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 1-8 of 8 results.

A071922 Unimodal analog of binomial coefficient, such that A071921(n,m) = a(n+m-1,n) for all (n,m) different from (0,0), arranged in a Pascal-like triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 9, 7, 1, 1, 5, 16, 22, 11, 1, 1, 6, 25, 50, 46, 16, 1, 1, 7, 36, 95, 130, 86, 22, 1, 1, 8, 49, 161, 295, 296, 148, 29, 1, 1, 9, 64, 252, 581, 791, 610, 239, 37, 1, 1, 10, 81, 372, 1036, 1792, 1897, 1163, 367, 46, 1, 1, 11, 100, 525, 1716, 3612
Offset: 0

Views

Author

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 14 2002

Keywords

Comments

Also, number of n-length k-ary words avoiding the pattern 1'-2-1". - Ralf Stephan, Apr 28 2004
The matrix inverse starts
1;
-1, 1;
1, -2, 1;
-2, 5, -4, 1;
8, -21, 19, -7, 1;
-56, 148, -137, 55, -11, 1;
608, -1608, 1493, -608, 130, -16, 1;
-9440, 24968, -23190, 9461, -2044, 266, -22, 1;
198272, -524416, 487088, -198761, 42997, -5642, 490, -29, 1; - R. J. Mathar, Mar 15 2013

Examples

			Triangle begins
  1;
  1,   1;
  1,   2,   1;
  1,   3,   4,   1;
  1,   4,   9,   7,   1;
  1,   5,  16,  22,  11,   1;
  1,   6,  25,  50,  46,  16,   1;
  1,   7,  36,  95, 130,  86,  22,   1;
		

Programs

  • GAP
    Flat(List([0..11], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial( 2*j+k-1, 2*j) )))); # G. C. Greubel, Aug 26 2019
  • Magma
    [&+[Binomial(2*j+k-1, 2*j): j in [0..n-k]]: k in [0..n], n in [0..11]]; // G. C. Greubel, Aug 26 2019
    
  • Maple
    A071922 := proc(n,k)
        add( binomial(2*j+k-1,2*j),j=0..n-k) ;
    end proc: # R. J. Mathar, Mar 15 2013
  • Mathematica
    a[n_, m_]:= Sum[Binomial[2k+m-1, 2k], {k, 0, n-m}]; Flatten[ Table[ a[n, m], {n, 0, 11}, {m, 0, n}]]
  • PARI
    a(n,k) = sum(j=0, n-k, binomial(2*j+k-1, 2*j));
    for(n=0,11, for(k=0,n, print1(a(n,k), ", "))) \\ G. C. Greubel, Aug 26 2019
    
  • Sage
    [[sum(binomial(2*j+k-1, 2*j) for j in (0..n-k)) for k in (0..n)] for n in (0..11)] # G. C. Greubel, Aug 26 2019
    

Formula

a(n, m) = Sum_{k=0..n-m} binomial(2*k+m-1, 2*k).
Sum_{m=0..n} a(n, m) = 1 + Fibonacci(2*n).
Sum_{m=0..n} (-1)^m*a(n, m) = 1 if 3 divides n, 0 otherwise.
G.f. for k-th row: 1/(1-x)^(2k-1) + Sum_{j=1..k-1} x/(1-x)^(2j). - Ralf Stephan, Apr 28 2004

Extensions

Edited by Robert G. Wilson v, Jun 17 2002

A059259 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 2, 0, 1, 3, 4, 2, 1, 1, 4, 7, 6, 3, 0, 1, 5, 11, 13, 9, 3, 1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 7, 22, 40, 46, 34, 16, 4, 1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, 1, 10, 46, 128, 239, 314, 296, 200
Offset: 0

