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A189281
Number of permutations p of 1,2,...,n satisfying p(i+2) - p(i) <> 2 for all 1 <= i <= n-2.
7
1, 1, 2, 5, 18, 75, 410, 2729, 20906, 181499, 1763490, 18943701, 222822578, 2847624899, 39282739034, 581701775369, 9202313110506, 154873904848803, 2762800622799362, 52071171437696453, 1033855049655584786, 21567640717569135515
OFFSET
0,3
COMMENTS
a(n) is also the number of ways to place n nonattacking pieces rook + semi-leaper (2,2) on an n X n chessboard.
Comments from Vaclav Kotesovec, Mar 05 2022: (Start)
The original submission had keyword hard because of the following running times (in 2012):
a(33) 39 hours
a(34) 78 hours
a(35) 147 hours
The conjectured recurrence would imply the asymptotic expansion for a(n)/n! ~
(1 + 3/n + 2/n^2 + 1/n^3 + 0/n^4 + 3/n^5 + 26/n^6 + 101/n^7 + 124/n^8 - 1409/n^9 - 13266/n^10)/e.
This exactly matches the formula from 2011. In addition, all coefficients are integers. It is highly probable that recurrence is correct.
(End)
There are good reasons to believe the conjecture is correct. (It has the expected form.) The problem is one of counting Hamiltonian cycles in the complement of some simple graph. There is a method for counting these efficiently (although I have not implemented in code). Similar to A242522 / A229430. - Andrew Howroyd, Mar 06 2022
See also Manuel Kauers's comments below. Since the four new terms took weeks of computation, the keyword "hard" continues to be justified. - N. J. A. Sloane, Mar 06 2022
a(40)-a(300) were computed using an independent solution (dynamic programming, O(N^4) per term), and the conjectured recurrence was further confirmed to be correct up to n=300. Consequently, the keyword "hard" is removed. - Rintaro Matsuo, Oct 18 2022
LINKS
Rintaro Matsuo, Table of n, a(n) for n = 0..300 (terms 0..35 from Vaclav Kotesovec, terms 36..39 from Christoph Koutschan, computed using a parallelization of Kotesovec's Mathematica program)
Robert Dougherty-Bliss, Experimental Methods in Number Theory and Combinatorics, Ph. D. Dissertation, Rutgers Univ. (2024). See p. 4.
Manuel Kauers and Christoph Koutschan, Guessing with Little Data, arXiv:2202.07966 [cs.SC], 2022.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 644.
FORMULA
Asymptotics: a(n)/n! ~ (1 + 3/n + 2/n^2)/e.
Conjectured recurrence of degree 11 and order 8: (262711*n + 1387742*n^2 - 824875*n^3 - 1855253*n^4 - 111530*n^5 + 680983*n^6 + 364242*n^7 + 84992*n^8 + 10332*n^9 + 640*n^10 + 16*n^11)*a(n) + (-1050844*n - 9705192*n^2 - 7414683*n^3 + 3536494*n^4 + 6459004*n^5 + 3326393*n^6 + 903534*n^7 + 144684*n^8 + 13756*n^9 + 720*n^10 + 16*n^11)*a(n+1) + (3492344 - 2212342*n - 8507169*n^2 - 11544227*n^3 - 12034116*n^4 - 8216995*n^5 - 3442049*n^6 - 890050*n^7 - 142300*n^8 - 13660*n^9 - 720*n^10 - 16*n^11)*a(n+2) + (19817984 + 45323852*n + 825228*n^2 - 57004661*n^3 - 57059306*n^4 - 28077270*n^5 - 8398637*n^6 - 1631510*n^7 - 207980*n^8 - 16828*n^9 - 784*n^10 - 16*n^11)*a(n+3) + (9586160 + 6680237*n - 13772613*n^2 - 27689586*n^3 - 22162455*n^4 - 9855085*n^5 - 2629562*n^6 - 427656*n^7 - 41332*n^8 - 2176*n^9 - 48*n^10)*a(n+4) + (22192864 + 44710768*n - 2924668*n^2 - 52385912*n^3 - 45161616*n^4 - 18784740*n^5 - 4549208*n^6 - 674256*n^7 - 60400*n^8 - 3008*n^9 - 64*n^10)*a(n+5) + (557152 - 2032472*n - 2937392*n^2 - 1594200*n^3 - 517688*n^4 - 122032*n^5 - 19856*n^6 - 1792*n^7 - 64*n^8)*a(n+6) + (3786960 + 7105324*n - 1191064*n^2 - 8059160*n^3 - 5938996*n^4 - 2073752*n^5 - 402736*n^6 - 44528*n^7 - 2624*n^8 - 64*n^9)*a(n+7) + (-598208 - 943004*n + 414196*n^2 + 1213772*n^3 + 728648*n^4 + 203584*n^5 + 29616*n^6 + 2176*n^7 + 64*n^8)*a(n+8) = 0. This recurrence correctly predicted the four new terms in the b-file. - Christoph Koutschan, Feb 19 2022
Comment from N. J. A. Sloane, Mar 12 2022: (Start)
The preceding conjectured recurrence is equivalent to the following, which has degree 3 and order 13, and was obtained by Doron Zeilberger and then reformatted by Manuel Kauers (it uses Mathematica syntax):
Conjecture: ((-1 + n)^2*n*a[n])/4 + (n*(-16 + 38*n + 11*n^2)*a[1 + n])/16 +
(3/2 + (139*n)/16 + (29*n^2)/8 + (3*n^3)/16)*a[2 + n] +
(-21/4 - (51*n)/4 - (79*n^2)/16 - (5*n^3)/8)*a[3 + n] +
(-15/2 - n/8 + (5*n^2)/4 + n^3/8)*a[4 + n] +
(603/4 + (307*n)/4 + (49*n^2)/4 + (11*n^3)/16)*a[5 + n] +
(-41 - (533*n)/16 - (49*n^2)/8 - (5*n^3)/16)*a[6 + n] +
(-911/2 - 161*n - (303*n^2)/16 - (3*n^3)/4)*a[7 + n] +
(-363 - (417*n)/4 - (37*n^2)/4 - n^3/4)*a[8 + n] +
(-993/4 - 53*n - (11*n^2)/4)*a[9 + n] + (-130 - (93*n)/4 - n^2)*a[10 + n] +
(-71/4 - 2*n)*a[11 + n] + (-10 - n)*a[12 + n] + a[13 + n] == 0.
(End)
From Mark van Hoeij, Jul 25 2012: (Start)
A compact way to write the order 13 recurrence is as follows:
Let b(n) = a(n+3) + a(n+2) + (n/2+2)*a(n+1) + (n-1)*a(n)/2
and c(n) = b(n+4) + (n/2+2)*b(n+2) - b(n+1)/2 + (1-n)*b(n)/2;
then c(n+6) - (n+11)*c(n+5) - (2*n+75/4)*c(n+4) + (3-n)*c(n+3)/4 - c(n+2)/2 - (7*n+22)*c(n+1)/4-n*c(n) = 0. (End)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Apr 19 2011
STATUS
approved