A058187 -id:A058187 - cp's OEIS frontend

A024862 a(n) = s(1)t(n) + s(2)t(n-1) + ... + s(k)*t(n-k+1), where k = floor(n/2), s = natural numbers, t = odd natural numbers.

3, 5, 17, 23, 50, 62, 110, 130, 205, 235, 343, 385, 532, 588, 780, 852, 1095, 1185, 1485, 1595, 1958, 2090, 2522, 2678, 3185, 3367, 3955, 4165, 4840, 5080, 5848, 6120, 6987, 7293, 8265, 8607, 9690, 10070, 11270, 11690, 13013, 13475, 14927, 15433, 17020, 17572, 19300

Offset: 2

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A058187.

[((2*n-1)*(2*n+1)*(2*n+3) +3*(-1)^n*(n^2+(n+1)^2))/48: n in [2..50]]; // G. C. Greubel, Apr 19 2023

CoefficientList[Series[(3+2x+3x^2)/((1+x)^3 (1-x)^4), {x,0,50}], x] (* Vincenzo Librandi, Sep 25 2013 *)

Vec(x^2*(3+2*x+3*x^2)/((1+x)^3*(x-1)^4) + O(x^100)) \\ Colin Barker, Jan 29 2016

[((2*n-1)*(2*n+1)*(2*n+3) +3*(-1)^n*(n^2+(n+1)^2))/48 for n in range(2,51)] # G. C. Greubel, Apr 19 2023

G.f.: x^2*(3+2*x+3*x^2) / ((1+x)^3*(x-1)^4). - R. J. Mathar, Sep 25 2013
a(n) = 3*A058187(n-2) + 2*A058187(n-3) + 3*A058187(n-4). - R. J. Mathar, Sep 25 2013
From Colin Barker, Jan 29 2016: (Start)
a(n) = (8*n^3 + 6*(-1)^n*n^2 + 12*n^2 + 6*(-1)^n*n - 2*n + 3*(-1)^n - 3)/48.
a(n) = (4*n^3 + 9*n^2 + 2*n)/24 for n even.
a(n) = (4*n^3 + 3*n^2 - 4*n - 3)/24 for n odd. (End)
E.g.f.: (1/48)*(3*(1 - 4*x + 2*x^2)*exp(-x) + (-3 + 18*x + 36*x^2 + 8*x^3)*exp(x)). - G. C. Greubel, Apr 19 2023

A024868 a(n) = 2(n+1) + 3n + ... + (k+1)*(n+2-k), where k = floor(n/2).

Original entry on oeis.org

6, 8, 22, 27, 52, 61, 100, 114, 170, 190, 266, 293, 392, 427, 552, 596, 750, 804, 990, 1055, 1276, 1353, 1612, 1702, 2002, 2106, 2450, 2569, 2960, 3095, 3536, 3688, 4182, 4352, 4902, 5091, 5700, 5909, 6580, 6810, 7546, 7798, 8602, 8877, 9752, 10051, 11000, 11324, 12350

Offset: 2

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A023855, A023856, A023857, A024305, A024854.

Apart from offset the same as A024306.

[19*n/24-9/16+n^3/12+11*n^2/16+(-1)^n*(3*n/8 +9/16+n^2/16): n in [2..50]]; // Vincenzo Librandi, Sep 26 2013

seq(sum((i+1)*(k-i+2), i=1..floor(k/2)), k=2..70); # Wesley Ivan Hurt, Sep 20 2013

Table[Floor[n/2] (-2Floor[n/2]^2 +3n*Floor[n/2] +9n +14)/6, {n, 2, 100}] (* Wesley Ivan Hurt, Sep 20 2013 *)
CoefficientList[Series[(6 +2x -4x^2 -x^3 +x^4)/((1+x)^3 (1-x)^4), {x, 0, 60}], x] (* Vincenzo Librandi, Sep 26 2013 *)
LinearRecurrence[{1,3,-3,-3,3,1,-1},{6,8,22,27,52,61,100},50] (* Harvey P. Dale, Aug 11 2023 *)

