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Number of ways to split an integer partition of n into consecutive subsequences.
26

%I #13 Jan 01 2023 19:21:25

%S 1,1,3,7,17,37,83,175,373,773,1603,3275,6693,13557,27447,55315,111397,

%T 223769,449287,900795,1805465,3615929,7240327,14491623,29001625,

%U 58027017,116093259,232237583,464558201,929224589,1858623819,3717475031,7435314013,14871103069

%N Number of ways to split an integer partition of n into consecutive subsequences.

%F a(n) = A070933(n)/2.

%F O.g.f.: (1/2)*Product_{n >= 1} 1/(1 - 2*x^n).

%F G.f.: 1 + Sum_{k>=1} 2^(k - 1) * x^k / Product_{j=1..k} (1 - x^j). - _Ilya Gutkovskiy_, Jan 28 2020

%e The a(3) = 7 ways to split an integer partition of 3 into consecutive subsequences are (3), (21), (2)(1), (111), (11)(1), (1)(11), (1)(1)(1).

%p b:= proc(n, i) option remember; `if`(n=0, 1/2, `if`(i<1, 0,

%p b(n, i-1) +`if`(i>n, 0, 2*b(n-i, i))))

%p end:

%p a:= n-> ceil(b(n$2)):

%p seq(a(n), n=0..33); # _Alois P. Heinz_, Jan 01 2023

%t Table[Sum[2^(Length[ptn]-1),{ptn,IntegerPartitions[n]}],{n,40}]

%t (* Second program: *)

%t (1/2) CoefficientList[1 - 1/QPochhammer[2, x] + O[x]^100 , x] (* _Jean-François Alcover_, Jan 02 2022, after _Vladimir Reshetnikov_ in A070933 *)

%Y Cf. A006951, A070933, A100883, A279784, A279786, A323433, A323582.

%K nonn

%O 0,3

%A _Gus Wiseman_, Jan 19 2019