A331406 Array read by antidiagonals: A(n,m) is the number of ways to place non-adjacent counters on the black squares of a 2n-1 X 2m-1 checker board.
1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 17, 8, 1, 1, 16, 73, 73, 16, 1, 1, 32, 314, 689, 314, 32, 1, 1, 64, 1351, 6556, 6556, 1351, 64, 1, 1, 128, 5813, 62501, 139344, 62501, 5813, 128, 1, 1, 256, 25012, 596113, 2976416, 2976416, 596113, 25012, 256, 1
Offset: 0
Examples
Array begins: =========================================================== n\m | 0 1 2 3 4 5 6 ----+------------------------------------------------------ 0 | 1 1 1 1 1 1 1 ... 1 | 1 2 4 8 16 32 64 ... 2 | 1 4 17 73 314 1351 5813 ... 3 | 1 8 73 689 6556 62501 596113 ... 4 | 1 16 314 6556 139344 2976416 63663808 ... 5 | 1 32 1351 62501 2976416 142999897 6888568813 ... 6 | 1 64 5813 596113 63663808 6888568813 748437606081 ... ... Case A(2,2): the checker board has 5 black squares as shown below. __ __ |__|__|__| __|__|__ |__| |__| If a counter is placed on the central square then a counter cannot be placed on the other 4 squares, otherwise counters can be placed in any combination. The total number of arrangements is then 1 + 2^4 = 17, so A(2, 2) = 17.
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..860
- Eric Weisstein's World of Mathematics, Independent Vertex Set
- Wikipedia, Independent set
- Z. Zhang, Merrifield-Simmons index of generalized Aztec diamond and related graphs, MATCH Commun. Math. Comput. Chem. 56 (2006) 625-636.
Crossrefs
Programs
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PARI
step1(v)={vector(#v/2, t, my(i=t-1); sum(j=0, #v-1, if(!bitand(i, bitor(j, j>>1)), v[1+j])))} step2(v)={vector(#v*2, t, my(i=t-1); sum(j=0, #v-1, if(!bitand(i, bitor(j, j<<1)), v[1+j])))} T(n,k)={if(n==0||k==0, 1, my(v=vector(2^k, i, 1)); for(i=2, n, v=step2(step1(v))); vecsum(v))} { for(n=0, 7, for(k=0, 7, print1(T(n,k), ", ")); print) }
Formula
A(n,m) = A(m,n).
Comments