A327642 - OEIS

%I #21 Sep 08 2022 08:46:24

%S 1,1,1,1,3,1,4,1,5,4,6,1,19,1,8,6,25,1,37,1,36,8,12,1,169,6,14,10,64,

%T 1,247,1,81,12,18,8,1072,1,20,14,405,1,512,1,144,82,24,1,2825,8,146,

%U 18,196,1,1000,12,743,20,30,1,19858,1,32,112,969,14,1728,1,324,24,1105

%N Number of partitions of n into divisors of n that are at most sqrt(n).

%C a(n) > n if n is in A058080 Union {0}, and, a(n) <= n if n is in A007964; indeed, a(n) = n only for n = 1. - _Bernard Schott_, Sep 22 2019

%H David A. Corneth, <a href="/A327642/b327642.txt">Table of n, a(n) for n = 0..9999</a> (first 5001 terms from Robert Israel)

%F a(n) = [x^n] Product_{d|n, d <= sqrt(n)} 1 / (1 - x^d).

%F a(p) = 1, where p is prime.

%F a(p*q) = q+1 if p <= q are primes. - _Robert Israel_, Sep 22 2019

%e The divisors of 6 are 1, 2, 3, 6 and sqrt(6) = 2.449..., so the possible partitions are 1+1+1+1+1+1 = 1+1+1+1+2 = 1+1+2+2 = 2+2+2; thus a(6) = 4. - _Bernard Schott_, Sep 22 2019

%p f:= proc(n) local x, t, S;

%p     S:= 1;

%p     for t in numtheory:-divisors(n) do

%p       if t^2 <= n then

%p         S:= series(S/(1-x^t),x,n+1);

%p       fi

%p     od;

%p     coeff(S,x,n);

%p end proc:

%p map(f, [$0..100]); # _Robert Israel_, Sep 22 2019

%t a[n_] := SeriesCoefficient[Product[1/(1 - Boole[d <= Sqrt[n]] x^d), {d, Divisors[n]}], {x, 0, n}]; Table[a[n], {n, 0, 70}]

%o (Magma) [1] cat [#RestrictedPartitions(n,{d:d in Divisors(n)| d le Sqrt(n)}):n in [1..70]]; // _Marius A. Burtea_, Sep 20 2019

%Y Cf. A018818, A210442, A211110, A293813.

%K nonn,easy

%O 0,5

%A _Ilya Gutkovskiy_, Sep 20 2019