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.

A348208 a(n) = Sum_{k=0..floor(n/2)} (-1)^(k-1)*(k-1)^2*A106828(n, k).

Original entry on oeis.org

-1, 0, 0, 0, -3, -20, -70, -84, 1267, 18824, 209484, 2284920, 26010369, 314864628, 4073158102, 56304102596, 830061867975, 13016975343184, 216535182535928, 3810394068301296, 70744547160678501, 1382375535029293500, 28364229790262962386, 609820072529413714012
Offset: 0

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Author

Mélika Tebni, Oct 07 2021

Keywords

Comments

For all p prime, a(p) == 0 (mod p*(p-1)).

Examples

			E.g.f.: -1 - 3*x^4/4! - 20*x^5/5! - 70*x^6/6! - 84*x^7/7! + 1267*x^8/8! + 18824*x^9/9! + ...
a(11) = Sum_{k=0..5} (-1)^(k-1)*(k-1)^2*A106828(11, k).
a(11) = (-1)*1*0 + (1)*0*3628800 + (-1)*1*6636960 + (1)*4*3678840 + (-1)*9*705320 + (1)*16*34650 = 2284920.
For k = 0, A106828(11,0) = 0.
For k = 1, (1-1)^2 = 0.
For 2 <= k <= 5, A106828(11, k) == 0 (mod 11*10).
Result a(11) == 0 (mod 11*10).
		

Crossrefs

Programs

  • Maple
    a := series((-1+2*x-2*x^2+x^3+(1-x)*(log((1-x)^(1-2*x))-(log(1-x))^2))*exp(x), x=0, 24):
    seq(n!*coeff(a, x, n), n=0..23);
    # second program:
    a := n -> add((-1)^(k-1)*(k-1)^2*A106828(n, k), k=0..iquo(n, 2)):
    seq(a(n), n=0..23);
  • Mathematica
    CoefficientList[Series[(-1+2*x-2*x^2+x^3+(1-x)*(Log[(1-x)^(1-2*x)]-(Log[1-x])^2))*Exp[x], {x, 0, 23}], x]*Range[0, 23]!
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace((-1 + 2*x - 2*x^2 + x^3 + (1 - x)*(log((1 - x)^(1 - 2*x)) - (log(1 - x))^2))*exp(x))) \\ Michel Marcus, Oct 07 2021

Formula

E.g.f.: (-1 + 2*x - 2*x^2 + x^3 + (1 - x)*(log((1 - x)^(1 - 2*x)) - (log(1 - x))^2))*exp(x).
a(n) ~ 2 * exp(1) * log(n) * n! / n^2 * (1 + (gamma - 3/2)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Dec 09 2021

A349959 a(n) = Sum_{k=0..floor(n/2)} (k-1)^2*A106828(n, k).

Original entry on oeis.org

1, 0, 0, 0, 3, 20, 190, 1764, 17773, 192632, 2250036, 28254600, 380304639, 5468906508, 83750505826, 1361579283596, 23431400945145, 425669127018416, 8142731710207432, 163636478165355408, 3447201944202849819, 75973975479088955460, 1748531872985454054246, 41951755708613404583732
Offset: 0

Views

Author

Mélika Tebni, Dec 07 2021

Keywords

Comments

For all p prime, a(p) == 0 (mod p*(p-1)).

Examples

			E.g.f.: 1 + 3*x^4/4! + 20*x^5/5! + 190*x^6/6! + 1764*x^7/7! + 17773*x^8/8! + 192632*x^9/9! + ...
a(13) = Sum_{k=0..6} (k-1)^2*A106828(13, k).
a(13) =  1*0 + 0*479001600 + 1*967524480 + 4*647536032 + 9*177331440 + 16*18858840 + 25*540540 = 5468906508.
For k = 0, A106828(13, 0) = 0.
For k = 1, (1-1)^2 = 0.
For 2 <= k <= 6, A106828(13, k) == 0 (mod 13*12).
Result a(13) == 0 (mod 13*12).
		