Views

Author

N. J. A. Sloane, Jan 23 2001

Keywords

Comments

This sequence provides the general solution to the recurrence a(n) = a(n-1) + k*(k+1)*a(n-2), a(0)=a(1)=1. The solution is (1, 1, k^2 + k + 1, 2*k^2 + 2*k + 1, ...) whose coefficients can be read from the rows of the triangle. The row sums of the triangle are given by the case k=1. These are the Jacobsthal numbers, A001045. Viewed as a square array, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows are successive partial sums given by 1/((1-x)^n * (1-x^2)). - Paul Barry, Mar 17 2003
Conjecture: every second column of this triangle is identical to a column in the square array A071921. For example, column 4 of A059259 (1, 3, 9, 22, 46, ...) appears to be the same as column 3 of A071921; column 6 of A059259 (1, 4, 16, 50, 130, 296, ...) appears to be the same as column 4 of A071921; and in general column 2k of A059259 appears to be the same as column k+1 of A071921. Furthermore, since A225010 is a transposition of A071921 (ignoring the latter's top row and two leftmost columns), there appears to be a correspondence between column 2k of A059259 and row k of A225010. - Mathew Englander, May 17 2014
T(n,k) is the number of n-tilings of a (one-dimensional) board that use k (1,1)-fence tiles and n-k squares. A (1,1)-fence is a tile composed of two pieces of width 1 separated by a gap of width 1. - Michael A. Allen, Jun 25 2020
See the Edwards-Allen 2020 paper, page 14, for proof of Englander's conjecture. - Michael De Vlieger, Dec 10 2020

Examples

			Triangle begins:
  1;
  1,  0;
  1,  1,  1;
  1,  2,  2,   0;
  1,  3,  4,   2,   1;
  1,  4,  7,   6,   3,   0;
  1,  5, 11,  13,   9,   3,   1;
  1,  6, 16,  24,  22,  12,   4,   0;
  1,  7, 22,  40,  46,  34,  16,   4,  1;
  1,  8, 29,  62,  86,  80,  50,  20,  5,  0;
  1,  9, 37,  91, 148, 166, 130,  70, 25,  5, 1;
  1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0;
...
		

Crossrefs

See A059260 for an explicit formula.
Diagonals of this triangle are given by A006498.
Similar to the triangles A035317, A080242, A108561, A112555.

Programs

  • Maple
    read transforms; 1/(1-x-x*y-y^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);
  • Mathematica
    T[n_, 0]:= 1; T[n_, n_]:= (1+(-1)^n)/2; T[n_, k_]:= T[n, k] = T[n-1, k] + T[n-1, k-1]; Table[T[n, k], {n, 0, 10} , {k, 0, n}]//Flatten (* G. C. Greubel, Jan 03 2017 *)
  • PARI
    {T(n,k) = if(k==0, 1, if(k==n, (1+(-1)^n)/2, T(n-1,k) +T(n-1,k-1)) )};
    for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Apr 29 2019
  • Sage
    def A059259_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return (-1)^n
            if k==0: return 0
            return prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n)]
    for n in (1..12): print(A059259_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 1/(1 - x - x*y - y^2).
As a square array read by antidiagonals, this is T(n, k) = Sum_{i=0..n} (-1)^(n-i)*C(i+k, k). - Paul Barry, Jul 01 2003
T(2*n,n) = A026641(n). - Philippe Deléham, Mar 08 2007
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = T(2,1) = T(2,2)=1, T(1,1)=0, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 24 2013
T(n,0) = 1, T(n,n) = (1+(-1)^n)/2, and T(n,k) = T(n-1,k) + T(n-1,k-1) for 0 < k < n. - Mathew Englander, May 24 2014
From Michael A. Allen, Jun 25 2020: (Start)
T(n,k) + T(n-1,k-1) = binomial(n,k) if n >= k > 0.
T(2*n-1,2*n-2) = T(2*n,2*n-1) = n, T(2*n,2*n-2) = n^2, T(2*n+1,2*n-1) = n*(n+1) for n > 0.
T(n,2) = binomial(n-2,2) + n - 1 for n > 1 and T(n,3) = binomial(n-3,3) + 2*binomial(n-2,2) for n > 2.
T(2*n-k,k) = A123521(n,k). (End)

A071920 Square array giving number of unimodal functions [n]->[m] for n>=0, m>=0, with a(0,m)=0 for all m>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 4, 1, 0, 0, 4, 9, 7, 1, 0, 0, 5, 16, 22, 11, 1, 0, 0, 6, 25, 50, 46, 16, 1, 0, 0, 7, 36, 95, 130, 86, 22, 1, 0, 0, 8, 49, 161, 295, 296, 148, 29, 1, 0, 0, 9, 64, 252, 581, 791, 610, 239, 37, 1, 0
Offset: 0

Views

Author

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 14 2002

Keywords

Comments

If one uses a definition of unimodality that involves existential quantifiers on the domain of a function then a(0,m)=0 a priori.

