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Written on the Wall II, conjecture 374  --  REFUTATION
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[1] Structure and closed forms for the spider S_k = S(2,2,1^k), n = k+5
  ok   k= 2 n= 7: gamma_t = 3  (brute force)
  ok   k= 2 n= 7: B = {2,4} (the two leg ends)
  ok   k= 2 n= 7: median of degree sequence = 1
  ok   k= 2 n= 7: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 2
  ok   k= 2 n= 7: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 10/7
  ok   k= 3 n= 8: gamma_t = 3  (brute force)
  ok   k= 3 n= 8: B = {2,4} (the two leg ends)
  ok   k= 3 n= 8: median of degree sequence = 1
  ok   k= 3 n= 8: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 13/6
  ok   k= 3 n= 8: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 13/8
  ok   k= 4 n= 9: gamma_t = 3  (brute force)
  ok   k= 4 n= 9: B = {2,4} (the two leg ends)
  ok   k= 4 n= 9: median of degree sequence = 1
  ok   k= 4 n= 9: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 16/7
  ok   k= 4 n= 9: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 16/9
  ok   k= 5 n=10: gamma_t = 3  (brute force)
  ok   k= 5 n=10: B = {2,4} (the two leg ends)
  ok   k= 5 n=10: median of degree sequence = 1
  ok   k= 5 n=10: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 19/8
  ok   k= 5 n=10: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 19/10
  ok   k= 6 n=11: gamma_t = 3  (brute force)
  ok   k= 6 n=11: B = {2,4} (the two leg ends)
  ok   k= 6 n=11: median of degree sequence = 1
  ok   k= 6 n=11: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 22/9
  ok   k= 6 n=11: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 2
  ok   k= 7 n=12: gamma_t = 3  (brute force)
  ok   k= 7 n=12: B = {2,4} (the two leg ends)
  ok   k= 7 n=12: median of degree sequence = 1
  ok   k= 7 n=12: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 5/2
  ok   k= 7 n=12: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 25/12
  ok   k= 8 n=13: gamma_t = 3  (proved below)
  ok   k= 8 n=13: B = {2,4} (the two leg ends)
  ok   k= 8 n=13: median of degree sequence = 1
  ok   k= 8 n=13: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 28/11
  ok   k= 8 n=13: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 28/13
  ok   k= 9 n=14: gamma_t = 3  (proved below)
  ok   k= 9 n=14: B = {2,4} (the two leg ends)
  ok   k= 9 n=14: median of degree sequence = 1
  ok   k= 9 n=14: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 31/12
  ok   k= 9 n=14: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 31/14
  ok   k=10 n=15: gamma_t = 3  (proved below)
  ok   k=10 n=15: B = {2,4} (the two leg ends)
  ok   k=10 n=15: median of degree sequence = 1
  ok   k=10 n=15: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 34/13
  ok   k=10 n=15: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 34/15
  ok   k=11 n=16: gamma_t = 3  (proved below)
  ok   k=11 n=16: B = {2,4} (the two leg ends)
  ok   k=11 n=16: median of degree sequence = 1
  ok   k=11 n=16: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 37/14
  ok   k=11 n=16: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 37/16
  ok   k=12 n=17: gamma_t = 3  (proved below)
  ok   k=12 n=17: B = {2,4} (the two leg ends)
  ok   k=12 n=17: median of degree sequence = 1
  ok   k=12 n=17: dist_avg(B) [nonzero-only] = (3k+4)/(k+3) = 8/3
  ok   k=12 n=17: dist_avg(B) [over all V]   = (3k+4)/(k+5) = 40/17

[2] The margin gamma_t - (dist_avg(B) + median), both normalisations
      k    n   margin (nonzero-only)   margin (over all V)
  ok   k=   2 n=    7           0 = (2-k)/(k+3)          4/7 = (6-k)/(k+5)
  ok   k=   3 n=    8        -1/6 = (2-k)/(k+3)          3/8 = (6-k)/(k+5)
  ok   k=   4 n=    9        -2/7 = (2-k)/(k+3)          2/9 = (6-k)/(k+5)
  ok   k=   5 n=   10        -3/8 = (2-k)/(k+3)         1/10 = (6-k)/(k+5)
  ok   k=   6 n=   11        -4/9 = (2-k)/(k+3)            0 = (6-k)/(k+5)
  ok   k=   7 n=   12        -1/2 = (2-k)/(k+3)        -1/12 = (6-k)/(k+5)
  ok   k=   8 n=   13       -6/11 = (2-k)/(k+3)        -2/13 = (6-k)/(k+5)
  ok   k=   9 n=   14       -7/12 = (2-k)/(k+3)        -3/14 = (6-k)/(k+5)
  ok   k=  10 n=   15       -8/13 = (2-k)/(k+3)        -4/15 = (6-k)/(k+5)
  ok   k=  11 n=   16       -9/14 = (2-k)/(k+3)        -5/16 = (6-k)/(k+5)
  ok   k=  12 n=   17        -2/3 = (2-k)/(k+3)        -6/17 = (6-k)/(k+5)
  ok   k=  13 n=   18      -11/16 = (2-k)/(k+3)        -7/18 = (6-k)/(k+5)
  ok   k=  20 n=   25      -18/23 = (2-k)/(k+3)       -14/25 = (6-k)/(k+5)
  ok   k=  50 n=   55      -48/53 = (2-k)/(k+3)         -4/5 = (6-k)/(k+5)
  ok   k= 200 n=  205    -198/203 = (2-k)/(k+3)     -194/205 = (6-k)/(k+5)
  ok   k=1000 n= 1005   -998/1003 = (2-k)/(k+3)    -994/1005 = (6-k)/(k+5)

  Both margins tend to -1 as k -> infinity, so the deficit is unbounded away from 0.
  ok   at k=1000 the nonzero-only deficit already exceeds 0.99
  ok   at k=1000 the over-all-V deficit already exceeds 0.98

[3] Smallest counterexamples
  ok   nonzero-only: S_3 (n=8) fails, 3 < 13/6 + 1 = 19/6
  ok   over all V:   S_7 (n=12) fails, 3 < 25/12 + 1 = 37/12
  ok   nonzero-only: even P4 fails, gamma_t = 2 < 1 + 3/2 = 5/2

[4] gamma_t(S_k) = 3 for every k >= 1  (proof check)
  ok   leg ends 2 and 4 have unique neighbours 1 and 3
  ok   every pendant leaf has unique neighbour c = 0
  ok   => any total dominating set contains 1, 3 and 0; and {0,1,3} is total dominating
  ok   {0,1,3} totally dominates S_6

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ALL CHECKS PASSED
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