[(1/48)*(4*n^3 +33*n^2 +38*n -27 +3*(-1)^n*(n+3)^2) for n in (2..60)] # G. C. Greubel, Jul 13 2022

a(n) = Sum_{i=1..floor(n/2)} (i+1)*(n-i+2) = floor(n/2)*(-2*floor(n/2)^2 + 3*n*floor(n/2) + 9*n + 14)/6, n>1. - Wesley Ivan Hurt, Sep 20 2013
G.f.: x^2*(6 + 2*x - 4*x^2 - x^3 + x^4) / ( (1+x)^3*(x-1)^4 ). - R. J. Mathar, Sep 25 2013
a(n) = 6*A058187(n-2) +2*A058187(n-3) -4*A058187(n-4) -A058187(n-5) +A058187(n-6). - R. J. Mathar, Sep 25 2013
a(n) = ( 4*n^3 + 33*n^2 + 38*n - 27 )/48 + (-1)^n*(n+3)^2/16. - R. J. Mathar, Sep 25 2013
E.g.f.: (1/24)*( x*(2*x^2 + 24*x + 27)*cosh(x) + (2*x^3 + 21*x^2 + 48*x - 27)*sinh(x) ). - G. C. Greubel, Jul 13 2022

A124458 Triangular array resulting from summing three repeated Pascal sequences; related to the generalized pentagonal sequence (A001318) and the classical modular tessellation (cf. A054886).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 3, 4, 2, 1, 3, 5, 5, 2, 1, 3, 7, 7, 6, 2, 1, 3, 8, 12, 9, 7, 2, 1, 3, 10, 15, 18, 11, 8, 2, 1, 3, 11, 22, 24, 25, 13, 9, 2, 1, 3, 13, 26, 40, 35, 33, 15, 10, 2, 1

Offset: 1

Alford Arnold, Nov 04 2006

The third diagonal is the generalized pentagonal sequence A001318

			Consider
1.......1.......6.......6.......21......21....
........1.......1.......6.......6.......21....
................1.......1.......6.......6.....
which sums to
1.....2....8.....13....33....48....,
a diagonal of A124458

Cf. A000012 A004526 A001651 A008805 A001318 A058187 A028724 A093039 A054886 (row sum).

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1

Offset: 0

Philippe Deléham, Feb 22 2012

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Gheorghe Coserea, Rows n = 0..200, flattened

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013

seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016

P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

A263794 Number of (n+1) X (3+1) 0..1 arrays with each row and column divisible by 3, read as a binary number with top and left being the most significant bits, and rows and columns lexicographically nonincreasing.

Original entry on oeis.org

3, 3, 7, 7, 14, 14, 25, 25, 41, 41, 63, 63, 92, 92, 129, 129, 175, 175, 231, 231, 298, 298, 377, 377, 469, 469, 575, 575, 696, 696, 833, 833, 987, 987, 1159, 1159, 1350, 1350, 1561, 1561, 1793, 1793, 2047, 2047, 2324, 2324, 2625, 2625, 2951, 2951, 3303, 3303

Offset: 1

R. H. Hardin, Oct 26 2015

nonn

Column 3 of A263799.

			Some solutions for n = 5:
  1 1 1 1    1 1 0 0    1 1 0 0    1 1 1 1    0 0 0 0
  1 1 1 1    1 1 0 0    1 1 0 0    1 1 1 1    0 0 0 0
  1 1 0 0    0 0 0 0    0 0 1 1    1 1 1 1    0 0 0 0
  1 1 0 0    0 0 0 0    0 0 1 1    1 1 1 1    0 0 0 0
  0 0 1 1    0 0 0 0    0 0 1 1    1 1 1 1    0 0 0 0
  0 0 1 1    0 0 0 0    0 0 1 1    1 1 1 1    0 0 0 0

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A058187, A263799.

Empirical: a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7).
Empirical: a(n) = A058187(n-1) + floor((n+3)/2). - Filip Zaludek, Dec 14 2016
Conjectures from Colin Barker, Dec 14 2016: (Start)
a(n) = (n^3 + 6*n^2 + 32*n + 48)/48 for n even.
a(n) = (n^3 + 9*n^2 + 47*n + 87)/48 for n odd.
G.f.: x*(3 - 5*x^2 + 4*x^4 - x^6) / ((1 - x)^4*(1 + x)^3).
(End)

A308401 Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.

