A006347 -id:A006347 - cp's OEIS frontend

A055596 Expansion of e.g.f. (2 - x - 2*exp(-x))/(1-x).

1, 0, 2, 6, 32, 190, 1332, 10654, 95888, 958878, 10547660, 126571918, 1645434936, 23036089102, 345541336532, 5528661384510, 93987243536672, 1691770383660094, 32143637289541788, 642872745790835758, 13500327661607550920, 297007208555366120238

Offset: 1

Gary Detlefs, Jul 10 2000

nonn

It appears that a(n) = n*a(n-1) + 2(-1)^(n+1) and that the n-th term of any sequence of the form {A(0) =a, A(1)= b, A(n) = (n-1)(A(n-1)+A(n-2))} is A(n) = b*A000166(n) + a*A055596(n). A(n) can also be expressed as A(n) = n*A(n-1) + (2a-b)(-1)^(n+1). - Gary Detlefs, Jun 13 2009
For n>1, sum over all permutations beginning with an ascent, ending with a descent, and without double ascents on n elements and each contributes 2 to the power of the number of double descents. By symmetry, also the sum over all permutations with a descent, ending with an ascent, and no double ascents and the sum is (again) over 2 to the power of the number of double descents. - Richard Ehrenborg, Oct 08 2013
It also appears related to a Secret Santa problem: given n people getting names from an urn to give presents and putting them back in the urn if they get their own name, this seems to be the number of ways that it may fail for the last person, as he/she has no other name to get from the urn. - João Batista Souza de Oliveira, Jan 25 2016

			a(4)=6 since the 3 permutations 1432, 2431, 3421 all have one double descent and hence each contributes 2 to the sum. - _Richard Ehrenborg_, Oct 08 2013
For the Secret Santa, a(3)=2 since person 1 will get the names of either person 2 or 3. Suppose it was person 2. Person 2 will then get either person 1 or person 3. If he/she gets person 1, the draw will fail for person 3. The other case occurs when person 1 draws person 3, person 3 draws person 1 and the draw fails for person 2. - _João Batista Souza de Oliveira_, Jan 25 2016

Alois P. Heinz, Table of n, a(n) for n = 1..200
R. Ehrenborg and J. Jung, Descent pattern avoidance, Adv. in Appl. Math., 49 (2012) 375-390.
Lapo Cioni and Luca Ferrari, Preimages under the Queuesort algorithm, arXiv preprint arXiv:2102.07628 [math.CO], 2021; Discrete Math., 344 (2021), #112561.

Cf. A000166, A002467, A230071, A227918.

A055596:= func< n | Factorial(n)*(1 -2*(&+[(-1)^j/Factorial(j): j in [0..n]]) ) >;
[A055596(n): n in [1..30]]; // G. C. Greubel, Sep 06 2022

Rest[CoefficientList[Series[(2-x-2*E^(-x))/(1-x), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Oct 07 2013 *)

PARI
```
a(n)=if(n<2, n>0, n*a(n-1)-2*(-1)^n)
```

a(n)=if(n<1,0,n!*polcoeff((2-x-2*exp(-x+x*O(x^n)))/(1-x),n))

def A055596(n): return factorial(n)*( 2*bool(n==0) + 1 - 2*sum((-1)^j/factorial(j) for j in (0..n)) )
[A055596(n) for n in (1..30)] # G. C. Greubel, Sep 06 2022

E.g.f.: (2-x-2*exp(-x))/(1-x).
a(n) = (n-1)*(a(n-1) + a(n-2)) = 2*A006347(n-1), n>2.
a(n) = n! - 2*A000166(n), n>0.
a(n) ~ n! * (1-2*exp(-1)). - Vaclav Kotesovec, Oct 07 2013
For n>2, a(n) = floor(n! * (1-2*exp(-1)) + 1/2). - Peter Bala, Oct 08 2013
a(n+1) = 2*A002467(n) - n!. - Vaclav Kotesovec, Oct 08 2013

More terms from James Sellers, Jul 11 2000

A176408 a(n) = (n+1)*(a(n-1) +a(n-2)) n>1, a(0)=1,a(1)=0.

Original entry on oeis.org

1, 0, 3, 12, 75, 522, 4179, 37608, 376083, 4136910, 49642923, 645357996, 9035011947, 135525179202, 2168402867235, 36862848742992, 663531277373859, 12607094270103318

Offset: 0

Gary Detlefs, Apr 16 2010

nonn

a(n) is one of two "basis" sequences for sequences of the form s(0)=a,s(1)=b,s(n)=(n+1)(s(n-1)+s(n-2)), n>1, the other being A006347.
s(n) = a*a(n) + b* A006347(n+1).
s(n) = 1/2*(b-2*a)(n+2)! +(3*a-b)*floor(((n+2)!+1)/e).

			a(2)= 3*9-24=3, a(3)= 3*44-120=12, a(4)= 3*265-720=75, ...

