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A181162
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Number of commuting functions: the number of ordered pairs (f,g) of functions from {1..n} to itself such that fg=gf (i.e., f(g(i))=g(f(i)) for all i).
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36
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1, 1, 10, 141, 2824, 71565, 2244096, 83982199, 3681265792, 186047433225, 10716241342240, 697053065658411, 50827694884298784, 4129325095108122637, 371782656333674104624, 36918345387693628911375, 4025196918605160943576576, 479796375191949916361466897
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OFFSET
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0,3
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COMMENTS
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Also, the total number of endomorphisms of all directed graphs on n labeled vertices with outdegree of each vertex equal 1. - Max Alekseyev, Jan 09 2015
Seems to be relatively hard to compute for large n. (a(n)-n^n)/2 is always an integer, since it gives the number of unordered pairs of distinct commuting functions.
a(n) is divisible by n as proved by Holloway and Shattuck (2015).
Multiply fg=gf from the right by f to obtain fgf=gff, and use f(gf)=f(fg)=ffg to see ffg=gff; iterate to see f^k g = g f^k for all k>=1; by symmetry g^k f = f g^k holds as well.
More generally, if X and Y are words of length w over the alphabet {f,g}, then X = Y (as functional composition) whenever both words contain j symbols f and k symbols g (and j+k=w). (End)
Functions with the same mapping pattern have the same number of commuting functions, so there is no need to check every pair. - Martin Fuller, Feb 01 2015
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LINKS
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EXAMPLE
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The a(2) = 10 pairs of maps [2] -> [2] are:
01: [ 1 1 ] [ 1 1 ]
02: [ 1 1 ] [ 1 2 ]
03: [ 1 2 ] [ 1 1 ]
04: [ 1 2 ] [ 1 2 ]
05: [ 1 2 ] [ 2 1 ]
06: [ 1 2 ] [ 2 2 ]
07: [ 2 1 ] [ 1 2 ]
08: [ 2 1 ] [ 2 1 ]
09: [ 2 2 ] [ 1 2 ]
10: [ 2 2 ] [ 2 2 ]
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MATHEMATICA
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(* This brute force code allows to get a few terms *)
a[n_] := a[n] = If[n == 0, 1, Module[{f, g, T}, T = Tuples[Range[n], n]; Table[f = T[[j, #]]&; g = T[[k, #]] &; Table[True, {n}] == Table[f[g[i]] == g[f[i]], {i, n}], {j, n^n}, {k, n^n}] // Flatten // Count[#, True]&]];
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CROSSREFS
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A053529 is a similar count for permutations. A254529 is for permutations commuting with functions.
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KEYWORD
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hard,nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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