 752, 1059, 1440, 1903, 2456, 3107, 3864, 4735, 5728, 6851, 8112, 9519, 11080, 12803, 14696, 16767, 19024, 21475, 24128, 26991, 30072, 33379, 36920, 40703, 44736, 49027, 53584, 58415, 63528, 68931, 74632, 80639, 86960, 93603

Offset: 0

Paul Curtz, Feb 04 2010

One of the bisections of the left central column in the Janet table A172002.
Row 1 of the convolution array A213844. - Clark Kimberling, Jul 05 2012
With offset 2, this is 4*n^3/3 - 3*n^2 + 8*n/3 - 1, the number of divisions of a 2 X n board into 3 pieces where the rightmost squares separate. See Jacob Brown article. - Michel Marcus, Jun 29 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Jacob Brown, Counting Divisions of a 2×n Rectangular Grid, arXiv:2106.14755 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A100178, A131941.

[(1+n)*(9+11*n+4*n^2)/3: n in [0..40]]; // Vincenzo Librandi, Aug 04 2011

CoefficientList[Series[(x + 3) (1 + x)/(x - 1)^4, {x, 0, 40}], x] (* Michael De Vlieger, Nov 02 2018 *)

a(n)=(1+n)*(9+11*n+4*n^2)/3 \\ Charles R Greathouse IV, Aug 04 2011

a(n) = A131941(2n+2), where A100178(n) = A131941(2n-1).
a(n) = 4*a(n) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) mod 10 = 3, 6, 7, 4, 5, 8, 1, 2, 9, 0 (and repeat periodically).
G.f.: (x+3)*(1+x)/(x-1)^4.
E.g.f.: exp(x)*(9 + 39*x + 27*x^2 + 4*x^3)/3. - Stefano Spezia, Mar 02 2025

Edited by R. J. Mathar, Feb 24 2010

A300437 Triangle T(nu,m) read by rows: The number of N-color odd self-inverse compositions of (2nu+1) into (2m+1) parts.

Original entry on oeis.org

1, 3, 1, 5, 3, 1, 7, 8, 3, 1, 9, 16, 11, 3, 1, 11, 29, 25, 14, 3, 1, 13, 47, 58, 34, 17, 3, 1, 15, 72, 110, 96, 43, 20, 3, 1, 17, 104, 206, 200, 143, 52, 23, 3, 1, 19, 145, 346, 442, 317, 199, 61, 26, 3, 1, 21, 195, 571, 822, 807, 461, 264, 70, 29, 3, 1, 23, 256, 881, 1565, 1613, 1328, 632, 338, 79, 32, 3, 1

Offset: 0

R. J. Mathar, Mar 05 2018

Table 1 of Guo contains several typos which are not compliant with the formula on page 4 for S_o(2k+1,2l+1). Also the formula has been modified to read S_o(2k+1,2l+1) = sum_{t=1..2k+1) sum_{i+j= (2k+1-t-2l)/4} t*binomial(2l+i-1,2l-1)*binomial(l,j). So the upper limit on t has been extended and a factor t has been inserted.

			The triangle starts in row nu=0 with columns 0<=m<=nu as:
1;
3,1;
5,3,1;
7,8,3,1;
9,16,11,3,1;
11,29,25,14,3,1;
13,47,58,34,17,3,1;
15,72,110,96,43,20,3,1;
17,104,206,200,143,52,23,3,1;
19,145,346,442,317,199,61,26,3,1;
21,195,571,822,807,461,264,70,29,3,1;
23,256,881,1565,1613,1328,632,338,79,32,3,1;
25,328,1337,2671,3478,2800,2032,830,421,88,35,3,1;
27,413,1939,4596,6402,6742,4464,2946,1055,513,97,38,3,1;

Y.-h. Guo, n-Color Odd Self-Inverse Compositions, J. Int. Seq. 17 (2014) # 14.10.5, Table 1.

Cf. A131941 (column 2?), A300438 (row sums), A292835.

A300437 := proc(k,l)
    local a,t,i,j ;
    a := 0 ;
    for t from 1 to 2*k+1 by 2 do
        for j from 0 to l do
            i := (2*k+1-t-2*l)/4-j ;
            if type(i,'integer') then
                a := a+t*binomial(2*l+i-1,2*l-1)*binomial(l,j) ;
            end if;
        end do:
    end do:
    a ;
end proc:
seq(seq(A300437(k,l),l=0..k),k=0..13) ;

A300437[k_, l_] := Module[{a, t, i, j }, a = 0; For[t = 1, t <= 2k + 1, t += 2, For[j = 0, j <= l, j++, i = (2k + 1 - t - 2*l)/4 - j; If[ IntegerQ[i], a = a + t*Binomial[2l + i - 1, 2l - 1]*Binomial[l, j]]]]; a];
Table[Table[A300437[k, l], {l, 0, k}], {k, 0, 13}] // Flatten (* Jean-François Alcover, Aug 15 2023, after Maple code *)

64*T(nu+2,2) = 51 +1306/15*nu +13*(-1)^nu +56/3*nu^3 +170/3*nu^2 +4/15*nu^5 +10*(-1)^nu*nu +2*(-1)^nu*nu^2 +10/3*nu^4 with g.f. (1+x^2)^2/[(1+x)^3*(1-x)^6], column 2.

