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A350770 Triangle read by rows: T(n, k) = 2^(n-k-1) + 2^k - 2, 0 <= k <= n-1.

%I A350770 #46 May 04 2024 19:41:49
%S A350770 0,1,1,3,2,3,7,4,4,7,15,8,6,8,15,31,16,10,10,16,31,63,32,18,14,18,32,
%T A350770 63,127,64,34,22,22,34,64,127,255,128,66,38,30,38,66,128,255,511,256,
%U A350770 130,70,46,46,70,130,256,511,1023,512,258,134,78,62,78,134,258,512,1023,2047,1024,514,262,142,94,94,142,262,514,1024,2047
%N A350770 Triangle read by rows: T(n, k) = 2^(n-k-1) + 2^k - 2, 0 <= k <= n-1.
%C A350770 T(n, k) is the number of player-reduced static games within an n-player two-strategy game scenario in which one player (the "standpoint") faces a specific combination of other players' individual strategies without the possibility of anti-coordination between them -- the total number of such combinations is 2^(n-1). The value of k represents the number of other players who (are expected to) agree on one of the two strategies in S, while the other n-k-1 choose the other strategy; the standpoint player is not included.
%C A350770 The sum of the products of T(n, k) and binomial(n-1,k) for 0 <= k <= n-1 equals 2*A001047(n-1). For instance, for n = 3, T(3, k) returns 3, 2, and 3 and binomial(3-1,k) returns 1, 2, and 1 for k = 0, 1, and 2, respectively. Then 3*1 + 2*2 + 3*1 = 2*A001047(3-1) = 2*5 = 10. Similarly, for n = 4, the result yields 7*1 + 4*3 + 4*3 + 7*1 = 2*A001047(4-1) = 2*19 = 38.
%H A350770 Andrew Howroyd, <a href="/A350770/b350770.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50).
%H A350770 Ambrosio Valencia-Romero, <a href="https://www.proquest.com/docview/2582196102">Strategy Dynamics in Collective Systems Design</a>, Ph.D. Thesis, Stevens Institute of Technology (Hoboken, 2021). [Table 5.4, page 67]
%H A350770 Ambrosio Valencia-Romero and P. T. Grogan, <a href="https://doi.org/10.1371/journal.pone.0301394">The strategy dynamics of collective systems: Underlying hindrances beyond two-actor coordination</a>, PLOS ONE 19(4): e0301394 (<a href="https://doi.org/10.1371/journal.pone.0301394.s001">S1 Appendix</a>).
%F A350770 T(n, k) = 2^(n-k-1) + 2^k - 2.
%e A350770 Triangle begins:
%e A350770      0;
%e A350770      1,    1;
%e A350770      3,    2,   3;
%e A350770      7,    4,   4,   7;
%e A350770     15,    8,   6,   8,  15;
%e A350770     31,   16,  10,  10,  16, 31;
%e A350770     63,   32,  18,  14,  18, 32, 63;
%e A350770    127,   64,  34,  22,  22, 34, 64, 127;
%e A350770    255,  128,  66,  38,  30, 38, 66, 128, 255;
%e A350770    511,  256, 130,  70,  46, 46, 70, 130, 256, 511;
%e A350770   1023,  512, 258, 134,  78, 62, 78, 134, 258, 512, 1023;
%e A350770   2047, 1024, 514, 262, 142, 94, 94, 142, 262, 514, 1024, 2047;
%e A350770   ...
%p A350770 T := n -> seq(2^(n - k - 1) + 2^k - 2, k = 0 .. n - 1);
%p A350770 seq(T(n), n=1..12);
%o A350770 (PARI) T(n, k) = 2^(n-k-1) + 2^k - 2 \\ _Andrew Howroyd_, May 06 2023
%Y A350770 Column k=0 gives A000225(n-1).
%Y A350770 Row sums give A145654.
%Y A350770 Cf. A001047.
%K A350770 nonn,easy,tabl
%O A350770 1,4
%A A350770 _Ambrosio Valencia-Romero_, Jan 14 2022