A328598 Number of compositions of n with no part circularly followed by a divisor.
1, 0, 0, 0, 0, 2, 0, 4, 2, 7, 12, 11, 22, 26, 55, 63, 99, 149, 215, 324, 458, 699, 1006, 1492, 2185, 3202, 4734, 6928, 10242, 14951, 22023, 32365, 47557, 69905, 102633, 150983, 221712, 325918, 478841, 703647, 1034103, 1519431, 2233061, 3281003, 4821790, 7085358
Offset: 0
Keywords
Examples
The a(5) = 2 through a(12) = 22 compositions (empty column not shown): (2,3) (2,5) (3,5) (2,7) (3,7) (2,9) (5,7) (3,2) (3,4) (5,3) (4,5) (4,6) (3,8) (7,5) (4,3) (5,4) (6,4) (4,7) (2,3,7) (5,2) (7,2) (7,3) (5,6) (2,7,3) (2,4,3) (2,3,5) (6,5) (3,2,7) (3,2,4) (2,5,3) (7,4) (3,4,5) (4,3,2) (3,2,5) (8,3) (3,5,4) (3,5,2) (9,2) (3,7,2) (5,2,3) (2,4,5) (4,3,5) (5,3,2) (4,5,2) (4,5,3) (2,3,2,3) (5,2,4) (5,3,4) (3,2,3,2) (5,4,3) (7,2,3) (7,3,2) (2,3,2,5) (2,3,4,3) (2,5,2,3) (3,2,3,4) (3,2,5,2) (3,4,3,2) (4,3,2,3) (5,2,3,2)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..200
Crossrefs
Programs
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Mathematica
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],And@@Not/@Divisible@@@Partition[#,2,1,1]&]],{n,0,10}]
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PARI
b(n, q, pred)={my(M=matrix(n, n)); for(k=1, n, M[k, k]=pred(q, k); for(i=1, k-1, M[i, k]=sum(j=1, k-i, if(pred(j, i), M[j, k-i], 0)))); M[q,]} seq(n)={concat([1], sum(k=1, n, b(n, k, (i,j)->i%j<>0)))} \\ Andrew Howroyd, Oct 26 2019
Formula
a(n > 0) = A318726(n) - 1.
Extensions
Terms a(26) and beyond from Andrew Howroyd, Oct 26 2019
Comments