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Number of partitions of n minus number of divisors of n.
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%I #27 Oct 16 2023 22:24:38

%S 0,0,1,2,5,7,13,18,27,38,54,71,99,131,172,226,295,379,488,621,788,998,

%T 1253,1567,1955,2432,3006,3712,4563,5596,6840,8343,10139,12306,14879,

%U 17968,21635,26011,31181,37330,44581,53166,63259,75169,89128,105554,124752

%N Number of partitions of n minus number of divisors of n.

%C a(n) is also the number of partitions of n with at least one distinct part (i.e., not all parts are equal).

%H Alois P. Heinz, <a href="/A144300/b144300.txt">Table of n, a(n) for n = 1..1000</a>

%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpatru.jpg">The shell model of partitions</a>

%F a(n) = p(n) - d(n) = A000041(n) - A000005(n).

%p with(numtheory): b:= proc(n) option remember; `if`(n=0, 1, add(add(d, d=divisors(j)) *b(n-j), j=1..n)/n) end: a:= n-> b(n)- tau(n):

%p seq(a(n), n=1..50); # _Alois P. Heinz_, Oct 07 2008

%t Table[PartitionsP[n]-DivisorSigma[0,n],{n,50}] (* _Harvey P. Dale_, Apr 10 2014 *)

%o (PARI) al(n)=vector(n,k,numbpart(k)-numdiv(k))

%o (Python)

%o from sympy import npartitions, divisor_count

%o def A144300(n): return npartitions(n)-divisor_count(n) # _Chai Wah Wu_, Oct 16 2023

%Y Cf. A000005, A000041, A135010, A138121, A195364.

%Y A182114(n,n-1) = a(n). - _Alois P. Heinz_, Nov 02 2012

%K easy,nonn

%O 1,4

%A _Omar E. Pol_, Sep 17 2008