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%I #11 Sep 10 2022 19:41:08
%S 1,1,1,2,2,2,2,3,4,5,6,7,8,9,10,12,14,17,19,23,25,31,33,38,42,51,57,
%T 66,75,86,97,109,122,138,155,177,200,230,253,287,320,363,405,456,507,
%U 572,639,707,785,877,971,1079,1198,1334,1476,1635,1802,2002,2213,2445,2700
%N Number of partitions of n whose reciprocal sum is an integer.
%C Also the number of ways to express an integer as the sum of unit fractions such that the sum of the denominators is n.
%D From a question posted to the news group comp.soft-sys.math.mathematica by "Juan" (erfa11(AT)hotmail.com) at Steven M. Christensen and Associates, Inc. and MathTensor, Inc. Jan 22, 2002 08:46:57 +0000 (UTC).
%H Seiichi Manyama, <a href="/A058360/b058360.txt">Table of n, a(n) for n = 1..80</a>
%e a(12) = 7 because the partitions of 12 whose reciprocal sum is an integer are: {{6, 3, 2, 1}, {4, 4, 2, 1, 1}, {3, 3, 3, 1, 1, 1}, {2, 2, 2, 2, 2, 2}, {2, 2, 2, 2, 1, 1, 1, 1}, {2, 2, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}. Individually their reciprocal sums are: 2, 3, 4, 3, 6, 9 and 12.
%t (* first do *) << "Combinatorica`"; (* then *) f[n_] := Block[{c = i = 0, k = PartitionsP@n, p = {n}}, While[i < k, If[ IntegerQ[ Plus @@ (1/p)], c++ ]; i++; p = NextPartition@ p]; c]; Array[f, 61]
%t Table[Count[IntegerPartitions[n],_?(IntegerQ[Total[1/#]]&)],{n,70}] (* _Harvey P. Dale_, Sep 10 2022 *)
%o (PARI) a(n)=my(s); forpart(v=n,if(type(sum(i=1,#v,1/v[i]))=="t_INT",s++)); s \\ _Charles R Greathouse IV_, Dec 15 2020
%o (PARI) b(n)=if(n<4,return(n==0)); my(s); forpart(v=n, if(type(sum(i=1, #v, 1/v[i]))=="t_INT", s++), [2,n]); s
%o a(n)=my(s=1); vector(n,i,s+=b(i)) \\ _Charles R Greathouse IV_, Dec 15 2020
%Y Cf. A066824, A051908.
%K nonn
%O 1,4
%A _Robert G. Wilson v_, Jan 25 2002, Sep 30 2009
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