cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A077042 Square array read by falling antidiagonals of central polynomial coefficients: largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^k = ((1-x^n)/(1-x))^k, i.e., the coefficient of x^floor(k*(n-1)/2) and of x^ceiling(k*(n-1)/2); also number of compositions of floor(k*(n+1)/2) into exactly k positive integers each no more than n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 6, 7, 4, 1, 1, 0, 1, 10, 19, 12, 5, 1, 1, 0, 1, 20, 51, 44, 19, 6, 1, 1, 0, 1, 35, 141, 155, 85, 27, 7, 1, 1, 0, 1, 70, 393, 580, 381, 146, 37, 8, 1, 1, 0, 1, 126, 1107, 2128, 1751, 780, 231, 48, 9, 1, 1, 0, 1, 252, 3139
Offset: 0

Views

Author

Henry Bottomley, Oct 22 2002

Keywords

Comments

From Michel Marcus, Dec 01 2012: (Start)
A pair of numbers written in base n are said to be comparable if all digits of the first number are at least as big as the corresponding digit of the second number, or vice versa. Otherwise, this pair will be defined as uncomparable. A set of pairwise uncomparable integers will be called anti-hierarchic.
T(n,k) is the size of the maximal anti-hierarchic set of integers written with k digits in base n.
For example, for base n=2 and k=4 digits:
- 0 (0000) and 15 (1111) are comparable, while 6 (0110) and 9 (1001) are uncomparable,
- the maximal antihierarchic set is {3 (0011), 5 (0101), 6 (0110), 9 (1001), 10 (1010), 12 (1100)} with 6 elements that are all pairwise uncomparable. (End)

Examples

			Rows of square array start:
  1,    0,    0,    0,    0,    0,    0, ...
  1,    1,    1,    1,    1,    1,    1, ...
  1,    1,    2,    3,    6,   10,   20, ...
  1,    1,    3,    7,   19,   51,  141, ...
  1,    1,    4,   12,   44,  155,  580, ...
  1,    1,    5,   19,   85,  381, 1751, ...
  ...
Read by antidiagonals:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 2, 1, 1;
  0, 1, 3, 3, 1, 1;
  0, 1, 6, 7, 4, 1, 1;
  ...
		

Crossrefs

Programs

Formula

By the central limit theorem, T(n,k) is roughly n^(k-1)*sqrt(6/(Pi*k)).
T(n,k) = Sum{j=0,h/n} (-1)^j*binomial(k,j)*binomial(k-1+h-n*j,k-1) with h=floor(k*(n-1)/2), k>0. - Michel Marcus, Dec 01 2012

A201552 Square array read by diagonals: T(n,k) = number of arrays of n integers in -k..k with sum equal to 0.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 19, 19, 1, 9, 37, 85, 51, 1, 11, 61, 231, 381, 141, 1, 13, 91, 489, 1451, 1751, 393, 1, 15, 127, 891, 3951, 9331, 8135, 1107, 1, 17, 169, 1469, 8801, 32661, 60691, 38165, 3139, 1, 19, 217, 2255, 17151, 88913, 273127, 398567, 180325, 8953, 1
Offset: 1

Views

Author

R. H. Hardin, Dec 02 2011

Keywords

Comments

Equivalently, the number of compositions of n*(k + 1) into n parts with maximum part size 2*k+1. - Andrew Howroyd, Oct 14 2017

Examples

			Some solutions for n=7, k=3:
..1...-2....1...-1....1...-3....0....0....1....2....3...-3....0....2....1....0
.-1....2...-2....2....2....2...-1....0....2....2...-2...-1...-2...-1....2...-1
.-3...-1....1...-3....2....1....0....1....3....0....2....0...-1....2...-2...-1
..0....3....3....3...-2...-2....3....3...-3...-3....0...-1...-1...-1....0....3
..2...-1...-1...-1...-3....0...-3...-2....1...-1...-1....1....1....0....3...-1
..2...-1...-3....0....2....3....0....1...-2....1....1....1....3...-2...-3...-3
.-1....0....1....0...-2...-1....1...-3...-2...-1...-3....3....0....0...-1....3
Table starts:
.   1,      1,       1,        1,        1,         1,...
.   3,      5,       7,        9,       11,        13,...
.   7,     19,      37,       61,       91,       127,...
.  19,     85,     231,      489,      891,      1469,...
.  51,    381,    1451,     3951,     8801,     17151,...
. 141,   1751,    9331,    32661,    88913,    204763,...
. 393,   8135,   60691,   273127,   908755,   2473325,...
.1107,  38165,  398567,  2306025,  9377467,  30162301,...
.3139, 180325, 2636263, 19610233, 97464799, 370487485,...
		

