1,1

Row sums with zero: {0, 2, 13, 33, 130, 459, 1533, 5266, 17884, 60532, 205129, ...};

This sequence uses the Markov substitution form that I have been using in my chord-geometry/ graph sequences.

This method of differentiating a substitution appears to be new.

f: 1->{1,2}, 2->{1,3}, 3->{3}); p(x,n)=Sum[Substitution[n,m]*t(m-1),{m,1,n}]; q(x,n)=dp(x,n)dx; out_n,m=Coefficients(q(x,n).

{2},

{2, 2, 9},

{2, 2, 9, 4, 10, 6},

{2, 2, 9, 4, 10, 6, 7, 16, 9, 30, 11, 24},

{2, 2, 9, 4, 10, 6, 7, 16, 9, 30, 11, 24, 13, 28, 15, 48, 17, 36, 19, 20, 42, 22, 69},

{2, 2, 9, 4, 10, 6, 7, 16, 9, 30, 11, 24, 13, 28, 15, 48, 17, 36, 19, 20, 42, 22, 69, 24, 50, 26, 81, 28, 58, 30, 31, 64, 33, 102, 35, 72, 37, 76, 39, 120, 41, 84, 43}

Clear[a, s, p, t, m, n]; (* substitution *); s[1] = {1, 2}; s[2] = {1, 3}; s[3] = {1}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; (*A103684*); a = Table[p[n], {n, 0, 10}]; Flatten[a]; b = Table[CoefficientList[D[Apply[Plus, Table[a[[n]][[m]]*x^( m - 1), {m, 1, Length[a[[n]]]}]], x], x], {n, 1, 11}]; Flatten[b] Table[Apply[Plus, CoefficientList[D[Apply[Plus, Table[a[[n]][[m]]* x^(m - 1), {m, 1, Length[a[[n]]]}]], x], x]], {n, 1, 11}];

Cf. A103684.

Sequence in context: A011140 A068718 A075097 this_sequence A022459 A060804 A086364

Adjacent sequences: A138053 A138054 A138055 this_sequence A138057 A138058 A138059

nonn,uned,tabf

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 02 2008