Original entry on oeis.org

3, 6, 16, 30, 56, 91, 150, 224, 336, 477, 672, 912, 1233, 1617, 2112, 2700, 3432, 4290, 5340, 6552, 8008, 9678, 11648, 13888, 16503, 19448, 22848, 26658, 31008, 35853, 41346, 47424, 54264, 61803, 70224, 79464, 89733, 100947, 113344, 126840, 141680, 157780, 175416, 194480, 215280, 237708

Offset: 9

Petros Hadjicostas, May 24 2019

Bracelets that have no reflection symmetry are also known as chiral bracelets.
Here, for n >= 6, a(n) is also the number of dihedral compositions of n with 6 parts that have no reflection symmetry. Taking the MacMahon conjugates of these dihedral compositions, we see that a(n) is also the number of dihedral compositions of n into n-6 parts that have no reflection symmetry.
A cyclic composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation. Such compositions were first studied extensively by Sommerville (1909).
A dihedral composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation or a reversal of order. Such compositions were studied, for example, by Knopfmacher and Robbins (2013).
Given a bracelet of length n with k white beads and n-k black beads, we may get the corresponding dihedral composition using MacMahon's correspondence: start with a white bead and count that bead and the black beads that follow (in one direction), and call that b_1; then start with the next white bead and count that one and the black beads that follow, and call that b_2; repeat this process until you reach the k-th white bead and count that one and the black beads that follow, and call that b_k. The corresponding dihedral composition is b_1 + b_2 + ... + b_k.
If in the previous paragraph (given a bracelet of length n with k white beads and n-k black beads), we replace the white beads with black beads and the black beads with white beads, we get a dihedral composition of n into n-k parts: c_1 + c_2 + ... + c_{n-k}. These two dihedral compositions (which correspond to the same bracelet) are called "conjugate" compositions. See p. 273 in Sommerville (1909) for an explanation of "conjugate" compositions in the context of cyclic compositions.
Symmetric cyclic compositions of a positive integer n were first studied by Sommerville (1909, pp. 301-304). It can be proved that the study of necklaces with reflection symmetry using beads of two colors is equivalent to the study of symmetric cyclic compositions of a positive integer. Clearly all the necklaces with reflection symmetry are all the bracelets (turnover necklaces) with reflection symmetry. See also the comments for sequences A119963, A292200, and A295925.

			Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry (and thus, a(9) = 3).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric (and thus, a(9) = 3).
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries (and thus, a(9) = 3). For example, dihedral composition 1 is symmetric because we can draw an axis of symmetry through one of the 1s and 4. In addition, dihedral composition 5 is symmetric because we may draw an axis of symmetry through the numbers 2 and 3.

Colin Barker, Table of n, a(n) for n = 9..1000
Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 964-999.
Petros Hadjicostas, The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry, 2019.
Arnold Knopfmacher and Neville Robbins, Some properties of dihedral compositions, Util. Math. 92 (2013), 207-220.
Richard H. Reis, A formula for C(T) in Gupta's paper, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 1000-1001.
Frank Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
Duncan M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,-1,1,4,-3,-3,4,1,-1,-3,1,2,-1).

Cf. A005513, A008804, A032246, A032247, A032248, A032249, A058187, A119963, A292200, A295925. Column k = 6 of A180472.

a(n) = (1/12)* (sumdiv(gcd(n, 6), d,  eulerphi(d)*binomial((n/d) - 1, (6/d) - 1))) - (1/2)*binomial(floor(n/2), 3); \\ Michel Marcus, May 28 2019

Vec(x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^50)) \\ Colin Barker, Jun 02 2019

G.f.: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)) with k = 6. (This formula is due to Herbert Kociemba.)
a(n) = A005513(n) - A058187(n-6) = A005513(n) - binomial(floor(n/2), 3) for n >= 6.
a(n) =  -(1/2)*binomial(floor(n/2), 3) + (1/12)* Sum_{d|gcd(n, 6)} phi(d)*binomial((n/d) - 1, (6/d) - 1) for n >= 6. (This is a modification of formulas found in Gupta (1979) and Shevelev (2004).)
From Colin Barker, May 26 2019: (Start)
G.f.: x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) + a(n-5) + 4*a(n-6) - 3*a(n-7) - 3*a(n-8) + 4*a(n-9) + a(n-10) - a(n-11) - 3*a(n-12) + a(n-13) + 2*a(n-14) - a(n-15) for n > 23. (End)

A339756 Mark each point on the n X n X n grid with the number of points that are visible from it; a(n) is the number of distinct values in the grid.

