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Table of coefficients in the iterations of Euler's tree function (A000169), as read by antidiagonals.
10

%I #33 Jul 05 2016 13:02:16

%S 1,1,0,1,2,0,1,4,9,0,1,6,30,64,0,1,8,63,332,625,0,1,10,108,948,4880,

%T 7776,0,1,12,165,2056,18645,89742,117649,0,1,14,234,3800,50680,454158,

%U 1986124,2097152,0,1,16,315,6324,112625,1537524,13221075,51471800,43046721,0,1,18,408,9772,219000,4090980,55494712,448434136,1530489744,1000000000,0,1,20,513,14288,387205,9266706,176238685,2325685632,17386204761,51395228090,25937424601,0,1,22,630,20016,637520,18704322,463975764,8793850560,111107380464,759123121050,1924687118684,743008370688,0,1,24,759,27100,993105,34617288,1067280319,26858490392,499217336145,5964692819140,36882981687519,79553145323940,23298085122481,0

%N Table of coefficients in the iterations of Euler's tree function (A000169), as read by antidiagonals.

%C See table A274391 for the coefficients in exp( T^n(x) ), n>=0, where T^n(x) is the e.g.f. of the n-th row of this table.

%H Paul D. Hanna, <a href="/A274390/b274390.txt">Table of n, a(n) for n = 0..1034 of rows 0..45 of the flattened table.</a>

%F Let T^n(x) denote the n-th iteration of Euler's tree function T(x), then the coefficients in T^n(x) form the n-th row of this table, and the functions satisfy:

%F (1) T^n(x) = x * exp( Sum_{i=1..n} T^i(x) ).

%F (2) T^n(x) = T^(n-1)(x) * exp( T^n(x) ).

%F (3) T^n(x) = T^(n+1)( x/exp(x) ).

%e This table begins:

%e 1, 0, 0, 0, 0, 0, 0, 0, ...;

%e 1, 2, 9, 64, 625, 7776, 117649, 2097152, ...;

%e 1, 4, 30, 332, 4880, 89742, 1986124, 51471800, ...;

%e 1, 6, 63, 948, 18645, 454158, 13221075, 448434136, ...;

%e 1, 8, 108, 2056, 50680, 1537524, 55494712, 2325685632, ...;

%e 1, 10, 165, 3800, 112625, 4090980, 176238685, 8793850560, ...;

%e 1, 12, 234, 6324, 219000, 9266706, 463975764, 26858490392, ...;

%e 1, 14, 315, 9772, 387205, 18704322, 1067280319, 70311813880, ...;

%e 1, 16, 408, 14288, 637520, 34617288, 2217367600, 163802295616, ...;

%e 1, 18, 513, 20016, 993105, 59879304, 4254311817, 348285415872, ...;

%e 1, 20, 630, 27100, 1480000, 98110710, 7656893020, 688058734520, ...;

%e ...

%e where the e.g.f.s of the rows are iterations of T(x) and begin:

%e T^0(x) = x;

%e T^1(x) = T(x) = x + 2*x^2/2! + 9*x^3/3! + 64*x^4/4! + 625*x^5/5! + 7776*x^6/6! + 117649*x^7/7! + 2097152*x^8/8! +...+ n^(n-1)*x^n/n! +...;

%e T^2(x) = T(T(x)) = x + 4*x^2/2! + 30*x^3/3! + 332*x^4/4! + 4880*x^5/5! + 89742*x^6/6! + 1986124*x^7/7! + 51471800*x^8/8! +...+ A207833(n)*x^n/n! +...;

%e T^3(x) = T(T(T(x))) = x + 6*x^2/2! + 63*x^3/3! + 948*x^4/4! + 18645*x^5/5! + 454158*x^6/6! + 13221075*x^7/7! + 448434136*x^8/8! +...+ A227278(n)*x^n/n! +...;

%e T^4(x) = T(T(T(T(x)))) = x + 8*x^2/2! + 108*x^3/3! + 2056*x^4/4! + 50680*x^5/5! + 1537524*x^6/6! + 55494712*x^7/7! + 2325685632*x^8/8! +...;

%e ...

%e where T^n(x)/exp( T^n(x) ) = T^n( x/exp(x) ) = T^(n-1)(x).

%e Also we have

%e T(x) = x*exp( T(x) );

%e T^2(x) = x*exp( T(x) + T^2(x) );

%e T^3(x) = x*exp( T(x) + T^2(x) + T^3(x) );

%e T^4(x) = x*exp( T(x) + T^2(x) + T^3(x) + T^4(x) ); ...

%o (PARI) {ITERATE(F,n,k) = my(G=x +x*O(x^k)); for(i=1,n,G=subst(G,x,F));G}

%o {T(n,k) = my(TREE = serreverse(x*exp(-x +x*O(x^k)))); k!*polcoeff(ITERATE(TREE,n,k),k)}

%o /* Print this table as a square array */

%o for(n=0,10,for(k=1,10,print1(T(n,k),", "));print(""))

%o /* Print this table as a flattened array */

%o for(n=0,12,for(k=1,n,print1(T(n-k,k),", "));)

%Y Cf. A274391, A000169, A207833, A227278; diagonals: A274389, A274392.

%Y Cf. A274570 (transforms diagonals).

%Y Cf. A274740 (same table, but read differently).

%K nonn,tabl

%O 0,5

%A _Paul D. Hanna_, Jun 19 2016