A289712 - OEIS

%I #22 Oct 13 2019 06:14:41

%S 1,3,15,22,12,36,24,66,126,420,90,364,270,264,240,210,672,780,864,

%T 1050,672,720,924,1092,1344,3240,3312,1260,3600,1200,8910,1080,27104,

%U 5940,1680,8568,8910,14280,6384,5670,5544,9600,43092,42900,5280,3360,9504,8580,21600,54288

%N Smallest integer such that the sum of its n smallest divisors is a square.

%C The first corresponding squares are 1, 4, 9, 36, 16, 25, 36, 144, 81, ...

%C The first squares in the sequence are 1, 36, 3600, ...

%H Robert Israel, <a href="/A289712/b289712.txt">Table of n, a(n) for n = 1..241</a>

%e a(4)=22 because the sum of the first 4 divisors of 22, i.e., 1 + 2 + 11 + 22 = 36, is a square, and 22 is the smallest integer with this property.

%p N:= 5*10^5: # to get terms before the first term > N

%p for k from 1 to N do

%p   d:= sort(convert(numtheory:-divisors(k),list));

%p   s:= ListTools:-PartialSums(d);

%p   for m from 1 to nops(d) do

%p     if not assigned(A[m]) and issqr(s[m]) then A[m]:= k fi

%p   od

%p od:

%p iA:= map(op,{indices(A)}):

%p seq(A[i],i=1..min({$1..max(iA)+1} minus iA)-1); # _Robert Israel_, Oct 01 2017

%t Table[k=1;While[Nand[Length@#>=n,IntegerQ[Sqrt[Total@Take[PadRight[#,n],n]]]]&@Divisors@k,k++];k,{n,1,50}] (* Program from _Michael De Vlieger_ adapted for this sequence. See A289776. *)

%o (PARI) isok(k, n) = {my(v = divisors(k)); if (#v < n, return(0)); issquare(sum(j=1, n, v[j]));}

%o a(n) = {my(k = 1); while(!isok(k,n), k++); k;} \\ _Michel Marcus_, Sep 04 2017

%Y Cf. A000290, A027750, A240698, A289776.

%K nonn

%O 1,2

%A _Michel Lagneau_, Sep 02 2017