A325517 a(n) = n((2n + 1)(2n^2 + 2n + 3) - 3(-1)^n)/24.

Original entry on oeis.org

0, 1, 6, 24, 64, 145, 282, 504, 832, 1305, 1950, 2816, 3936, 5369, 7154, 9360, 12032, 15249, 19062, 23560, 28800, 34881, 41866, 49864, 58944, 69225, 80782, 93744, 108192, 124265, 142050, 161696, 183296, 207009, 232934, 261240, 292032, 325489, 361722, 400920, 443200

Offset: 0

Stefano Spezia, May 07 2019

For n > 0, a(n) is the n-th row sum of the triangle A325516.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-5,5,1,-3,1).

Cf. A131941, A317614, A322277, A325516.

Flat(List([0..50], n->n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24));

[n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24: n in [0..50]];

a:=n->n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24: seq(a(n), n=0..50);

a[n_]:=n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24; Array[a,50,0]

a(n) = n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24;

O.g.f.: x*(1 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/((1 - x)^5*(1 + x)^2).
E.g.f.: (1/24)*exp(-x)*x*(3 + 21*exp(2*x) + 54*exp(2*x)*x + 30*exp(2*x)*x^2 + 4*exp(2*x)*x^3).
a(n) = 3*a(n-1) - a(n-2) - 5*a(n-3) + 5*a(n-4) + a(n-5) - 3*a(n-6) + a(n-7) for n > 6.
a(n) = n^2*(2*n^2 + 3*n + 4)/12 if n is even.
a(n) = n*(n + 1)*(2*n^2 + n + 3)/12 if n is odd.
a(n) = n*A131941(n). - Stefano Spezia, Dec 21 2021

A179783 a(n) = 2n(n+1)*(n+2)/3 + (-1)^n.

Original entry on oeis.org

1, 3, 17, 39, 81, 139, 225, 335, 481, 659, 881, 1143, 1457, 1819, 2241, 2719, 3265, 3875, 4561, 5319, 6161, 7083, 8097, 9199, 10401, 11699, 13105, 14615, 16241, 17979, 19841, 21823, 23937, 26179, 28561, 31079

Offset: 0

Bruno Berselli, Jul 29 2010 - Sep 07 2010

First differences in 2*A081352.

Second differences in 4*A004442.

B. Berselli, Table of n, a(n) for n = 0..10000.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A005744, A026035, A175109, A131941.

[(2/3)*n*(n+1)*(n+2)+(-1)^n: n in [0..35]];

LinearRecurrence[{3,-2,-2,3,-1},{1,3,17,39,81},40] (* Harvey P. Dale, Mar 04 2023 *)

for(n=0, 35, print1((2/3)*n*(n+1)*(n+2)+(-1)^n", "));

G.f.: (1+10*x^2-4*x^3+x^4)/((1+x)*(1-x)^4); exp(-x)+(2/3)*exp(x)*x*(6+6*x+x^2).
a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5) for n>4.
a(n) = 4*A000292(n)+(-1)^n.

cp's OEIS Frontend

A171218 a(n) = Sum_{k=0..n} A109613(k)*A005843(n-k).

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A172482 a(n) = (1+n)*(9 + 11*n + 4*n^2)/3.

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A300437 Triangle T(nu,m) read by rows: The number of N-color odd self-inverse compositions of (2*nu+1) into (2*m+1) parts.

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A325517 a(n) = n*((2*n + 1)*(2*n^2 + 2*n + 3) - 3*(-1)^n)/24.

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Magma

Maple

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A179783 a(n) = 2*n*(n+1)*(n+2)/3 + (-1)^n.

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Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A172482 a(n) = (1+n)(9 + 11n + 4*n^2)/3.

A300437 Triangle T(nu,m) read by rows: The number of N-color odd self-inverse compositions of (2nu+1) into (2m+1) parts.

A325517 a(n) = n((2n + 1)(2n^2 + 2n + 3) - 3(-1)^n)/24.

A179783 a(n) = 2n(n+1)*(n+2)/3 + (-1)^n.