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Number of partitions of n into distinct divisors > 1 of n.
8

%I #23 Nov 18 2021 12:13:36

%S 1,0,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,3,1,1,1,

%T 1,1,6,1,1,1,2,1,2,1,1,1,1,1,6,1,1,1,1,1,3,1,2,1,1,1,19,1,1,1,1,1,3,1,

%U 1,1,1,1,16,1,1,1,1,1,2,1,4,1,1,1,14,1

%N Number of partitions of n into distinct divisors > 1 of n.

%C a(A136446(n)) > 1.

%H Alois P. Heinz, <a href="/A211111/b211111.txt">Table of n, a(n) for n = 0..10000</a> (terms n=1..1000 from Reinhard Zumkeller)

%e n=12: the divisors > 1 of 12 are {2,3,4,6,12}, there are exactly two subsets which sum up to 12, namely {12} and {2,4,6}, therefore a(12) = 2;

%e a(13) = #{13} = 1, because 13 is prime, having no other divisor > 1;

%e n=14: the divisors > 1 of 14 are {2,7,14}, {14} is the only subset summing up to 14, therefore a(14) = 1.

%p with(numtheory):

%p a:= proc(n) local b, l; l:= sort([(divisors(n) minus {1})[]]):

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

%p b(m, i-1)+`if`(l[i]>m, 0, b(m-l[i], i-1))))

%p end; forget(b):

%p b(n, nops(l))

%p end:

%p seq(a(n), n=0..100); # _Alois P. Heinz_, Nov 18 2021

%t a[n_] := Count[IntegerPartitions[n, All, Divisors[n] // Rest], P_ /; Reverse[P] == Union[P]];

%t Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Nov 18 2021 *)

%o (Haskell)

%o a211111 n = p (tail $ a027750_row n) n where

%o p _ 0 = 1

%o p [] _ = 0

%o p (k:ks) m | m < k = 0

%o | otherwise = p ks (m - k) + p ks m

%Y Cf. A211110, A033630, A027750.

%Y Cf. A065205, A136446.

%K nonn

%O 0,13

%A _Reinhard Zumkeller_, Apr 01 2012

%E a(0)=1 prepended by _Alois P. Heinz_, Nov 18 2021