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A319411 Triangle read by rows: T(n,k) = number of binary vectors of length n with runs-resistance k (1 <= k <= n).

%I A319411 #18 Nov 27 2019 08:34:42
%S A319411 2,2,2,2,2,4,2,4,6,4,2,2,12,12,4,2,6,30,18,8,0,2,2,44,44,32,4,0,2,6,
%T A319411 82,76,74,16,0,0,2,4,144,138,172,52,0,0,0,2,6,258,248,350,156,4,0,0,0,
%U A319411 2,2,426,452,734,404,28,0,0,0,0,2,10,790,752,1500,938,104,0,0,0,0,0
%N A319411 Triangle read by rows: T(n,k) = number of binary vectors of length n with runs-resistance k (1 <= k <= n).
%C A319411 "Runs-resistance" is defined in A318928.
%C A319411 Row sums are 2,4,8,16,... (the binary vectors may begin with 0 or 1).
%C A319411 This is similar to A329767 but without the k = 0 column and with a different row n = 1. - _Gus Wiseman_, Nov 25 2019
%H A319411 Hiroaki Yamanouchi, <a href="/A319411/b319411.txt">Table of n, a(n) for n = 1..1830</a>
%H A319411 Claude Lenormand, <a href="/A318921/a318921.pdf">Deux transformations sur les mots</a>, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003.
%e A319411 Triangle begins:
%e A319411 2,
%e A319411 2, 2,
%e A319411 2, 2, 4,
%e A319411 2, 4, 6, 4,
%e A319411 2, 2, 12, 12, 4,
%e A319411 2, 6, 30, 18, 8, 0,
%e A319411 2, 2, 44, 44, 32, 4, 0,
%e A319411 2, 6, 82, 76, 74, 16, 0, 0,
%e A319411 2, 4, 144, 138, 172, 52, 0, 0, 0,
%e A319411 2, 6, 258, 248, 350, 156, 4, 0, 0, 0,
%e A319411 2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0,
%e A319411 2, 10, 790, 752, 1500, 938, 104, 0, 0, 0, 0, 0,
%e A319411 ...
%e A319411 Lenormand gives the first 20 rows.
%e A319411 The calculation of row 4 is as follows.
%e A319411 We may assume the first bit is a 0, and then double the answers.
%e A319411 vector / runs / steps to reach a single number:
%e A319411 0000 / 4 / 1
%e A319411 0001 / 31 -> 11 -> 2 / 3
%e A319411 0010 / 211 -> 12 -> 11 -> 2 / 4
%e A319411 0011 / 22 -> 2 / 2
%e A319411 0100 / 112 -> 21 -> 11 -> 2 / 4
%e A319411 0101 / 1111 -> 4 / 2
%e A319411 0110 / 121 -> 111 -> 3 / 3
%e A319411 0111 / 13 -> 11 -> 2 / 3
%e A319411 and we get 1 (once), 2 (twice), 3 (three times) and 4 (twice).
%e A319411 So row 4 is: 2,4,6,4.
%t A319411 runsresist[q_]:=If[Length[q]==1,1,Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1];
%t A319411 Table[Length[Select[Tuples[{0,1},n],runsresist[#]==k&]],{n,10},{k,n}] (* _Gus Wiseman_, Nov 25 2019 *)
%Y A319411 Row sums are A000079.
%Y A319411 Column k = 2 is 2 * A032741 = A319410.
%Y A319411 Column k = 3 is 2 * A329745 (because runs-resistance 2 for compositions corresponds to runs-resistance 3 for binary words).
%Y A319411 The version for compositions is A329744.
%Y A319411 The version for partitions is A329746.
%Y A319411 The number of nonzero entries in row n > 0 is A319412(n).
%Y A319411 The runs-resistance of the binary expansion of n is A318928.
%Y A319411 Cf. A096365, A225485, A245563, A319413, A319414, A319415, A325280, A329747, A329750, A329767, A329870.
%K A319411 nonn,tabl,base
%O A319411 1,1
%A A319411 _N. J. A. Sloane_, Sep 20 2018