Original entry on oeis.org

1, 4, 4, 8, 4, 17, 12, 15, 14, 33, 12, 58, 28, 43, 52, 113, 39, 140, 57, 124, 129, 240, 66, 241, 173, 270, 217, 362, 58, 388, 292, 454, 351, 539, 166, 783, 471, 723, 463, 880, 229, 1134, 642, 843, 763, 1441, 311, 1415, 740, 1295, 987, 1888, 357, 1629, 1063, 1750, 1231, 2381, 289, 2652

Offset: 1

Torlach Rush, Dec 15 2020

nonn

a(n) <= A058187(n). This is because A058187(n) is the maximum number of points required to calculate a(n).

			a(1) = 1 because there are 7 visible points from every point on the grid.
a(2) = 4 because 19 points are visible from every vertex of the grid, 23 points are visible from the midpoint of every edge of the grid, 25 points are visible from the midpoint of every face of the grid, and 26 points are visible from the middle of the grid.
a(3) = 4 because 49 points are visible from every vertex of the grid, 53 points are visible from the inner points of every edge of the grid, 55 points are visible from the inner points of every face of the grid, and 56 points are visible from the inner points of the grid.

Eric Weisstein's World of Mathematics, Visible Point

Cf. A049687, A049691, A058187, A090025, A339400.

\\ n = side length, d = dimension
cdvps(n, d) ={my(m=Map());
  forvec(u=vector(d, i, [0, n\2]),
    my(c=0); forvec(v=[[t-n, t]|t<-u], c+=(gcd(v)==1));
    mapput(m, c, 1), 1);
  #m; }
a(n) = cdvps(n, 3)

A024869 a(n) = s(1)t(n) + s(2)t(n-1) + ... + s(k)t(n-k+1), where k = floor( n/2 ), s = natural numbers >= 2, t = natural numbers >= 3.

Original entry on oeis.org

8, 10, 27, 32, 61, 70, 114, 128, 190, 210, 293, 320, 427, 462, 596, 640, 804, 858, 1055, 1120, 1353, 1430, 1702, 1792, 2106, 2210, 2569, 2688, 3095, 3230, 3688, 3840, 4352, 4522, 5091, 5280, 5909, 6118, 6810, 7040, 7798, 8050, 8877, 9152, 10051, 10350, 11324, 11648

Offset: 2

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1)

CoefficientList[Series[(8 + 2 x - 7 x^2 - x^3 + 2 x^4)/((1 + x)^3 (x - 1)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 25 2013 *)

Vec(x^2*(8+2*x-7*x^2-x^3+2*x^4)/((1+x)^3*(x-1)^4) + O(x^100)) \\ Colin Barker, Jan 29 2016

G.f.: x^2*(8+2*x-7*x^2-x^3+2*x^4) / ((1+x)^3*(x-1)^4). - R. J. Mathar, Sep 25 2013
a(n) = 8*A058187(n-2) +2*A058187(n-3) -7*A058187(n-4) -A058187(n-5) +2*A058187(n-6). - R. J. Mathar, Sep 25 2013
From Colin Barker, Jan 29 2016: (Start)
a(n) = (4*n^3+3*((-1)^n+13)*n^2+4*(6*(-1)^n+17)*n+42*((-1)^n-1))/48.
a(n) = (2*n^3+21*n^2+46*n)/24 for n even.
a(n) = (2*n^3+18*n^2+22*n-42)/24 for n odd.
(End)

A024875 a(n) = s(1)s(n) + s(2)s(n-1) + ... + s(k)s(n-k+1), where k = floor( n/2 ), s = natural numbers >= 3.

Original entry on oeis.org

12, 15, 38, 45, 82, 94, 148, 166, 240, 265, 362, 395, 518, 560, 712, 764, 948, 1011, 1230, 1305, 1562, 1650, 1948, 2050, 2392, 2509, 2898, 3031, 3470, 3620, 4112, 4280, 4828, 5015, 5622, 5829, 6498, 6726, 7460, 7710, 8512, 8785, 9658, 9955, 10902, 11224, 12248, 12596

Offset: 2

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1)

CoefficientList[Series[(12 + 3 x - 13 x^2 - 2 x^3 + 4 x^4)/((1 + x)^3 (x - 1)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 25 2013 *)
LinearRecurrence[{1,3,-3,-3,3,1,-1},{12,15,38,45,82,94,148},50] (* Harvey P. Dale, Jul 21 2015 *)

