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A232636
The second largest value of permanent of (0,1) square matrices of order n with row and column sums equal to 3.
0
34012224, 53747712, 131010048, 204073344, 322486272, 786060288, 1224440064, 1934917632, 4716361728, 7346640384, 11609505792, 28298170368, 44079842304, 69657034752, 169789022208, 264479053824, 417942208512
OFFSET
30,1
COMMENTS
The interval (a(n), A232553(n)) contains no permanents of the matrices under consideration. For n=3 and n=4, the permanent takes only one value: 6 and 9 respectively. Our method (see Shevelev link) allows us to find a simple regularity of the sequence only beginning with n=30 (see formula). However, several values for 5<=n<30 are known: a(5)=13, a(6)=20, a(7)=32, a(8)=78, a(9)=120, a(12)=729, a(15)=4374, a(18)=26244, a(21)=157464, a(24)=944784, a(27)=5668704 (Bolshakov) and a(28)=8957952.
We conjecture that formula below also holds for other values.
In cases n==0 (nod 3), n>=6, and n==2 (mod 3), at least, for n=8 and n>=32, the second largest value of permanent attains on subset of symmetric matrices with the main diagonal of 1's, and does not attain, if n==1 (mod 3), at least, for n=7 and n>=28.
All terms for n>30 not divisible by 3 are conjectural since our method is based on the following very plausible but yet not proved conjecture with confirms in many cases and without counterexample since 1992 (when it was posed, see ref. [11] in Shevelev link). Let L_n be the set of considered n X n (0,1)-matrices with all row and column sums equal to 3. Let S_n be the subset of completely indecomposable such matrices, i.e., not containing submatrices from L_m (m<n). Let M(n) be the maximal value of permanent on S_n. Then M(n_1+n_2) <= M(n_1)*M(n_2). V. I. Bolshakov proved the result for n divisible by 3, using quite different arguments which, generally speaking, do not work for other n and for third, fourth, etc. largest values of permanent, where our (yet conditional) method does work successfully, at least for sufficiently large n.
REFERENCES
V. I. Bolshakov, On spectrum of permanent on Lambda_n^k, Proc. of Seminar on Discrete Math. and Appl., Moscow State Univ. (1986), 65-73 (in Russian).
FORMULA
a(n) = 9/16 * 6^(n/3), if n>=12, n==0 (mod 3) (Bolshakov); a(n) = 8/9 * 6^((n-1)/3), if n>=28, n== 1 (mod 3); a(n) = 13/6 * 6^((n-2)/3), if n>=32, n==2 (mod 3).
Conjectures from Colin Barker, May 27 2016: (Start)
a(n) = 6*a(n-3) for n>32.
G.f.: 419904*x^30*(81+128*x+312*x^2) / (1-6*x^3).
(End)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Nov 27 2013
EXTENSIONS
Edited by M. F. Hasler, Nov 30 2013
STATUS
approved