Examples

			Square array a(n,m) begins:
  0,    0,    0,    0,    0,    0,     0,     0,      0, ...
  0,    1,    2,    3,    4,    5,     6,     7,      8, ...
  0,    1,    4,    9,   16,   25,    36,    49,     64, ...
  0,    1,    7,   22,   50,   95,   161,   252,    372, ...
  0,    1,   11,   46,  130,  295,   581,  1036,   1716, ...
  0,    1,   16,   86,  296,  791,  1792,  3612,   6672, ...
  0,    1,   22,  148,  610, 1897,  4900, 11088,  22716, ...
  0,    1,   29,  239, 1163, 4166, 12174, 30738,  69498, ...
  0,    1,   37,  367, 2083, 8518, 27966, 78354, 194634, ...
		

Crossrefs

Main diagonal is A088536.

Programs

Formula

a(n,m) = Sum_{k=0..m-1} binomial(n+2k-1, 2k) if n>0.

A225010 T(n,k) = number of n X k 0..1 arrays with rows unimodal and columns nondecreasing.

Original entry on oeis.org

2, 4, 3, 7, 9, 4, 11, 22, 16, 5, 16, 46, 50, 25, 6, 22, 86, 130, 95, 36, 7, 29, 148, 296, 295, 161, 49, 8, 37, 239, 610, 791, 581, 252, 64, 9, 46, 367, 1163, 1897, 1792, 1036, 372, 81, 10, 56, 541, 2083, 4166, 4900, 3612, 1716, 525, 100, 11, 67, 771, 3544, 8518, 12174, 11088, 6672, 2685, 715, 121, 12
Offset: 1

Views

Author

R. H. Hardin, Apr 23 2013

Keywords

Comments

Table starts
..2...4...7...11....16.....22.....29......37......46.......56.......67
..3...9..22...46....86....148....239.....367.....541......771.....1068
..4..16..50..130...296....610...1163....2083....3544.....5776.....9076
..5..25..95..295...791...1897...4166....8518...16414....30086....52834
..6..36.161..581..1792...4900..12174...27966...60172...122464...237590
..7..49.252.1036..3612..11088..30738...78354..186142...416394...884236
..8..64.372.1716..6672..22716..69498..194634..505912..1233584..2845492
..9..81.525.2685.11517..43065.144111..439791.1241383..3276559..8157227
.10.100.715.4015.18832..76714.278707..920491.2803658..7963384.21280337
.11.121.946.5786.29458.129844.508937.1808521.5911763.17978389.51325352
From Charles A. Lane, Aug 22 2013: (Start)
The first column is also the coefficients of a in y''[x] - a*x^n*y[x] + b*en*y[x] = 0 where n = 0. The recursion yields coefficients of a, a*b*en, a*b^2*en^2 etc.
The second column is obtained when n=1, the third column when n=2. The final column is for n=10.
Example: Write a normal recursion for n=4. For convenience set x to 1. Running the recursion yields
1-(b en)/2+(b^2 en^2)/24+1/30 (a-(b^3 en^3)/24)+(-384 a b en+b^4 en^4)/40320+(2064 a b^2 en^2-b^5 en^5)/3628800+(120960 a^2-7104 a b^3 en^3+b^6 en^6)/479001600+(-4682880 a^2 b en+18984 a b^4 en^4-b^7 en^7)/87178291200+(54268416 a^2 b^2 en^2-43008 a b^5 en^5+b^8 en^8)/20922789888000.
The coefficient of a is 24, the coefficient of a b en is 384 and the coefficient of a b^2 en^2 is 2064. Dividing by 4! yields a sequence of 1,16,86... , the same as column 5 without the leading 1. There is a hint of unity among the oscillators. (End)

Examples

			Some solutions for n=3 k=4
..0..0..0..0....0..1..0..0....0..0..0..0....1..1..1..1....0..0..0..0
..0..0..0..0....0..1..1..0....0..0..0..0....1..1..1..1....1..1..0..0
..0..0..0..1....1..1..1..0....1..1..0..0....1..1..1..1....1..1..1..1
		

Crossrefs

Column 2 is A000290(n+1).
Column 3 is A002412(n+1).
Column 4 is A006324(n+1).
Row 1 is A000124.
Row 2 is A223718.
Row 3 is A223659.
Cf. A071920, A071921 (larger and reflected versions of table). - Alois P. Heinz, Sep 22 2013

Programs

  • Maple
    T:= (n, k)-> add(binomial(k+2*j-1, 2*j), j=0..n):
    seq(seq(T(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Sep 22 2013
  • Mathematica
    T[n_, k_] := Sum[Binomial[k + 2*j - 1, 2*j], {j, 0, n}]; Table[T[n - k + 1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Apr 07 2016, after Alois P. Heinz *)

Formula

Empirical: columns k=1..7 are polynomials of degree k.
Empirical: rows n=1..7 are polynomials of degree 2n.
T(n,k) = Sum_{j=0..n} C(k+2*j-1,2*j). - Alois P. Heinz, Sep 22 2013

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

Views

Author

Tom Copeland, Mar 19 2014

Keywords

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1
		

Crossrefs

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Programs

  • Magma
    [[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
    
  • Maple
    A239473 := proc(n,k)
        add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
    end proc; # R. J. Mathar, Jul 21 2016
  • Mathematica
    Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
    
  • Sage
    Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Extensions

Inverse array added by Tom Copeland, Mar 26 2014
Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A227402 Number of unimodal functions f:[n]->[n^2].

