A282467 - OEIS

%I #33 Feb 27 2020 15:49:00

%S 0,1,1,4,5,9,13,21,27,40,54,75,99,133,172,230,295,382,488,625,788,

%T 1000,1253,1573,1955,2434,3006,3716,4563,5600,6840,8348,10139,12308,

%U 14879,17974,21635,26013,31181,37336,44581,53170,63259,75173,89128,105556,124752,147271,173522,204223,239939,281587

%N Number of partitions of n which are not the partitions into (one or more) consecutive parts.

%C Also number of partitions of n minus the number of odd divisors of n.

%H G. C. Greubel, <a href="/A282467/b282467.txt">Table of n, a(n) for n = 1..5000</a>

%H M. D. Hirschhorn and P. M. Hirschhorn, <a href="http://www.maa.org/sites/default/files/3004420056860.pdf.bannered.pdf">Partitions into Consecutive Parts</a>, Mathematics Magazine: 2003, Volume 76, Number 4, Pages: 306-308.

%F a(n) = A000041(n) - A001227(n).

%e For n = 6, the number of partitions of 6 is A000041(6) = 11. There are two partitions of 6 into (one or more) consecutive parts: [6] and [3, 2, 1], so a(6) = 11 - 2 = 9. On the other hand, 6 has two odd divisors: 1 and 3, so a(6) = 11 - 2 = 9.

%e For n = 15, the number of partitions of 15 is A000041(15) = 176. There are four partitions of 15 into (one or more) consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1], so a(15) = 176 - 4 = 172. On the other hand, 15 has four odd divisors: 1, 3, 5 and 15, so a(15) = 176 - 4 = 172.

%p a:= n-> combinat[numbpart](n)-mul(`if`(i[1]=2, 1, i[2]+1), i=ifactors(n)[2]):

%p seq(a(n), n=1..100);  # _Alois P. Heinz_, Feb 27 2020

%t Table[PartitionsP@ n - DivisorSum[n, Boole[# > 0] &, OddQ@ # &], {n, 52}] (* _Michael De Vlieger_, Feb 27 2017 *)

%o (Sage)

%o A282467 = lambda n: number_of_partitions(n) - len(list(filter(is_odd, divisors(n))))

%o [A282467(n) for n in (1..52)] # _Peter Luschny_, Feb 28 2017

%o (PARI) a(n)=numbpart(n) - numdiv(n>>valuation(n,2)) \\ _Charles R Greathouse IV_, Mar 01 2017

%Y Cf. A000041, A001227.

%K nonn

%O 1,4

%A _Omar E. Pol_, Feb 25 2017