A005491 -id:A005491 - cp's OEIS frontend

A108087 Array, read by antidiagonals, where A(n,k) = exp(-1)*Sum_{i>=0} (i+k)^n/i!.

1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 15, 15, 10, 4, 1, 52, 52, 37, 17, 5, 1, 203, 203, 151, 77, 26, 6, 1, 877, 877, 674, 372, 141, 37, 7, 1, 4140, 4140, 3263, 1915, 799, 235, 50, 8, 1, 21147, 21147, 17007, 10481, 4736, 1540, 365, 65, 9, 1, 115975, 115975, 94828, 60814, 29371, 10427, 2727, 537, 82, 10, 1

Offset: 0

Gerald McGarvey, Jun 05 2005

The column for k=0 is A000110 (Bell or exponential numbers). The column for k=1 is A000110 starting at offset 1. The column for k=2 is A005493 (Sum_{k=0..n} k*Stirling2(n,k).). The column for k=3 is A005494 (E.g.f.: exp(3*z+exp(z)-1).). The column for k=4 is A045379 (E.g.f.: exp(4*z+exp(z)-1).). The row for n=0 is 1's sequence, the row for n=1 is the natural numbers. The row for n=2 is A002522 (n^2 + 1.). The row for n=3 is A005491 (n^3 + 3n + 1.). The row for n=4 is A005492.
Number of ways of placing n labeled balls into n+k boxes, where k of the boxes are labeled and the rest are indistinguishable. - Bradley Austin (artax(AT)cruzio.com), Apr 24 2006
The column for k = -1 (not shown) is A000296 (Number of partitions of an n-set into blocks of size >1. Also number of cyclically spaced (or feasible) partitions.). - Gerald McGarvey, Oct 08 2006
Equals antidiagonals of an array in which (n+1)-th column is the binomial transform of n-th column, with leftmost column = the Bell sequence, A000110. - Gary W. Adamson, Apr 16 2009
Number of partitions of [n+k] where at least k blocks contain their own index element.  A(2,2) = 10: 134|2, 13|24, 13|2|4, 14|23, 1|234, 1|23|4, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4. - Alois P. Heinz, Jan 07 2022

			Array A(n,k) begins:
   1,   1,   1,    1,    1,     1,     1,     1,     1,      1, ... A000012;
   1,   2,   3,    4,    5,     6,     7,     8,     9,     10, ... A000027;
   2,   5,  10,   17,   26,    37,    50,    65,    82,    101, ... A002522;
   5,  15,  37,   77,  141,   235,   365,   537,   757,   1031, ... A005491;
  15,  52, 151,  372,  799,  1540,  2727,  4516,  7087,  10644, ... A005492;
  52, 203, 674, 1915, 4736, 10427, 20878, 38699, 67340, 111211, ... ;
Antidiagonal triangle, T(n, k), begins as:
     1;
     1,    1;
     2,    2,    1;
     5,    5,    3,    1;
    15,   15,   10,    4,   1;
    52,   52,   37,   17,   5,   1;
   203,  203,  151,   77,  26,   6,  1;
   877,  877,  674,  372, 141,  37,  7,  1;
  4140, 4140, 3263, 1915, 799, 235, 50,  8,  1;

F. Ruskey, Combinatorial Generation, preprint, 2001.

Alois P. Heinz, Antidiagonals n = 0..140, flattened
I. Mezo, The r-Bell numbers, J. Int. Seq. 14 (2011) # 11.1.1, Figure 1.
J. Riordan, Letter, Oct 31 1977, The array is on the second page.
F. Ruskey, Combinatorial Generation, 2003.
F. Ruskey, Lexicographic Algorithms [Broken link]

Cf. A000012, A000027, A000110, A002522, A005490, A005491.
Cf. A005492, A005493, A005494, A045379, A086659, A124824.
Main diagonal gives A134980.
Antidiagonal sums give A347420.