Crossrefs

Programs

  • Maple
    seq(print(seq(add((-1)^i*binomial(n, i)*binomial((k+1)*n-(2*k+1)*i-1, n-1), i = 0..floor((1/2)*n)), k = 1..10)), n = 1..10); # Peter Bala, Oct 16 2024
  • Mathematica
    comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r - 1 - i*m, k - 1]*Binomial[k, i], {i, 0, Floor[(r - k)/m]}];  T[n_, k_] := comps[n*(k + 1), 2*k + 1, n]; Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)
  • PARI
    comps(r, m, k)=sum(i=0, floor((r-k)/m), (-1)^i*binomial(r-1-i*m, k-1)*binomial(k, i));
    T(n,k) = comps(n*(k+1), 2*k+1, n); \\ Andrew Howroyd, Oct 14 2017

Formula

Empirical: T(n,k) = Sum_{i=0..floor(k*n/(2*k+1))} (-1)^i*binomial(n,i)* binomial((k+1)*n-(2*k+1)*i-1,n-1).
The above empirical formula is true and can be derived from the formula for the number of compositions with given number of parts and maximum part size. - Andrew Howroyd, Oct 14 2017
Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k + 1
T(3,k) = 3*k^2 + 3*k + 1
T(4,k) = (16/3)*k^3 + 8*k^2 + (14/3)*k + 1
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (185/12)*k^2 + (35/6)*k + 1
T(6,k) = (88/5)*k^5 + 44*k^4 + 46*k^3 + 25*k^2 + (37/5)*k + 1
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (2275/18)*k^4 + (357/4)*k^3 + (6643/180)*k^2 + (259/30)*k + 1
T(m,k) = (1/Pi)*integral_{x=0..Pi} (sin((k+1/2)x)/sin(x/2))^m dx; for the proof see Dirichlet Kernel link; so f(m,n) = (1/Pi)*integral_{x=0..Pi} (Sum_{k=-n..n} exp(I*k*x))^m dx = sum(integral(exp(I(k_1+...+k_m).x),x=0..Pi)/Pi,{k_1,...,k_m=-n..n}) = sum(delta_0(k1+...+k_m),{k_1,...,k_m=-n..n}) = number of arrays of m integers in -n..n with sum zero. - Yalcin Aktar, Dec 03 2011
T(n, k) = the constant term in the expansion of (x^(-k) + ... + x^(-1) + 1 + x + ... + x^k)^n = the coefficient of x^(k*n) (i.e., the central coefficient) in the expansion of (1 + x + ... + x^(2*k))^n = the coefficient of x^(k*n) in the expansion of ( (1 - x^(2*k+1))/(1 - x) )^n. Expanding the binomials and collecting terms gives the empirical formula above. - Peter Bala, Oct 16 2024

A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10
Offset: 1

Views

Author

Andrey Zabolotskiy, Nov 10 2016

Keywords

Comments

Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
From Juan Pablo Herrera P., Nov 21 2016: (Start)
T(k,n,h) is the number of possible ways of randomly selecting h cards from k-1 sets, each with n different playing cards. It is also the number of lattice paths from (0,0) to (n,h) using steps (1,0), (1,1), (1,2), ..., (1,k-1).
Shallow diagonal sums of each triangle with fixed k give the k-bonacci numbers. (End)
T(k,n,h) is the number of n-dimensional grid points of a k X k X ... X k grid, which are lying in the (n-1)-dimensional hyperplane which is at an L1 distance of h from one of the grid's corners, and normal to the corresponding main diagonal of the grid. - Eitan Y. Levine, Apr 23 2023

Examples

			For first few k and for first few n, the rows with h = 0..n*(k-1) are given:
k=1:  1;  1;  1;  1;  1; ...
k=2:  1;  1, 1;  1, 2, 1;  1, 3, 3, 1;  1, 4, 6, 4, 1; ...
k=3:  1;  1, 1, 1;  1, 2, 3, 2, 1;  1, 3, 6, 7, 6, 3, 1; ...
k=4:  1;  1, 1, 1, 1;  1, 2, 3, 4, 3, 2, 1; ...
For example, (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, hence T(3,3,2) = T(3,3,4) = 6.
From _Eitan Y. Levine_, Apr 23 2023: (Start)
Example for the repeated cumulative sum formula, for (k,n)=(3,3) (each line is the cumulative sum of the previous line, and the first line is the padded, alternating 3rd row from Pascal's triangle):
  1  0  0 -3  0  0  3  0  0 -1
  1  1  1 -2 -2 -2  1  1  1
  1  2  3  1 -1 -3 -2 -1
  1  3  6  7  6  3  1
which is T(3,3,h). (End)
		

