OFFSET
0,3
COMMENTS
For solving the differential equation A'(x)=G(A(x)), where G(0)!=0,
a(n) = 1/n!*sum(pi(i) in P(2*n-1,n), T(n,i)*prod(j=1..n, g(k_j-1))),
where pi(i) is the partition of 2*n-1 into n parts in lexicographic order P(2*n-1,n).
G(x) = g(0)+g(1)*x+g(2)*x^2+...
Examples
A003422 A'(x)=A(x)+1/(1-x)
A000108 A'(x)=1/(1-2*A(x)),
A001147 A'(x)=1/(1-A(x))
A007489 A'(x)=A(x)+x/(1-x)^2+1.
A006351 B'(x)=(1+B(x))/(1-B(x))
A029768 A'(x)=log(1/(1-A(x)))+1.
A001662 B'(x)=1/(1+B(x))
A180254 A'(x)=(1-sqrt(1-4*A(x)))/2
Compare with A145271. There (j')^k = [(d/dx)^j g(x)]^k evaluated at x=0 gives formulas expressed in terms of the coefficients of the Taylor series g(x). If, instead, we express the formulas in terms of the coefficients of the power series of g(x), we obtain a row reversed array for A190015 since the partitions there are in reverse order to the ones here. Simply exchange (j!)^k * (j")^k for (j')^k, where (j")^k = [(d/dx)^j g(x) / j!]^k, to transform from one array to the other. E.g., R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1 = 1 (O")^1 (1")^3 + 4 (0")^2 (1")^1 2*(2")^1 + 1 (0")^1 3!*(3")^1 = 1 (O")^1 (1")^3 + 8 (0")^2 (1")^1 (2")^1 + 6 (0")^1 (3")^1, the fourth partition polynomial here. - Tom Copeland, Oct 17 2014
EXAMPLE
Triangle begins:
1;
1;
2,1;
6,8,1;
24,42,16,22,1;
120,264,180,192,136,52,1;
720,1920,1248,540,1824,2304,272,732,720,114,1;
5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1;
40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816, 101844,142704,7936,51048,110448,34304,8334,11616,494,1;
Example for n=5:
partitions of number 9 into 5 parts in lexicographic order:
[1,1,1,1,5]
[1,1,1,2,4]
[1,1,1,3,3]
[1,1,2,2,3]
[1,2,2,2,2]
a(5) = (24*g(0)^4*g(4) +42*g(0)^3*g(1)*g(3) +16*g(0)^3*g(2)^2 +22*g(0)^2*g(1)^2*g(2) +g(0)*g(1)^4)/5!.
PROG
(Maxima)
/* array of triangle */
M:[1, 1, 2, 1, 6, 8, 1, 24, 42, 16, 22, 1, 120, 264, 180, 192, 136, 52, 1, 720, 1920, 1248, 540, 1824, 2304, 272, 732, 720, 114, 1, 5040, 15840, 10080, 8064, 18720, 22752, 9612, 7056, 10224, 17928, 3968, 2538, 3072, 240, 1, 40320, 146160, 92160, 70560, 32256, 207360, 249120, 193536, 73728, 61560, 144720, 246816, 101844, 142704, 7936, 51048, 110448, 34304, 8334, 11616, 494, 1];
/* function of triangle */
T(n, k):=M[sum(num_partitions(i), i, 0, n-1)+k+1];
/* count number of partitions of n into m parts */
b(n, m):=if n<m then 0 else if m=1 then 1 else b(n-1, m-1)+b(n-m, m);
/* unranking partitions(n, m) , num - numbers partitions of lexicographic order */
array(pa, 10);
gen_partitions(n, m, num, pos):= if n<m then return else
if m=1 then pa[pos]:n else
if num<b(n-1, m-1) then (pa[pos]:1, gen_partitions(n-1, m-1, num, pos+1)) else
if num<b(n-m, m)+b(n-1, m-1) then
(gen_partitions(n-m, m, num-b(n-1, m-1), pos),
for i:0 thru m-1 do pa[i+pos]:pa[i+pos]+1);
/* solve differential equation A'(x)=G(A(x)), G(x)=g(0)+g(1)*x+g(2)^x^2+...*/
/* gcoeff(n)=g(n) */
Solve(n, gcoeff):=block([s, h, num], s:0, for num:0 thru b(2*n-1, n)-1 do (
gen_partitions(2*n-1, n, num, 0), s:s+T(n-1, num+1)*prod(gcoeff(pa[i]-1), i, 0, n-1)), s/n!);
/*Test */
one(n):=1;
makelist(n!*Solve(n, one), n, 1, 9);
g(n):=2^n;
makelist(Solve(n, g), n, 1, 9);
(Maxima) /* Find triangle */
Co(n, k):=if k=1 then a(n) else sum(a(i+1)*Co(n-i-1, k-1), i, 0, n-k);
a(n):=if n=1 then 1 else 1/n*sum(Co(n-1, k)*x(k), k, 1, n-1);
makelist(ratsimp(n!*a(n)), n, 1, 5);
/* Vladimir Kruchinin, Jun 15 2012 */
(PARI) serlaplace( serreverse( intformal( 1 / sum(n=0, 9, eval(Str("g"n)) * x^n, x * O(x^9))))) /* Michael Somos, Oct 22 2014 */
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Vladimir Kruchinin, May 04 2011
STATUS
approved