Displaying 1-10 of 13 results found.
Number of circular permutations of length n without 3-sequences.
+10
15
1, 5, 20, 102, 627, 4461, 36155, 328849, 3317272, 36757822, 443846693, 5800991345, 81593004021, 1228906816941, 19733699436636, 336554404751966, 6075478765948135, 115734570482611885, 2320148441078578447, 48827637296350480457, 1076313671861962141616
COMMENTS
Circular permutations are permutations whose indices are from the ring of integers modulo n. 3-sequences are of the form i,i+1,i+2. Sequence gives number of permutations of [n] starting with 1 and having no 3-sequences.
a(n) is also the number of permutations of length n-1 without consecutive fixed points (cf. A180187). - David Scambler, Mar 27 2011
REFERENCES
Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, http://math.ku.edu/~ilambert/CN.pdf, 2012. - From N. J. A. Sloane, Sep 15 2012 [broken link]
FORMULA
Let b(n) be the sequence A002628. Then for n > 5, this sequence satisfies a(n) = b(n-1) - b(n-3) + a(n-3). a(n) = Sum_{k=0..n/2} binomial(n-k,k)*d(n-k-1), where d(j)= A000166(j) are the derangement numbers. - Emeric Deutsch, Sep 07 2010
EXAMPLE
For n=4 the a(4)=5 solutions are (0,1,3,2), (0,2,1,3), (0,2,3,1), (0,3,1,2) and (0,3,2,1).
MAPLE
d[0] := 1: for n to 51 do d[n] := n*d[n-1]+(-1)^n end do: a := proc (n) options operator, arrow: sum(binomial(n-k, k)*d[n-k-1], k = 0 .. floor((1/2)*n)) end proc: seq(a(n), n = 3 .. 23); # Emeric Deutsch, Sep 07 2010
MATHEMATICA
a[n_] := Sum[Binomial[n-k, k] Subfactorial[n-k-1], {k, 0, n/2}];
Number of circular permutations of length n without modular 3-sequences
+10
15
1, 5, 18, 95, 600, 4307, 35168, 321609, 3257109, 36199762, 438126986, 5736774126, 80808984725, 1218563180295, 19587031966352, 334329804347219, 6039535339644630, 115118210694558105, 2308967760171049528, 48613722701436777455, 1072008447320752890459
COMMENTS
Circular permutations are permutations whose indices are from the ring of integers modulo n. Modular 3-sequences are of the following form: i,i+1,i+2, where arithmetic is modulo n.
REFERENCES
Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, http://math.ku.edu/~ilambert/CN.pdf, 2012. - N. J. A. Sloane, Sep 15 2012
FORMULA
This sequence can be related to A165961 by the use of auxiliary sequences (and the auxiliary sequences can themselves be calculated by recurrence relations).
EXAMPLE
For n=4 the a(4)=5 solutions are (0,1,3,2), (0,2,1,3), (0,2,3,1), (0,3,1,2) and (0,3,2,1).
MATHEMATICA
f[i_, n_, k_]:=If[i==0&&k==0, 1, If[i==n&&n==k, 1, Binomial[k-1, k-i]*Binomial[n-k-1, k-i-1]+2*Binomial[k-1, k-i-1]*Binomial[n-k-1, k-i-1]+Binomial[k-1, k-i-1]*Binomial[n-k-1, k-i]]];
w1[i_, n_, k_]:=If[n-2k+i<0, 0, If[n-2k+i==0, 1, (n-2k+i-1)!]];
a[n_, k_]:=Sum[f[i, n, k]*w1[i, n, k], {i, 0, k}];
A165962[n_]:=(n-1)!+Sum[(-1)^k*a[n, k], {k, 1, n}];
Number of permutations of length n with no consecutive triples i,i+2,i+4.
+10
9
1, 1, 2, 6, 24, 114, 674, 4714, 37754, 340404, 3412176, 37631268, 452745470, 5900431012, 82802497682, 1244815252434, 19958707407096, 339960096280062, 6130407887839754, 116675071758609742, 2337186717333367706, 49153251967227002616, 1082860432463176004544
COMMENTS
Note for n<5 there are no such subsequences, so those values are trivially n!.
EXAMPLE
For n=5 (0,2,4,1,3) is an example of a permutation with an i,i+2,i+4 triple. If we look at 0,2,4 as a block, then we have 3! ways to permute the triple with the remaining 1 & 3. Hence a(5) = 5! - 3! = 114.
MAPLE
b:= proc(s, x, y) option remember; `if`(s={}, 1, add(
`if`(x=0 or x-y<>2 or y-j<>2, b(s minus {j}, y, j), 0), j=s))
end:
a:= n-> b({$1..n}, 0$2):
MATHEMATICA
b[s_, x_, y_] := b[s, x, y] = If[s == {}, 1, Sum[
If[x == 0 || x - y != 2 || y - j != 2,
b[s ~Complement~ {j}, y, j], 0], {j, s}]];
a[n_] := b[Range[n], 0, 0];
EXTENSIONS
a(0)-a(4) and a(10)-a(11) moved from a duplicate entry based on the Dymacek et al. paper on Apr 13 2021
Number of permutations of length n without modular 3-sequences.
+10
5
1, 1, 2, 3, 20, 100, 612, 4389, 35688, 325395, 3288490, 36489992, 441093864, 5770007009, 81213878830, 1223895060315, 19662509071056, 335472890422812, 6057979285535388, 115434096553014565, 2314691409652237700, 48723117262650147387, 1074208020519710570054
COMMENTS
Modular 3-sequences are of the following form: i,i+1,i+2, where arithmetic is modulo n.
