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Numbers k such that A255217(k) divides A007504(k).
2

%I #13 Nov 29 2021 01:20:21

%S 2,5,15,17,20,25,26,33,37,45,49,51,71,87,88,91,105,111,121,127,173,

%T 175,199,203,213,221,262,271,287,305,307,319,324,329,368,377,410,411,

%U 415,439,445,455,463,467,468,473,547,558,561,567,585,589,591,614,651,661,663,665,670,673,743,761,765

%N Numbers k such that A255217(k) divides A007504(k).

%H Robert Israel, <a href="/A349734/b349734.txt">Table of n, a(n) for n = 1..10000</a>

%e a(2) = 5 is a term because A255217(5) = 2*3*5*7*11 mod (2+3+5+7+11) = 14 divides 2+3+5+7+11 = 28.

%p P:= 1: S:= 0: p:= 1:

%p count:= 0: R:= NULL:

%p for n from 1 while count < 100 do

%p p:= nextprime(p);

%p P:= P*p; S:= S+p;

%p r:= P mod S;

%p if r = 0 then next fi;

%p v:= S mod r;

%p if v = 0 then

%p count:= count+1; R:= R,n;

%p fi

%p od:

%p R;

%t Select[Range[1000], (m = Mod[Times @@ (p = Prime[Range[#]]), Plus @@ p]) > 0 && Divisible[Plus @@ p, m] &] (* _Amiram Eldar_, Nov 28 2021 *)

%Y Cf. A002110, A007504, A255217, A349738.

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Nov 28 2021