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G.f.: A(x) = Sum_{n>=0} x^n / Product_{k=1..n} (1 - k*x^k).
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%I #25 Jun 18 2019 03:58:33

%S 1,1,2,3,6,9,19,29,57,94,172,280,519,833,1472,2433,4185,6800,11666,

%T 18816,31686,51340,84929,136561,225476,359746,586133,936243,1511650,

%U 2397400,3856698,6084186,9711492,15299490,24247456,38016261,60079125,93752706,147284928

%N G.f.: A(x) = Sum_{n>=0} x^n / Product_{k=1..n} (1 - k*x^k).

%C Number of rooted ordered trees with n non-root nodes such that both successive branch heights and the lengths of the branches are weakly increasing; see example. - _Joerg Arndt_, Aug 27 2014

%H Robert Israel, <a href="/A193196/b193196.txt">Table of n, a(n) for n = 0..5627</a>

%H Joerg Arndt, <a href="/A193196/a193196.txt">trees as in comment for 1<=n<=7</a>

%F G.f.: G(0) - 1 where G(k) = 1 + (1-x)/(1-x^k*k)/(1-x/(x+(1-x)/G(k+1) )); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Jan 22 2013

%F a(n) = sum( prod(j=2..m, min(C[j-1], C[j]))) where the sum is over all partitions C[1..m] (m parts) of n, see example. - _Joerg Arndt_, Sep 03 2014

%F From _Vaclav Kotesovec_, Jun 18 2019: (Start)

%F a(n) ~ c * 3^(n/3), where

%F c = 9390.8440644933535486959046639452060731482141... if mod(n,3)=0

%F c = 9390.7389359914729419715573277079935321683397... if mod(n,3)=1

%F c = 9390.7321933046037554603013237581369727858708... if mod(n,3)=2

%F (End)

%e G.f.: A(x) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 19*x^6 + 29*x^7 +...

%e where:

%e A(x) = 1 + x/(1-x) + x^2/((1-x)*(1-2*x^2)) + x^3/((1-x)*(1-2*x^2)*(1-3*x^3)) + x^4/((1-x)*(1-2*x^2)*(1-3*x^3)*(1-4*x^4)) +...

%e From _Joerg Arndt_, Aug 27 2014: (Start)

%e The a(4) = 5 trees described in the comment are:

%e :

%e : 1:

%e : [ 1 1 1 1 ] <--= branch lengths

%e : [ 0 0 0 0 ] <--= branch heights

%e :

%e : O--o

%e : .--o

%e : .--o

%e : .--o

%e :

%e :

%e : 2:

%e : [ 1 1 2 ]

%e : [ 0 0 0 ]

%e :

%e : O--o

%e : .--o

%e : .--o--o

%e :

%e :

%e : 3:

%e : [ 1 3 ]

%e : [ 0 0 ]

%e :

%e : O--o

%e : .--o--o--o

%e :

%e :

%e : 4:

%e : [ 2 2 ]

%e : [ 0 0 ]

%e :

%e : O--o--o

%e : .--o--o

%e :

%e :

%e : 5:

%e : [ 2 2 ]

%e : [ 0 1 ]

%e :

%e : O--o--o

%e : .--o--o

%e :

%e :

%e : 6:

%e : [ 4 ]

%e : [ 0 ]

%e :

%e : O--o--o--o--o

%e :

%e See the Arndt link for all examples for 1 <= n <= 7.

%e (End)

%e a(6) = 19 because the 11 partitions of 6 with the products as in the comment are

%e 01: [ 1 1 1 1 1 1 ] 1*1*1*1*1 = 1

%e 02: [ 1 1 1 1 2 ] 1*1*1*1 = 1

%e 03: [ 1 1 1 3 ] 1*1*1 = 1

%e 04: [ 1 1 2 2 ] 1*1*2 = 2

%e 05: [ 1 1 4 ] 1*1 = 1

%e 06: [ 1 2 3 ] 1*2 = 1

%e 07: [ 1 5 ] 1 = 1

%e 08: [ 2 2 2 ] 2*2 = 4

%e 09: [ 2 4 ] 2 = 2

%e 10: [ 3 3 ] 3 = 3

%e 11: [ 6 ] (empty prod.) = 1

%e and the sum of the products is 19. - _Joerg Arndt_, Sep 03 2014

%p N:= 100: # to get all terms up to a(N)

%p gN:= add(x^n/mul(1-k*x^k,k=1..n),n=0..N):

%p S:= series(gN,x,N+1):

%p seq(coeff(S,x,n), n=0..N); # _Robert Israel_, Aug 28 2014

%o (PARI) {a(n)=my(A=1);polcoeff(sum(m=0,n,x^m/prod(k=1,m,1-k*x^k +x*O(x^n))),n)}

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 17 2011