Original entry on oeis.org

1, 1, 16, 525, 24616, 1505205, 113772114, 10253539205, 1073769343504, 128165285630637, 17177527372642000, 2554518029816653175, 417444979902876203656, 74358489250362053975095, 14340040595865309129453250, 2976703788777987140216622005
Offset: 0

Views

Author

Alois P. Heinz, Sep 20 2013

Keywords

Crossrefs

Main diagonal of A226031.

Programs

  • Maple
    a:= n-> `if`(n=0, 1, add(binomial(n+2*j-1, 2*j), j=0..n^2-1)):
    seq(a(n), n=0..20);
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n+2*j-1, 2*j], {j,0,n^2-1}],{n,1,20}]}] (* Vaclav Kotesovec, Aug 29 2014 *)

Formula

a(n) = Sum_{j=0..n^2-1} C(n+2*j-1,2*j), a(0) = 1.
a(n) = A071921(n,n^2).
a(n) ~ 2^(n-3/2) * n^(n-1/2) * exp(n+1/4) / sqrt(Pi). - Vaclav Kotesovec, Aug 29 2014

A226031 Number A(n,k) of unimodal functions f:[n]->[k*n]; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 4, 0, 1, 3, 16, 22, 0, 1, 4, 36, 161, 130, 0, 1, 5, 64, 525, 1716, 791, 0, 1, 6, 100, 1222, 8086, 18832, 4900, 0, 1, 7, 144, 2360, 24616, 128248, 210574, 30738, 0, 1, 8, 196, 4047, 58730, 510664, 2072862, 2385644, 194634, 0
Offset: 0

Views

Author

Alois P. Heinz, May 23 2013

Keywords

Examples

			Square array A(n,k) begins:
  1,   1,     1,      1,      1,       1, ...
  0,   1,     2,      3,      4,       5, ...
  0,   4,    16,     36,     64,     100, ...
  0,  22,   161,    525,   1222,    2360, ...
  0, 130,  1716,   8086,  24616,   58730, ...
  0, 791, 18832, 128248, 510664, 1505205, ...
		

Crossrefs

Columns k=0-2 give: A000007, A088536, A226012.
Rows n=0-2 give: A000012, A001477, A016742.
Main diagonal gives: A227402.
Cf. A071920.

Programs

  • Maple
    A:= (n, k)-> `if`(n=0, 1, add(binomial(n+2*j-1, 2*j), j=0..k*n-1)):
    seq(seq(A(n, d-n), n=0..d), d=0..10);
  • Mathematica
    A[n_, k_] := If[n==0, 1, Sum[Binomial[n + 2j - 1, 2j], {j, 0, k n - 1}]];
    Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 20 2020, after Alois P. Heinz *)

Formula

A(n,k) = Sum_{j=0..k*n-1} C(n+2*j-1,2*j), A(0,k) = 1.
A(n,k) = A071921(n,k*n).

A227406 Number of unimodal functions f:[n]->[2^n].

Original entry on oeis.org

1, 2, 16, 372, 24616, 5014592, 3349471840, 7649590386464, 61356625102897216, 1758844330913892684288, 182379122144778004351027200, 69026760045145802122822210022400, 96048744530120196897251255933762037760, 494360393380904255996973467025921794482614272
Offset: 0

Views

Author

Alois P. Heinz, Sep 21 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> sum(binomial(n+2*j-1, 2*j), j=0..2^n-1):
    seq(a(n), n=0..20);
  • Mathematica
    Table[Sum[Binomial[n+2*j-1,2*j],{j,0,2^n-1}],{n,0,15}] (* Vaclav Kotesovec, Sep 22 2013 *)

Formula

a(n) = Sum_{j=0..2^n-1} C(n+2*j-1,2*j).
a(n) = A071921(n,2^n).
a(n) ~ 2^(n^2+n-1)/n!. - Vaclav Kotesovec, Sep 22 2013
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