Vec(x^2*(12+3*x-13*x^2-2*x^3+4*x^4)/((1+x)^3*(x-1)^4) + O(x^100)) \\ Colin Barker, Jan 29 2016

G.f.: x^2*(12+3*x-13*x^2-2*x^3+4*x^4) / ((1+x)^3*(x-1)^4). - R. J. Mathar, Sep 25 2013
a(n) = 12*A058187(n-2) +3*A058187(n-3) -13*A058187(n-4) -2*A058187(n-5) +4*A058187(n-6). - R. J. Mathar, Sep 25 2013
From Colin Barker, Jan 29 2016: (Start)
a(n) = (4*n^3+3*((-1)^n+19)*n^2+2*(15*(-1)^n+61)*n+75*((-1)^n-1))/48.
a(n) = (2*n^3+30*n^2+76*n)/24 for n even.
a(n) = (2*n^3+27*n^2+46*n-75)/24 for n odd. (End)

A093039 Sequence resulting from a sum of three repeated binomial(n+3,4) sequences.

Original entry on oeis.org

1, 2, 7, 11, 25, 35, 65, 85, 140, 175, 266, 322, 462, 546, 750, 870, 1155, 1320, 1705, 1925, 2431, 2717, 3367, 3731, 4550, 5005, 6020, 6580, 7820, 8500, 9996, 10812, 12597, 13566, 15675, 16815, 19285, 20615, 23485, 25025, 28336, 30107, 33902

Offset: 1

Alford Arnold, May 08 2004

Euler transform of length 3 sequence [2,k,-1] with k=4 (cf. A028724 for k=3). - Georg Fischer, Nov 28 2020

			b(n) = 1,  1,  5,  5, 15, 15, 35, 35, 70, 70,126,126
     + 0,  1,  1,  5,  5, 15, 15, 35, 35, 70, 70,126
     + 0,  0,  1,  1,  5,  5, 15, 15, 35, 35, 70, 70
     -----------------------------------------------
a(n) = 1,  2,  7, 11, 25, 35, 65, 85,140,175,266,322

Cf. A001651(k=1), A001318(k=2), A028724(k=3).

Cf. repeated binomial coefficients: A008805(k=2), A058187(k=3), A189976(k=4).

k := 4; nmax := 32; a := Flatten[Table[{Binomial[n,k], Binomial[n,k]},{n,k,nmax}]];
a + Flatten[Join[{0}, Drop[a,-1]]] + Flatten[Join[{0,0}, Drop[a,-2]]] (* Georg Fischer, Nov 29 2020 *)

a(1) = b(1), a(2) = b(2), a(n) = b(n) + b(n-1) + b(n-2) for n > 2, where k = 4 and b(n) = binomial(floor((n+7)/2), k) = A189976(n-7).

More terms from and edited by Georg Fischer, Nov 28 2020

cp's OEIS Frontend

A024862 a(n) = s(1)*t(n) + s(2)*t(n-1) + ... + s(k)*t(n-k+1), where k = floor(n/2), s = natural numbers, t = odd natural numbers.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A024868 a(n) = 2*(n+1) + 3*n + ... + (k+1)*(n+2-k), where k = floor(n/2).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A124458 Triangular array resulting from summing three repeated Pascal sequences; related to the generalized pentagonal sequence (A001318) and the classical modular tessellation (cf. A054886).

Views

Author

Keywords

Comments

Examples

Crossrefs

A207974 Triangle related to A152198.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

PARI

Formula

A263794 Number of (n+1) X (3+1) 0..1 arrays with each row and column divisible by 3, read as a binary number with top and left being the most significant bits, and rows and columns lexicographically nonincreasing.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A308401 Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A339756 Mark each point on the n X n X n grid with the number of points that are visible from it; a(n) is the number of distinct values in the grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A024869 a(n) = s(1)t(n) + s(2)t(n-1) + ... + s(k)t(n-k+1), where k = floor( n/2 ), s = natural numbers >= 2, t = natural numbers >= 3.

Views

Author

Keywords

A024862 a(n) = s(1)t(n) + s(2)t(n-1) + ... + s(k)*t(n-k+1), where k = floor(n/2), s = natural numbers, t = odd natural numbers.

A024868 a(n) = 2(n+1) + 3n + ... + (k+1)*(n+2-k), where k = floor(n/2).