A240559
a(n) = -2^n*(E(n, 1/2) + E(n, 1) + (n mod 2)*2*(E(n+1, 1/2) + E(n+1, 1))), where E(n, x) are the Euler polynomials.
Original entry on oeis.org
0, 0, 1, -3, -5, 45, 61, -1113, -1385, 42585, 50521, -2348973, -2702765, 176992725, 199360981, -17487754833, -19391512145, 2195014332465, 2404879675441, -341282303124693, -370371188237525, 64397376340013805, 69348874393137901, -14499110277050234553
Offset: 0
G.f. = x^2 - 3*x^3 - 5*x^4 + 45*x^5 + 61*x^6 - 1113*x^7 - 1385*x^8 + ...
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A240559 := proc(n) euler(n,1/2) + euler(n,1); if n mod 2 = 1 then % + 2*(euler(n+1,1/2)+euler(n+1,1)) fi; -2^n*% end: seq(A240559(n),n=0..19);
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skp[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; skp[n_, x0_?NumericQ] := skp[n, x] /. x -> x0; a[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*(skp[k, 0] + skp[k+1, -1]), {k, 0, n}]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Dec 09 2014, after Peter Luschny *)
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# Efficient computation with L. Seidel's boustrophedon transformation.
def A240559_list(n) :
A = [0]*(n+1); A[0] = 1; R = [0]
k = 0; e = 1; x = -1; s = -1
for i in (0..n):
Am = 0; A[k + e] = 0; e = -e;
for j in (0..i): Am += A[k]; A[k] = Am; k += e
if e == 1: x += 1; s = -s
v = -A[-x] if e == 1 else A[-x] - A[x]
if i > 1: R.append(s*v)
return R
A240559_list(24)
A291677
Number of permutations p of [2n] such that 0p has exactly n alternating runs.
Original entry on oeis.org
1, 1, 7, 148, 6171, 425976, 43979902, 6346283560, 1219725741715, 301190499710320, 92921064554444490, 35025128774218944648, 15838288022236083603486, 8462453158197423495502224, 5274234568391796228927038748, 3792391176672742840187796835728
Offset: 0
a(2) = 7: 1243, 1342, 1432, 2341, 2431, 3421, 4321.
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b:= proc(n, k) option remember; `if`(k=0,
`if`(n=0, 1, 0), `if`(k<0 or k>n, 0,
k*b(n-1, k)+b(n-1, k-1)+(n-k+1)*b(n-1, k-2)))
end:
a:= n-> b(2*n, n):
seq(a(n), n=0..20);
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b[n_, k_] := b[n, k] = If[k == 0, If[n == 0, 1, 0], If[k < 0 || k > n, 0, k*b[n - 1, k] + b[n - 1, k - 1] + (n - k + 1)*b[n - 1, k - 2]]];
a[n_] := b[2*n, n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 30 2019, after Alois P. Heinz *)
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from sympy.core.cache import cacheit
@cacheit
def b(n, k): return (1 if n==0 else 0) if k==0 else 0 if k<0 or k>n else k*b(n - 1, k) + b(n - 1, k - 1) + (n - k + 1)*b(n - 1, k - 2)
def a(n): return b(2*n, n)
print([a(n) for n in range(31)]) # Indranil Ghosh, Aug 30 2017
A303159
Number of permutations p of [2n+1] such that 0p has exactly n+1 alternating runs.
Original entry on oeis.org
1, 3, 43, 1344, 74211, 6384708, 789649750, 132789007200, 29145283614115, 8092186932120060, 2772830282722806978, 1149343084932146388144, 566844242187778610648334, 328043720353943611689811272, 220147053200818211779539712908, 169580070210721829547034445169024
Offset: 0
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b:= proc(n, k) option remember; `if`(k=0,
`if`(n=0, 1, 0), `if`(k<0 or k>n, 0,
k*b(n-1, k)+b(n-1, k-1)+(n-k+1)*b(n-1, k-2)))
end:
a:= n-> b(2*n+1, n+1):
seq(a(n), n=0..20);
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b[n_, k_] := b[n, k] = If[k == 0,
If[n == 0, 1, 0], If[k < 0 || k > n, 0,
k b[n-1, k] + b[n-1, k-1] + (n-k+1) b[n-1, k-2]]];
a[n_] := b[2n+1, n+1];
a /@ Range[0, 20] (* Jean-François Alcover, Dec 21 2020, after Alois P. Heinz *)
A303160
Number of permutations p of [n] such that 0p has exactly ceiling(n/2) alternating runs.
Original entry on oeis.org
1, 1, 1, 3, 7, 43, 148, 1344, 6171, 74211, 425976, 6384708, 43979902, 789649750, 6346283560, 132789007200, 1219725741715, 29145283614115, 301190499710320, 8092186932120060, 92921064554444490, 2772830282722806978, 35025128774218944648, 1149343084932146388144
Offset: 0
a(2) = 1: 12.
a(3) = 3: 132, 231, 321.
a(4) = 7: 1243, 1342, 1432, 2341, 2431, 3421, 4321.
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b:= proc(n, k) option remember; `if`(k=0,
`if`(n=0, 1, 0), `if`(k<0 or k>n, 0,
k*b(n-1, k)+b(n-1, k-1)+(n-k+1)*b(n-1, k-2)))
end:
a:= n-> b(n, ceil(n/2)):
seq(a(n), n=0..25);
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b[n_, k_] := b[n, k] = If[k == 0,
If[n == 0, 1, 0], If[k < 0 || k > n, 0,
k*b[n-1, k] + b[n-1, k-1] + (n-k+1)*b[n-1, k-2]]];
a[n_] := b[n, Ceiling[n/2]];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 31 2021, after Alois P. Heinz *)
A241242
a(n) = -2^(2*n+1)*(E(2*n+1, 1/2) + E(2*n+1, 1) + 2*(E(2*n+2, 1/2) + E(2*n+2, 1))), where E(n,x) are the Euler polynomials.
Original entry on oeis.org
0, -3, 45, -1113, 42585, -2348973, 176992725, -17487754833, 2195014332465, -341282303124693, 64397376340013805, -14499110277050234553, 3840151029102915908745, -1182008039799685905580413, 418424709061213506712209285, -168805428822414120140493978273
Offset: 0
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A241242 := proc(n) e := n -> euler(n,1/2) + euler(n,1); -2^(2*n+1)*(e(2*n+1) + 2*e(2*n+2)) end: seq(A241242(n),n=0..15);
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Array[-2^(2 # + 1)*(EulerE[2 # + 1, 1/2] + EulerE[2 # + 1, 1] + 2 (EulerE[2 # + 2, 1/2] + EulerE[2 # + 2, 1])) &, 16, 0] (* Michael De Vlieger, May 24 2018 *)
A186371
Number of up-down runs in all permutations of {1,2,...,n}.
Original entry on oeis.org
0, 1, 3, 13, 68, 420, 3000, 24360, 221760, 2237760, 24796800, 299376000, 3911846400, 55005350400, 828193766400, 13294689408000, 226663557120000, 4090405423104000, 77895546753024000, 1561112121913344000, 32844177110384640000, 723788347432550400000
Offset: 0
a(3)=13 because the permutations 123, 132, 213, 231, 312, and 321 have a total of 1 + 2 + 3 + 2 + 3 + 2 = 13 up-down runs.
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[0,1] cat [Factorial(n)*(4*n+1)/6: n in [2..30]]; // Vincenzo Librandi, Sep 11 2015
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0, 1, seq((1/6)*factorial(n)*(4*n+1), n = 2 .. 20);
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Join[{0, 1}, Table[n! (4 n + 1)/6, {n, 2, 20}]] (* Vincenzo Librandi, Sep 11 2015 *)
Showing 1-6 of 6 results.
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