Crossrefs

k-nomial arrays for fixed k=1..10: A000012, A007318, A027907, A008287, A035343, A063260, A063265, A171890, A213652, A213651.
Arrays for fixed n=0..6: A000012, A000012, A004737, A109439, A277949, A277950, A277951.
Central n-nomial coefficients for n=1..9, i.e., sequences with h=floor(n*(k-1)/2) and fixed n: A000012, A000984 (A001405), A002426, A005721 (A005190), A005191, A063419 (A018901), A025012, (A025013), A025014, A174061 (A025015), A201549, (A225779), A201550. Arrays: A201552, A077042, see also cfs. therein.
Triangle n=k-1: A181567. Triangle n=k: A163181.

Programs

  • Mathematica
    a = Table[CoefficientList[Sum[x^(h-1),{h,k}]^n,x],{k,10},{n,0,9}];
    Flatten@Table[a[[s-n,n+1]],{s,10},{n,0,s-1}]
    (* alternate program *)
    row[k_, n_] := Nest[Accumulate,Upsample[Table[((-1)^j)*Binomial[n,j],{j,0,n}],k],n][[;;n*(k-1)+1]] (* Eitan Y. Levine, Apr 23 2023 *)

Formula

T(k,n,h) = Sum_{i = 0..floor(h/k)} (-1)^i*binomial(n,i)*binomial(n+h-1-k*i,n-1). [Corrected by Eitan Y. Levine, Apr 23 2023]
From Eitan Y. Levine, Apr 23 2023: (Start)
(T(k,n,h))_{h=0..n*(k-1)} = f(f(...f(g(P))...)), where:
(x_i)_{i=0..m} denotes a tuple (in particular, the LHS contains the values for 0 <= h <= n*(k-1)),
f repeats n times,
f((x_i){i=0..m}) = (Sum{j=0..i} x_j)_{i=0..m} is the cumulative sum function,
g((x_i){i=0..m}) = (x(i/k) if k|i, otherwise 0)_{i=0..m*k} is adding k-1 zeros between adjacent elements,
and P=((-1)^i*binomial(n,i))_{i=0..n} is the n-th row of Pascal's triangle, with alternating signs. (End)
From Eitan Y. Levine, Jul 27 2023: (Start)
Recurrence relations, the first follows from the sequence's defining polynomial as mentioned in the Smarandache link:
T(k,n+1,h) = Sum_{i = 0..s-1} T(k,n,h-i)
T(k+1,n,h) = Sum_{i = 0..n} binomial(n,i)*T(k,n-i,h-i*k) (End)

A208595 Number of n-bead necklaces labeled with numbers -6..6 not allowing reversal, with sum zero.

Original entry on oeis.org

1, 7, 43, 371, 3431, 34153, 353333, 3770475, 41165305, 457714497, 5164908167, 58997692301, 680874861687, 7926902673655, 92986983743513, 1097999648804923, 13040634990748733, 155677447454317639, 1866995100779692627, 22482675584863229261
Offset: 1

Views

Author

R. H. Hardin, Feb 29 2012

Keywords

Examples

			Some solutions for n=4:
.-4...-5...-4...-6...-5...-3...-4...-1...-4...-6...-6...-4...-6...-1...-5...-4
..4....2...-3....5....0....1....0....0....2...-1....3....2....5...-1....4....2
..0...-1....4....1....0....2...-1....0...-3....1....2...-2...-4....0....2....4
..0....4....3....0....5....0....5....1....5....6....1....4....5....2...-1...-2
		

Crossrefs

Column 6 of A208597.

Programs

  • Mathematica
    comps[r_, m_, k_] := Sum[(-1)^i*Binomial[r - 1 - i*m, k - 1]*Binomial[k, i], {i, 0, Floor[(r - k)/m]}]; a[n_Integer, k_] := DivisorSum[n, EulerPhi[n/#] comps[#*(k + 1), 2 k + 1, #] &]/n; a[n_] = a[n, 6]; Array[a, 20] (* Jean-François Alcover, Nov 01 2017, after Andrew Howroyd *)

Formula

a(n) = (1/n) * Sum_{d | n} totient(n/d) * A201550(d). - Andrew Howroyd, Mar 02 2017

Extensions

a(15)-a(20) from Andrew Howroyd, Mar 02 2017
Showing 1-4 of 4 results.