EXAMPLE
For n=3 the a(3) = 3 solutions are (0,2,1), (1,0,2) and (2,1,0).
Number of permutations of length n with no consecutive triples i,i+d,i+2d for all d>0.
+10
5
1, 1, 2, 5, 21, 100, 597, 4113, 32842, 292379, 2925367, 31983248, 383514347, 4966286235, 69508102006, 1039315462467, 16627618496319, 282023014602100, 5075216962675445, 96263599713301975, 1925002914124917950
EXAMPLE
For n=4, there are 4!-a(4)=3 permutations with some consecutive triple i,i+d,i+2d. Note for n=4, only d=1 applies. Hence those three permutations are (0,1,2,3), (1,2,3,0), and (3,0,1,2). Since here only d=1, this is the same value of a(4) in A002628.
MAPLE
b:= proc(s, x, y) option remember; `if`(s={}, 1, add(
`if`(x=0 or x<y or x-y<>y-j,
b(s minus {j}, y, j), 0), j=s))
end:
a:= n-> b({$1..n}, 0$2):
MATHEMATICA
b[s_, x_, y_] := b[s, x, y] = If[s == {}, 1, Sum[If[x == 0 || x < y || x-y != y-j, b[s~Complement~{j}, y, j], 0], {j, s}]];
a[n_] := b[Range[n], 0, 0];
Triangle of numbers a(n,k) = number of permutations on n letters containing k 3-sequences (n >= 0, 0<=k<=max(0,n-2)).
+10
4
1, 1, 2, 5, 1, 21, 2, 1, 106, 11, 2, 1, 643, 62, 12, 2, 1, 4547, 406, 71, 13, 2, 1, 36696, 3046, 481, 80, 14, 2, 1, 332769, 25737, 3708, 559, 89, 15, 2, 1, 3349507, 242094, 32028, 4414, 640, 98, 16, 2, 1, 37054436, 2510733, 306723, 38893, 5164, 724, 107, 17, 2, 1
FORMULA
Riordan gives a recurrence.
EXAMPLE
Triangle begins:
1;
1;
2;
5, 1;
21, 2, 1;
106, 11, 2, 1;
643, 62, 12, 2, 1;
4547, 406, 71, 13, 2, 1;
36696, 3046, 481, 80, 14, 2, 1;
332769, 25737, 3708, 559, 89, 15, 2, 1;
...
Number of permutations of [n] having no substring [k,k+1,k+2,k+3].
+10
4
1, 1, 2, 6, 23, 117, 706, 4962, 39817, 359171, 3597936, 39630372, 476066277, 6194080387, 86776390796, 1302376048620, 20847721870931, 354549730559949, 6384006047649910, 121330369923079290, 2427196999663678987, 50981866833670160201, 1121806937829102793662
FORMULA
G.f.: Sum_{n>=0} n!*((x^m-x)/(x^m-1))^n where m = 4.
a(n) ~ n! * (1 - 1/n^2 + 1/n^3 + 9/(2*n^4) + 7/n^5 - 55/(6*n^6) - 114/n^7 - 11419/(24*n^8) - 970/n^9 + 345199/(120*n^10) + ...). - Vaclav Kotesovec, Feb 17 2024
PROG
(PARI) seq(n)={Vec(sum(k=0, n, k!*((x^4-x)/(x^4-1) + O(x*x^n))^k))} \\ Andrew Howroyd, Aug 31 2018
AUTHOR
Ivana Jovovic (ivana121(AT)EUnet.yu), Nov 14 2007
Number of permutations of length n with no consecutive triples i,...i+r,...i+2r for all r, and for all equal spacings d.
+10
3
21, 94, 544, 3509, 26799, 223123
COMMENTS
Here we count both the sequence 1,2,3 (r=1) as a progression in 1,2,3,0,4,5, (note d=1) and in 1,0,2,4,3,5 (here, d=2).
EXAMPLE
For n=4 there are 4!-a(4)=3 with some progression. These are (0,1,2,3), (1,2,3,0), and (3,0,1,2). Here for all the progressions, r=1 and d=1, hence this term is the same as a(4) in A002628.
Number of permutations of [n] whose longest block is of length 2. A block of a permutation is a maximal sequence of consecutive integers which appear in consecutive positions.
+10
3
0, 0, 1, 2, 10, 53, 334, 2428, 20009, 184440, 1881050, 21034905, 255967940, 3367720736, 47641219569, 721160081974, 11631770791362, 199159952915293, 3607908007376418, 68946510671942892, 1386140583681969289, 29247292475233307612, 646231776371742321826
FORMULA
G.f.: Sum_{k>=0} k! * x^k * ( ((1-x^2)/(1-x^3))^k - ((1-x)/(1-x^2))^k ).
PROG
(PARI) my(N=30, x='x+O('x^N)); concat([0, 0], Vec(sum(k=0, N, k!*x^k*(((1-x^2)/(1-x^3))^k-((1-x)/(1-x^2))^k))))
Number of permutations of [n] having no substring [k,k+1,k+2,k+3,k+4].
+10
2
1, 1, 2, 6, 24, 119, 717, 5026, 40242, 362376, 3625081, 39885851, 478714416, 6224078292, 87145277160, 1307271652917, 20917481850667, 355612235468396, 6401234296266540, 121626707638142280, 2432586885636105251, 51085230669413519349, 1123891538655073251190
FORMULA
G.f.: Sum_{k>=0} k! * ( (x-x^5)/(1-x^5) )^k.
PROG
(PARI) my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, k!*((x-x^5)/(1-x^5))^k))
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