Indranil Ghosh, Table of n, a(n) for n = 0..447
Michael Wallner, A bijection of plane increasing trees with relaxed binary trees of right height at most one, arXiv:1706.07163 [math.CO], 2017, Table 2 on p. 13.

Cf. A000166, A006347.

seq(3*floor(((n+2)!+1)/E) - (n+2)!,n=1..20);

a(n) = 3*floor(((n+2)!+1)/e) - (n+2)!.

a(n) = 3* A000166(n+1) - (n+2)!, where A000166 are the subfactorial numbers.

Data section corrected by Indranil Ghosh, Feb 15 2017

A352650 Triangle read by rows: T(n,k) = n * T(n-1,k) + (-1)^(n-k) for 0 <= k <= n with initial values T(n,k) = 0 if n < 0 or k < 0 or k > n.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 4, 2, 1, 9, 15, 9, 3, 1, 44, 76, 44, 16, 4, 1, 265, 455, 265, 95, 25, 5, 1, 1854, 3186, 1854, 666, 174, 36, 6, 1, 14833, 25487, 14833, 5327, 1393, 287, 49, 7, 1, 133496, 229384, 133496, 47944, 12536, 2584, 440, 64, 8, 1, 1334961, 2293839, 1334961, 479439, 125361, 25839, 4401, 639, 81, 9, 1

Offset: 0

Werner Schulte, Apr 04 2022

Conjecture 1: T(n,k) = Sum_{i=0..n-k} (-1)^(n+k+i) * A326326(n-k,i) * n^i for 0 <= k <= n.

Conjecture 2: T(n,k) = T(n-k,0) + Sum_{i=1..n-k} T(n-k,i) * T(i+k,k) * k / (i + k - 1) for 0 < k <= n.

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :       0       1       2      3      4     5    6   7  8  9
================================================================
  0 :       1
  1 :       0       1
  2 :       1       1       1
  3 :       2       4       2      1
  4 :       9      15       9      3      1
  5 :      44      76      44     16      4     1
  6 :     265     455     265     95     25     5    1
  7 :    1854    3186    1854    666    174    36    6   1
  8 :   14833   25487   14833   5327   1393   287   49   7  1
  9 :  133496  229384  133496  47944  12536  2584  440  64  8  1
  etc.

Cf. A000166 (column 0 and 2), A002467 (column 1), A006347 (column 3), A006348 (column 4), A009179 (row sums, signed), A352988 (matrix inverse).

Cf. A094587, A097807, A326326.

T := proc(n,k) option remember;
if k > n then 0 else n * T(n-1,k) + (-1)^(n-k) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, Apr 11 2022

T(n,n) = 1 for n >= 0.
T(n,n-1) = n - 1 for n > 0.
T(n,n-2) = (n - 1)^2 for n > 1.
T(n,0) = A000166(n) for n >= 0.
T(n,1) = A002467(n) for n > 0.
T(n,2) = A000166(n) for n > 1.
T(n,k) + T(n,k+1) = (n!) / (k!) for 0 <= k <= n.
T(n,k) = (n - 1) * (T(n-1,k) + T(n-2,k)) for 0 <= k < n-1.
T(n,k) = (T(n,k-2) - (k - 2) * T(n,k-1)) / (k - 1) for 1 < k <= n.
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k) * x^k satisfy recurrence equation p(n,x) = (n - 1) * (p(n-1,x) + p(n-2,x)) + x^n for n > 0 with initial value p(0,x) = 1.
Row sums are p(n,1) = abs(A009179(n)) for n >= 0.
Alternating row sums are p(n,-1) = (-1)^n for n >= 0.
T(n,k) * T(n+1,k+1) - T(n+1,k) * T(n,k+1) = (-1)^(n-k) * A094587(n,k) for 0 <= k <= n.
Define 3x3-matrices T(i,j) with n <= i <= n+2 and k <= j <= k+2. Then we have: det(T(i,j)) = 0^(n-k) for 0 <= k <= n.
E.g.f. of column k >= 0: Sum_{n>=k} T(n,k) * t^n / (n!) = (Sum_{n>=k} (-t)^n / (n!)) * (-1)^k / (1 - t).
E.g.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n / (n!) = (x * exp(x * t) + exp(-t)) / ((1 + x) * (1 - t)).
p(n,x) = Sum_{k=0..n} ((n!)/(k!))*(x^(k+1) + (-1)^k)/(x + 1) for n >= 0.
T(n,k) = Sum_{i=0..n-k} (-1)^i * (n!) / ((k+i)!) for 0 <= k <= n.
T(n,k) equals matrix product of A094587 and A097807.

cp's OEIS Frontend

A055596 Expansion of e.g.f. (2 - x - 2*exp(-x))/(1-x).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

SageMath

Formula

Extensions

A176408 a(n) = (n+1)*(a(n-1) +a(n-2)) n>1, a(0)=1,a(1)=0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

Extensions

A352650 Triangle read by rows: T(n,k) = n * T(n-1,k) + (-1)^(n-k) for 0 <= k <= n with initial values T(n,k) = 0 if n < 0 or k < 0 or k > n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula