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A301272
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Number of derangements of S_n with exactly one peak.
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2
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0, 0, 0, 1, 6, 33, 152, 663, 2778, 11413, 46332, 186867, 750878, 3011025, 12060480, 48277423, 193186146, 772908429, 3091983236, 12368675691, 49476275622, 197908422985, 791640682440, 3166577409831, 12666340397546, 50665425902853, 202661837829132, 810647630936803
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OFFSET
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0,5
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LINKS
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FORMULA
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G.f.: x^3*(2*x^2+1)/((1-4*x)*(x+1)^2*(2*x-1)^2). - Alois P. Heinz, Apr 29 2018
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EXAMPLE
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a(3) = 1: 231.
a(4) = 6: 2143, 2341, 2413, 3142, 3412, 3421.
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MAPLE
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a:= n-> floor((<<0|1|0|0|0>, <0|0|1|0|0>, <0|0|0|1|0>, <0|0|0|0|1>,
<16|12|-16|-5|6>>^n. <<1/8, 0, 0, 1, 6>>)[1, 1]):
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MATHEMATICA
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Join[{0}, LinearRecurrence[{6, -5, -16, 12, 16}, {0, 0, 1, 6, 33}, 30]] (* Jean-François Alcover, May 31 2019 *)
CoefficientList[Series[x^3(2x^2+1)/((1-4x)(x+1)^2(2x-1)^2), {x, 0, 40}], x] (* Harvey P. Dale, Sep 01 2021 *)
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PROG
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(Python)
def count_peaks(pi):
count = 0
for i in range(i, len(pi)-1):
if pi[i] > pi[i+1] and pi[i] > pi[i-1]:
count += 1
return count
def main(args):
n = int(args[0])
set = {1, 2, ..., n}
drmts = []
for pi in itertools.permutations(set):
drmts.append(pi)
for i in range(n):
if pi[i] == i+1:
drmts.remove(pi)
break
num = 0
for pi in drmts:
if count_peaks(pi) == 1:
num += 1
print('number of 1 peak derangements: ', num)
(PARI) A301272(n)={my(c=0, v, t, ok); for(k=0, n!-1, v=numtoperm(n, k); ok=1; for(i=1, n, if((v[i]==i), ok=0; break)); if(ok, t=0; for(i=2, n-1, if((v[i]>v[i-1])&&(v[i]>v[i+1]), t++; if(t>1, break))); if(t==1, c++))); c} \\ R. J. Cano, Apr 25 2018
(PARI) \\ See Cano link.
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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