Crossrefs

Programs

  • Maple
    a := n -> add((k-1)^2*A106828(n, k), k=0..iquo(n, 2)):
    seq(a(n), n=0..23);
    # second program:
    a := series((-2-x+(3+log((1-x)^(1+2*x))+(log(1-x))^2)/(1-x))/exp(x), x=0, 24):
    seq(n!*coeff(a, x, n), n=0..23);
  • Mathematica
    CoefficientList[Series[(-2-x+(3+Log[(1-x)^(1+2*x)]+(Log[1-x])^2)/(1-x))/Exp[x], {x, 0, 23}], x]*Range[0, 23]!
  • PARI
    E2(n, m) = sum(k=0, n-m, (-1)^(n+k)*binomial(2*n+1, k)*stirling(2*n-m-k+1, n-m-k+1, 1)); \\ A008517
    ast1(n, k) = if ((n==0) && (k==0), 1, sum(j=0, n-k, binomial(j, n-2*k)*E2(n-k, j+1))); \\ A106828
    a(n) = sum(k=0, n\2, (k-1)^2*ast1(n, k)); \\ Michel Marcus, Dec 07 2021

Formula

E.g.f.: (-2 - x + (3 + log((1 - x)^(1 + 2*x)) + (log(1 - x))^2) / (1 - x)) / exp(x).
a(n) ~ n! * exp(-1) * log(n)^2 * (1 + (2*gamma - 3)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Dec 09 2021

A008306 Triangle T(n,k) read by rows: associated Stirling numbers of first kind (n >= 2, 1 <= k <= floor(n/2)).

Original entry on oeis.org

1, 2, 6, 3, 24, 20, 120, 130, 15, 720, 924, 210, 5040, 7308, 2380, 105, 40320, 64224, 26432, 2520, 362880, 623376, 303660, 44100, 945, 3628800, 6636960, 3678840, 705320, 34650, 39916800, 76998240, 47324376, 11098780, 866250, 10395
Offset: 2

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Author

Keywords

Comments

Also, T(n,k) is the number of derangements (permutations with no fixed points) of {1..n} with k cycles.
The sum of the n-th row is the n-th subfactorial: A000166(n). - Gary Detlefs, Jul 14 2010

Examples

			Rows 2 through 7 are:
    1;
    2;
    6,   3;
   24,  20;
  120, 130,  15;
  720, 924, 210;
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 256.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 75.

Crossrefs

Cf. A000166, A106828 (another version), A079510 (rearranged triangle), A235706 (specializations).
Diagonals: A000142, A000276, A000483.
Diagonals give reversed rows of A111999.

Programs

  • Haskell
    a008306 n k = a008306_tabf !! (n-2) !! (k-1)
    a008306_row n = a008306_tabf !! (n-2)
    a008306_tabf = map (fst . fst) $ iterate f (([1], [2]), 3) where
       f ((us, vs), x) =
         ((vs, map (* x) $ zipWith (+) ([0] ++ us) (vs ++ [0])), x + 1)
    -- Reinhard Zumkeller, Aug 05 2013
  • Maple
    A008306 := proc(n,k) local j;
    add(binomial(j,n-2*k)*A008517(n-k,j),j=0..n-k) end;
    seq(print(seq(A008306(n,k),k=1..iquo(n,2))),n=2..12):
    # Peter Luschny, Apr 20 2011
  • Mathematica
    t[0, 0] = 1; t[n_, 0] = 0; t[n_, k_] /; k > n/2 = 0; t[n_, k_] := t[n, k] = (n - 1)*(t[n - 1, k] + t[n - 2, k - 1]); A008306 = Flatten[ Table[ t[n, k], {n, 2, 12}, {k, 1, Quotient[n, 2]}]] (* Jean-François Alcover, Jan 25 2012, after David Callan *)
  • PARI
    { A008306(n,k) = (-1)^(n+k) * sum(i=0,k, (-1)^i * binomial(n,i) * stirling(n-i,k-i,1) ); } \\ Max Alekseyev, Sep 08 2018
    