A108087:= func< n,k | (&+[Binomial(n-k,j)*k^j*Bell(n-k-j): j in [0..n-k]]) >;
[A108087(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 02 2022

with(combinat):
A:= (n, k)-> add(binomial(n, i) * k^i * bell(n-i), i=0..n):
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Jul 18 2012

Unprotect[Power]; 0^0 = 1; A[n_, k_] := Sum[Binomial[n, i] * k^i * BellB[n - i], {i, 0, n}]; Table[Table[A[d - k, k], {k, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Nov 05 2015, after Alois P. Heinz *)

f(n,k)=round (suminf(i=0,(i+k)^n/i!)/exp(1));
g(n,k)=for(k=0,k,print1(f(n,k),",")) \\ prints k+1 terms of n-th row

def A108087(n,k): return sum( k^j*bell_number(n-k-j)*binomial(n-k,j) for j in range(n-k+1))
flatten([[A108087(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 02 2022

For n> 1, A(n, k) = k^n + sum_{i=0..n-2} A086659(n, i)*k^i. (A086659 is set partitions of n containing k-1 blocks of length 1, with e.g.f: exp(x*y)*(exp(exp(x)-1-x)-1).)
A(n, k) = k * A(n-1, k) + A(n-1, k+1), A(0, k) = 1. - Bradley Austin (artax(AT)cruzio.com), Apr 24 2006
A(n,k) = Sum_{i=0..n} C(n,i) * k^i * Bell(n-i). - Alois P. Heinz, Jul 18 2012
Sum_{k=0..n-1} A(n-k,k) = A005490(n). - Alois P. Heinz, Jan 05 2022
From G. C. Greubel, Dec 02 2022: (Start)
T(n, n) = A000012(n).
T(n, n-1) = A000027(n).
T(n, n-2) = A002522(n-1).
T(n, n-3) = A005491(n-2).
T(n, n-4) = A005492(n+1).
T(2*n, n) = A134980(n).
T(2*n, n+1) = A124824(n), n >= 1.
Sum_{k=0..n} T(n, k) = A347420(n). (End)

A271025 A(n, k) is the n-th binomial transform of the Catalan sequence (A000108) evaluated at k. Array read by descending antidiagonals for n >= 0 and k >= 0.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 14, 15, 10, 4, 1, 42, 51, 37, 17, 5, 1, 132, 188, 150, 77, 26, 6, 1, 429, 731, 654, 371, 141, 37, 7, 1, 1430, 2950, 3012, 1890, 798, 235, 50, 8, 1, 4862, 12235, 14445, 10095, 4706, 1539, 365, 65, 9, 1, 16796, 51822, 71398, 56040, 28820, 10392, 2726, 537, 82, 10, 1

Offset: 0

John M. Campbell, Mar 28 2016

Interestingly, the determinant of the n X n array of entries of the form A(i,j) is equal to the (n-1)-th superfactorial number (see A000178).
As indicated in A104455, the k-th binomial transform of A000108 will have:
   o.g.f.: (1-sqrt((1-(k+4)*x)/(1-k*x)))/(2*x),
   e.g.f.: exp((k+2)*x)*(BesselI(0,2x) - BesselI(1,2x)) and
   a(n) = Sum_{i=0..n} binomial(n, i)*CatalanNumber(i)*k^(n-i).
The columns of this array are polynomial integer sequences. The successive polynomials corresponding to the columns of this array are: p0(n) = 1, p1(n) = n + 1, p2(n) = n^2 + 2n + 2, p3(n) = n^3 + 3*n^2 + 6*n + 5, p4(n) = n^4 + 4*n^3 + 12*n^2 + 20*n + 14, and so forth. The coefficients of these successive polynomials form a number triangle, which is given by A098474.

			The array given by integers of the form A(n,k) is illustrated below:
[0] 1, 1,  2,   5,    14,    42,     132,     429,      1430, ...
[1] 1, 2,  5,   15,   51,    188,    731,     2950,     12235, ...
[2] 1, 3,  10,  37,   150,   654,    3012,    14445,    71398, ...
[3] 1, 4,  17,  77,   371,   1890,   10095,   56040,    320795, ...
[4] 1, 5,  26,  141,  798,   4706,   28820,   182461,   1188406, ...
[5] 1, 6,  37,  235,  1539,  10392,  72267,   516474,   3783115, ...
[6] 1, 7,  50,  365,  2726,  20838,  162996,  1303485,  10642310, ...
[7] 1, 8,  65,  537,  4515,  38654,  337007,  2991340,  27013723, ...
[8] 1, 9,  82,  757,  7086,  67290,  648420,  6340365,  62893270, ...
[9] 1, 10, 101, 1031, 10643, 111156, 1174875, 12568686, 136080971, ...
Seen as a triangle:
                          1
                         1, 1
                       2, 2, 1
                      5, 5, 3, 1
                   14, 15, 10, 4, 1
                 42, 51, 37, 17, 5, 1
             132, 188, 150, 77, 26, 6, 1
          429, 731, 654, 371, 141, 37, 7, 1
      1430, 2950, 3012, 1890, 798, 235, 50, 8, 1

Cf. A098474, A000178. Rows 0-5: A000108, A007317, A064613, A104455, A104498, A154623. Columns 0-3: A000012, A000027, A002522, A005491.

A := (n, k) -> (2/Pi)*int((k+4*x^2)^(n-k)*sqrt(1 - x^2), x=-1..1):
for n from 0 to 9 do seq(A(n,k), k=0..n) od; # Peter Luschny, Jan 27 2020

A000108[n_]:= Binomial[2*n,n]/(n+1) ;
T[i_,j_]: Sum[Binomial(j,k)*A000108(k)*i^(j-k), {k,0,j}] ;
A[0, k_] := CatalanNumber[k]; A[n_, k_] := n^k*Hypergeometric2F1[1/2, -k, 2, -4/n];
Table[A[n, k], {n, 0, 6}, {k, 0, 8}] (* Peter Luschny, Jan 27 2020 *)

def A000108(n): return binomial(2*n,n)/(n+1) ;
def T(i,j): return sum(binomial(j,k)*A000108(k)*i^(j-k) for k in range(j+1))

A(0,j) = A000108(j).
A(i,j) = Sum_{k=0..j} binomial(j,k)*A(i-1,k) for i >= 1.
A(i,j) = Sum_{k=0..j} binomial(j,k)*A000108(k)*i^(j-k).
From Peter Luschny, Jan 27 2020: (Start)
A(n,k) = n^k*hypergeom([1/2, -k], [2], -4/n) for n >= 1.
A(n,k) = (2/Pi)*Integral_{x=-1..1}(k + 4*x^2)^(n - k)*sqrt(1 - x^2). (End)

A197773 Ceiling((n+1/n)^3).

Original entry on oeis.org

8, 16, 38, 77, 141, 235, 365, 537, 757, 1031, 1365, 1765, 2237, 2787, 3421, 4145, 4965, 5887, 6917, 8061, 9325, 10715, 12237, 13897, 15701, 17655, 19765, 22037, 24477, 27091, 29885, 32865, 36037, 39407, 42981, 46765, 50765, 54987, 59437, 64121, 69045, 74215

Offset: 1

Vincenzo Librandi, Oct 19 2011

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Cf. A005491, A014058

Magma
```
[Ceiling((n+1/n)^3): n in [1..60]]
```

A375577 Array read by ascending antidiagonals: A(n,k) = k^n + k*n + 1.

Original entry on oeis.org

2, 1, 2, 1, 3, 2, 1, 4, 5, 2, 1, 5, 9, 7, 2, 1, 6, 15, 16, 9, 2, 1, 7, 25, 37, 25, 11, 2, 1, 8, 43, 94, 77, 36, 13, 2, 1, 9, 77, 259, 273, 141, 49, 15, 2, 1, 10, 143, 748, 1045, 646, 235, 64, 17, 2, 1, 11, 273, 2209, 4121, 3151, 1321, 365, 81, 19, 2

Offset: 0

Stefano Spezia, Aug 19 2024

			Array begins:
  2, 2,  2,   2,    2,     2, ...
  1, 3,  5,   7,    9,    11, ...
  1, 4,  9,  16,   25,    36, ...
  1, 5, 15,  37,   77,   141, ...
  1, 6, 25,  94,  273,   646, ...
  1, 7, 43, 259, 1045,  3151, ...
  1, 8, 77, 748, 4121, 15656, ...
  ...

Cf. A000290, A004247, A004248, A005408 (n=1), A005491 (n=3), A007395 (n=0), A054977 (k=0), A176691 (k=2), A176805 (k=3), A176916 (k=5), A176972 (k=7), A214647.

Cf. A375578 (antidiagonal sums).

A[0,0]=2; A[n_,k_]:=k^n+k*n+1;Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

G.f. for the k-th column: (2*x^2 - 3*x - k^2 + k + 1)/((x - 1)^2*(x - k)).
E.g.f. for the k-th column: exp(x)*(1 + exp((k-1)*x) + k*x).
A(n,1) = n + 2.
A(2,n) = A000290(n+1).
A(n,n) = 2*A214647(n) + 1.

cp's OEIS Frontend

A108087 Array, read by antidiagonals, where A(n,k) = exp(-1)*Sum_{i>=0} (i+k)^n/i!.

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A271025 A(n, k) is the n-th binomial transform of the Catalan sequence (A000108) evaluated at k. Array read by descending antidiagonals for n >= 0 and k >= 0.

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A197773 Ceiling((n+1/n)^3).

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A375577 Array read by ascending antidiagonals: A(n,k) = k^n + k*n + 1.

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A180275 Primes of the form n^3+3*n+1.

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