Formula

T(n,k) = Sum_{i=0..k} (-1)^i * binomial(n,i) * |stirling1(n-i,k-i)| = (-1)^(n+k) * Sum_{i=0..k} (-1)^i * binomial(n,i) * A008275(n-i,k-i). - Max Alekseyev, Sep 08 2018
E.g.f.: 1 + Sum_{1 <= 2*k <= n} T(n, k)*t^n*u^k/n! = exp(-t*u)*(1-t)^(-u).
Recurrence: T(n, k) = (n-1)*(T(n-1, k) + T(n-2, k-1)) for 1 <= k <= n/2 with boundary conditions T(0,0) = 1, T(n,0) = 0 for n >= 1, and T(n,k) = 0 for k > n/2. - David Callan, May 16 2005
E.g.f. for column k: B(A(x)) where A(x) = log(1/1-x)-x and B(x) = x^k/k!.
From Tom Copeland, Jan 05 2016: (Start)
The row polynomials of this signed array are the orthogonal NL(n,x;x-n) = n! Sum_{k=0..n} binomial(x,n-k)*(-x)^k/k!, the normalized Laguerre polynomials of order (x-n) as discussed in Gautschi (the Temme, Carlitz, and Karlin and McGregor references come from this paper) in regard to asymptotic expansions of the upper incomplete gamma function--Tricomi's Cinderella of special functions.
e^(x*t)*(1-t)^x = Sum_{n>=0} NL(n,x;x-n)*x^n/n!.
The first few are
NL(0,x) = 1
NL(1,x) = 0
NL(2,x) = -x
NL(3,x) = 2*x
NL(4,x) = -6*x + 3*x^2.
With D=d/dx, :xD:^n = x^n D^n, :Dx:^n = D^n x^n, and K(a,b,c), the Kummer confluent hypergeometric function, NL(n,x;y-n) = n!*e^x binomial(xD+y,n)*e^(-x) = n!*e^x Sum_{k=0..n} binomial(k+y,n) (-x)^k/k! = e^x x^(-y+n) D^n (x^y e^(-x)) = e^x x^(-y+n) :Dx:^n x^(y-n)*e^(-x) = e^x*x^(-y+n)*n!*L(n,:xD:,0)*x^(y-n)*e^(-x) = n! binomial(y,n)*K(-n,y-n+1,x) = n!*e^x*(-1)^n*binomial(-xD-y+n-1,n)*e^(-x). Evaluate these expressions at y=x after the derivative operations to obtain NL(n,x;x-n). (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 16 2001

A340556 E2(n, k), the second-order Eulerian numbers with E2(0, k) = δ_{0, k}. Triangle read by rows, E2(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 22, 58, 24, 0, 1, 52, 328, 444, 120, 0, 1, 114, 1452, 4400, 3708, 720, 0, 1, 240, 5610, 32120, 58140, 33984, 5040, 0, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 0

Views

Author

Peter Luschny, Feb 05 2021

Keywords

Comments

The second-order Eulerian number E2(n, k) is the number of Stirling permutations of order n with exactly k descents; here the last index is defined to be a descent. More formally, let Q_n denote the set of permutations of the multiset {1,1,2,2, ..., n,n} in which, for all j, all entries between two occurrences of j are larger than j, then E2(n, k) = card({s in Q_n with des(s) = k}), where des(s) = card({j: s(j) > s(j+1)}) is the number of descents of s.
Also the number of Riordan trapezoidal words of length n with k distinct letters (see Riordan 1976, p. 9).
Also the number of rooted plane trees on n + 1 vertices with k leaves (see Janson 2008, p. 543).
Let b(n) = (1/2)*Sum_{k=0..n-1} (-1)^k*E2(n-1, k+1) / C(2*n-1, k+1). Apparently b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1}F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See Rzadkowski and Urlinska, example 4.)

Examples

			Triangle starts:
  [0] 1;
  [1] 0, 1;
  [2] 0, 1, 2;
  [3] 0, 1, 8,    6;
  [4] 0, 1, 22,   58,    24;
  [5] 0, 1, 52,   328,   444,     120;
  [6] 0, 1, 114,  1452,  4400,    3708,    720;
  [7] 0, 1, 240,  5610,  32120,   58140,   33984,    5040;
  [8] 0, 1, 494,  19950, 195800,  644020,  785304,   341136,   40320;
  [9] 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880.
To illustrate the generating function for row 3: The expansion of (1 - x)^7*(x*exp(-x) + 16*x^2*exp(-x)^2 + (243*x^3*exp(-x)^3)/2) gives the polynomial x + 8*x^2 + 6*x^3. The coefficients of this polynomial give row 3.
.
Stirling permutations of order 3 with exactly k descents: (When counting the descents one may assume an invisible '0' appended to the permutations.)
  T[3, k=0]:
  T[3, k=1]: 112233;
  T[3, k=2]: 331122; 223311; 221133; 133122; 122331; 122133; 113322; 112332;
  T[3, k=3]: 332211; 331221; 233211; 221331; 133221; 123321.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.

Crossrefs

Indexing the second-order Eulerian numbers comes in three flavors: A008517 (following Riordan and Comtet), A201637 (following Graham, Knuth, and Patashnik) and this indexing, extending the definition of Gessel and Stanley. (A008517 is the main entry of the numbers.) The corresponding triangles of the first-order Eulerian numbers are A008292, A173018, and A123125.
Row reversed: A163936 (with offset = 0).
Values: E2poly(n, 1) = A001147(n), E2poly(n, -1) ~ -A001662(n+1), E2poly(n, 2) = A112487(n), 2^n*E2poly(n, 1/2) = A000311(n+1), 2^n*E2poly(n, -1/2) = A341106(n).

Programs

  • Maple
    # Using the recurrence:
    E2 := proc(n, k) option remember;
    if k = 0 and n = 0 then return 1 fi; if n < 0 then return 0 fi;
    E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) end: seq(seq(E2(n, k), k = 0..n), n = 0..9);
    # Using the row generating function:
    E2egf := n -> (1-x)^(2*n+1)*add(k^(n+k)/k!*(x*exp(-x))^k, k=0..n);
    T := (n, k) -> coeftayl(E2egf(n), x=0, k): seq(print(seq(T(n, j),j=0..n)), n=0..7);
    # Using the built-in function:
    E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1)):
    # Using the compositional inverse (series reversion):
    E2triangle := proc(N) local r, s, C; Order := N + 2;
    s := solve(y = series(x - t*(exp(x) - 1), x), x):
    r := n -> -n!*(t - 1)^(2*n - 1)*coeff(s, y, n); C := [seq(expand(r(n)), n = 1..N)];
    seq(print(seq(coeff(C[n+1], t, k), k = 0..n)), n = 0..N-1) end: E2triangle(10);
  • Mathematica
    T[n_, k_] := T[n, k] = If[k == 0, Boole[n == 0], If[n < 0, 0, k T[n - 1, k] + (2 n - k) T[n - 1, k - 1]]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
    (* Via row polynomials: *)
    E2poly[n_] := If[n == 0, 1,
      Expand[Simplify[x (x - 1)^(2 n) D[((1 - x)^(1 - 2 n) E2poly[n - 1]), x]]]];
    Table[CoefficientList[E2poly[n], x], {n, 0, 9}] // Flatten
    (* Series reversion *)
    Revert[gf_, len_] := Module[{S = InverseSeries[Series[gf, {x, 0, len + 1}], x]},
    Table[CoefficientList[(n + 1)! (1 - t)^(2 n + 1) Coefficient[S, x, n + 1], t],
    {n, 0, len}] // Flatten]; Revert[x + t - t Exp[x], 6]
  • PARI
    E2poly(n) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1)));
    { for(n = 0, 9, print(Vecrev(E2poly(n)))) }
    
  • PARI
    T(n, k) = sum(j=0, n-k, (-1)^(n-j)*binomial(2*n+1, j)*stirling(2*n-k-j+1, n-k-j+1, 1)); \\ Michel Marcus, Feb 11 2021
    
  • SageMath
    # See also link to notebook.
    @cached_function
    def E2(n, k):
        if n < 0: return 0
        if k == 0: return k^n
        return k * E2(n - 1, k) + (2*n - k) * E2(n - 1, k - 1)  # Peter Luschny, Mar 08 2025

Formula

E2(n, k) = E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) for n > 0 and 0 <= k <= n, and E2(0, 0) = 1; in all other cases E(n, k) = 0.
E2(n, k) = Sum_{j=0..n-k}(-1)^(n-j)*binomial(2*n+1, j)*Stirling1(2*n-k-j+1, n-k-j+1).
E2(n, k) = Sum_{j=0..k}(-1)^(k-j)*binomial(2*n + 1, k - j)*Stirling2(n + j, j).
Stirling1(x, x - n) = (-1)^n*Sum_{k=0..n} E2(n, k)*binomial(x + k - 1, 2*n).
Stirling2(x, x - n) = Sum_{k=0..n} E2(n, k)*binomial(x + n - k, 2*n).
E2poly(n, x) = Sum_{k=0..n} E2(n, k)*x^k, as row polynomials.
E2poly(n, x) = x*(x-1)^(2*n)*d_{x}((1-x)^(1-2*n)*E2poly(n-1)) for n>=1 and E2poly(0)=1.
E2poly(n, x) = (1 - x)^(2*n + 1)*Sum_{k=0..n}(k^(n + k)/k!)*(x*exp(-x))^k.
W(n, k) = [x^k] (1+x)^n*E2poly(n, x/(1 + x)) are the Ward numbers A269939.
E2(n, k) = [x^k] (1-x)^n*Wpoly(n, x/(1 - x)); Wpoly(n, x) = Sum_{k=0..n}W(n, k)*x^k.
W(n, k) = Sum_{j=0..k} E2(n, j)*binomial(n - j, n - k).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*W(n, j)*binomial(n - j, k - j).
The compositional inverse with respect to x of x - t*(exp(x) - 1) (see B. Drake):
T(n, k) = [t^k](n+1)!*(1-t)^(2*n+1)*[x^(n+1)] InverseSeries(x - t*(exp(x)-1), x).
AS1(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, j+1), where AS1(n, k) are the associated Stirling numbers of the first kind (A008306, A106828).
E2(n, k) = Sum_{j=0..n-k+1} (-1)^(n-k-j+1)*AS1(n+j, j)*binomial(n-j, n-k-j+1), for n >= 1.
AS2(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, n-k-j) for n >=1, where AS2(n, k) are the associated Stirling numbers of the second kind (A008299, A137375).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*AS2(n + j, j)*binomial(n - j, k - j).

A345652 Expansion of the e.g.f. exp(-1 + (x + 1)*exp(-x)).

Original entry on oeis.org

1, 0, -1, 2, 0, -16, 65, -78, -749, 6232, -22068, -28920, 1004685, -7408740, 22263215, 157632230, -2874256740, 21590948480, -53087332675, -956539294506, 16344490525835, -132605481091060, 294656170409328, 9113173803517344, -167298122286332823
Offset: 0

Views

Author

Mélika Tebni, Jun 21 2021

Keywords

Comments

For all p prime, a(p)/(p-1) == 1 (mod p). - Mélika Tebni, Mar 21 2022

Examples

			exp(-1+(x+1)*exp(-x)) = 1 - x^2/2! + 2*x^3/3! - 16*x^5/5! + 65*x^6/6! - 78*x^7/7! - 749*x^8/8! + 6232*x^9/9! + ...
		

Crossrefs

Cf. A292935 (without 1+x: EGF e^(e^(-x)-1)), A000110 (absolute values: Bell numbers, EGF e^(e^x-1))

Programs

  • Maple
    a := series(exp(-1+(x+1)*exp(-x)), x=0, 25): seq(n!*coeff(a, x, n), n=0..24);
    a := proc(n) option remember; `if`(n=0, 1, add((n-1)*binomial(n-2, k)*(-1)^(n-1-k)*a(k), k=0..n-2)) end: seq(a(n), n=0..24);
    # third program:
    A345652 := n -> add((-1)^(n-k)*combinat[bell](k)*A106828(n, k), k=0..iquo(n, 2)):
    seq(A345652(n), n=0..24); # Mélika Tebni, Sep 21 2021
  • Mathematica
    nmax = 24; CoefficientList[Series[Exp[-1+(x+1)*Exp[-x]], {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    seq(n) = {Vec(serlaplace(exp(-1+(x+1)*exp(-x + O(x*x^n)))))} \\ Andrew Howroyd, Jun 21 2021
    
  • PARI
    a(n) = if(n==0, 1, sum(k=2, n, (-1)^(k-1)*(k-1)*binomial(n-1, k-1)*a(n-k))); \\ Seiichi Manyama, Mar 15 2022

Formula

The e.g.f. y(x) satisfies y' = -x*y*exp(-x).
a(n) = Sum_{k=0..n-2} (n-1)*binomial(n-2, k)*(-1)^(n-1-k)*a(k) for n > 0.
Conjecture: a(n) = 0 for only n = 1 and n = 4.
Conjecture: For all p prime, a(p)^2 == 1 (mod p).
Stronger conjecture: For n > 1, a(n) == -1 (mod n) iff n is a prime or 6. - M. F. Hasler, Jun 23 2021
a(n) = Sum_{k=0..floor(n/2)} (-1)^(n-k)*Bell(k)*A106828(n, k). - Mélika Tebni, Sep 21 2021
a(n) = Sum_{k=0..n} (-1)^k*A003725(n-k)*Bell(k)*binomial(n, k). - Mélika Tebni, Mar 21 2022

A347210 Expansion of the e.g.f. (1 - 2*x - 2*log(1 - x) - exp(2*x)*(1 - x)^2) / 4 - 1.

Original entry on oeis.org

-1, 0, 1, 2, 3, 4, 20, 216, 2072, 18880, 177984, 1805440, 19935872, 239445504, 3113377280, 43588830208, 653836446720, 10461393240064, 177843710148608, 3201186844016640, 60822550184493056, 1216451004043755520, 25545471085755629568, 562000363888584687616
Offset: 0

Views

Author

Mélika Tebni, Aug 23 2021

Keywords

Comments

For all p prime, a(p) == -1 (mod p).
For n > 1, a(n) == 0 (mod (n-1)).

Examples

			E.g.f.: -1 + x^2/2! + 2*x^3/3! + 3*x^4/4! + 4*x^5/5! + 20*x^6/6! + 216*x^7/7! + 2072*x^8/8! + 18880*x^9/9! + ...
a(19) = Sum_{k=1..9} (-1)^(k-1)*ceiling(2^(k-2))*A106828(19, k) = 3201186844016640.
For k = 1, (-1)^(1-1)*ceiling(2^(1-2))*A106828(19, 1) == -1 (mod 19), because (-1)^(1-1)*ceiling(2^(1-2)) = 1 and A106828(19, 1) = (19-1)!
For k >= 2, (-1)^(k-1)*ceiling(2^(k-2))*A106828(19, k) == 0 (mod 19), because A106828(19, k) == 0 (mod 19), result a(19) == -1 (mod 19).
a(10) = Sum_{k=1..5}  (-1)^(k-1)*ceiling(2^(k-2))*A106828(10, k) = 177984.
a(10) == 0 (mod (10-1)), because for k >= 1, A106828(10, k) == 0 (mod 9).
		

Crossrefs

Programs

  • Maple
    a := series((1-2*x-2*log(1-x)-exp(2*x)*(1-x)^2)/4-1, x=0, 25):
    seq(n!*coeff(a, x, n), n=0..23);
    # second program:
    a := n -> add((-1)^(k-1)*ceil(2^(k-2))*A106828(n, k), k=0..iquo(n, 2)):
    seq(a(n), n=0..23);
  • Mathematica
    CoefficientList[Series[(1 - 2*x - 2*Log[1 - x] - E^(2*x)*(1 - x)^2)/4 - 1, {x, 0, 23}], x]*Range[0, 23]!
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace((1-2*x-2*log(1-x)-exp(2*x)*(1-x)^2)/4 - 1)) \\ Michel Marcus, Aug 23 2021

Formula

a(n) = Sum_{k=0..floor(n/2)} (-1)^(k-1)*ceiling(2^(k-2))*A106828(n, k).
a(n) ~ (n-1)!/2. - Vaclav Kotesovec, Dec 09 2021

A350452 Number T(n,k) of endofunctions on [n] with exactly k connected components and no fixed points; triangle T(n,k), n>=0, 0<=k<=floor(n/2), read by rows.

Original entry on oeis.org

1, 0, 0, 1, 0, 8, 0, 78, 3, 0, 944, 80, 0, 13800, 1810, 15, 0, 237432, 41664, 840, 0, 4708144, 1022252, 34300, 105, 0, 105822432, 27098784, 1286432, 10080, 0, 2660215680, 778128336, 47790540, 648900, 945, 0, 73983185000, 24165049920, 1815578160, 36048320, 138600
Offset: 0

Views

Author

Alois P. Heinz, Dec 31 2021

Keywords

Comments

For k >= 2 and p prime, T(p,k) == 0 (mod 4*p*(p-1)). - Mélika Tebni, Jan 20 2023

Examples

			Triangle T(n,k) begins:
  1;
  0;
  0,          1;
  0,          8;
  0,         78,         3;
  0,        944,        80;
  0,      13800,      1810,       15;
  0,     237432,     41664,      840;
  0,    4708144,   1022252,    34300,    105;
  0,  105822432,  27098784,  1286432,  10080;
  0, 2660215680, 778128336, 47790540, 648900, 945;
  ...
		

Crossrefs

Columns k=0-1 give: A000007, A000435.
Row sums give A065440.
T(2n,n) gives A001147.

Programs

  • Maple
    c:= proc(n) option remember; add(n!*n^(n-k-1)/(n-k)!, k=2..n) end:
    b:= proc(n) option remember; expand(`if`(n=0, 1, add(
          b(n-i)*binomial(n-1, i-1)*x*c(i), i=1..n)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n/2))(b(n)):
    seq(T(n), n=0..12);
  • Mathematica
    c[n_] := c[n] = Sum[n!*n^(n - k - 1)/(n - k)!, {k, 2, n}];
    b[n_] := b[n] = Expand[If[n == 0, 1, Sum[
         b[n - i]*Binomial[n - 1, i - 1]*x*c[i], {i, 1, n}]]];
    T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n/2}]][b[n]];
    Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Mar 18 2022, after Alois P. Heinz *)
  • PARI
    \\ here AS1(n,k) gives associated Stirling numbers of 1st kind.
    AS1(n,k)={(-1)^(n+k)*sum(i=0, k, (-1)^i * binomial(n, i) * stirling(n-i, k-i, 1) )}
    T(n,k) = {if(n==0, k==0, sum(j=k, n, n^(n-j)*binomial(n-1, j-1)*AS1(j,k)))} \\ Andrew Howroyd, Jan 20 2023

Formula

From Mélika Tebni, Jan 20 2023: (Start)
E.g.f. column k: (LambertW(-x) - log(1 + LambertW(-x)))^k / k!.
-Sum_{k=1..n/2} (-1)^k*T(n,k) = A071720(n+1), for n > 0.
-Sum_{k=1..n/2} (-1)^k*T(n,k) / (n-1) = A007830(n-2), for n > 1.
T(n,k) = Sum_{j=k..n} n^(n-j)*binomial(n-1, j-1)*A106828(j, k) for n > 0. (End)

A347571 Expansion of the e.g.f. (-1 - 2*x - 2*log(1 - x) + exp(-2*x) / (1 - x)^2) / 4 + 1.

Original entry on oeis.org

1, 0, 1, 2, 9, 44, 280, 2064, 17528, 167488, 1777536, 20721920, 263055232, 3610443264, 53256280064, 839974309888, 14103897738240, 251146689069056, 4726795773018112, 93746994502828032, 1954053073794596864, 42702893781890498560, 976276451410488066048, 23303485413254033309696
Offset: 0

Views

Author

Mélika Tebni, Sep 07 2021

Keywords

Comments

For all p prime, a(p) == -1 (mod p).
For n > 1, a(n) == 0 (mod (n-1)).

Examples

			E.g.f.: 1 + x^2/2! + 2*x^3/3! + 9*x^4/4! + 44*x^5/5! + 280*x^6/6! + 2064*x^7/7! + 17528*x^8/8! + 167488*x^9/9! + ...
a(11) = Sum_{k=0..5} ceiling(2^(k-2))*A106828(11, k) = 20721920.
For k = 0, A106828(11,0) = 0.
For k = 1, ceiling(2^(1-2))*A106828(11, 1) == -1 (mod 11), because ceiling(2^(1-2)) = 1 and A106828(11, 1) = (11-1)!
For k >= 2, ceiling(2^(k-2))*A106828(11, k) == 0 (mod 11), because A106828(11, k) == 0 (mod 11), result a(11) == -1 (mod 11).
a(10) = Sum_{k=0..5} ceiling(2^(k-2))*A106828(10, k) = 1777536.
a(10) == 0 (mod (10-1)), because for k >= 0, A106828(10, k) == 0 (mod 9).
		

Crossrefs

Programs

  • Maple
    a := series((-1-2*x-2*log(1-x)+exp(-2*x)/(1-x)^2)/4+1, x=0, 24):
    seq(n!*coeff(a, x, n), n=0..23);
    # second program:
    a := n -> add(ceil(2^(k-2))*A106828(n, k), k=0..iquo(n, 2)):
    seq(a(n), n=0..23);
  • Mathematica
    CoefficientList[Series[(-1 - 2*x - 2*Log[1 - x] + Exp[-2*x]/(1 - x)^2)/4 + 1, {x, 0, 23}], x]*Range[0, 23]!
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace((-1-2*x-2*log(1-x)+exp(-2*x)/(1-x)^2)/4 + 1)) \\ Michel Marcus, Sep 07 2021

Formula

a(n) = Sum_{k=0..floor(n/2)} ceiling(2^(k-2))*A106828(n, k).
a(n) ~ n * n! / (4*exp(2)). - Vaclav Kotesovec, Sep 10 2021
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