A376336 Least positive integer m such that tau(m) is divisible by n, where the Ramanujan tau function is given by A000594.

1, 2, 2, 2, 5, 2, 3, 2, 3, 5, 8, 2, 7, 3, 5, 4, 239, 3, 89, 8, 3, 8, 4, 2, 25, 7, 6, 3, 13, 5, 139, 4, 8, 239, 5, 3, 191, 89, 11, 8, 257, 3, 19, 8, 10, 4, 67, 6, 15, 40, 239, 7, 107, 6, 8, 6, 89, 13, 61, 8, 9, 139, 3, 4, 35, 8, 31, 239, 5, 5, 137, 6, 2069, 191, 40, 178, 19, 11, 25, 8, 9, 257, 227, 3, 239, 19, 26, 8, 59, 10, 7, 4, 278, 67, 89, 6, 863, 15, 24, 40

Offset: 1

Author

Zhi-Wei Sun, Dec 22 2024

Keywords

• nonn

Comments

Conjecture: a(n) exists for any positive integer n. Moreover, a(n) <= n^2 for all n > 0.

It seems that for some primes p the value of a(p) is relatively large. For example, 4327 is a prime with a(4327) = 316717, 9133 is a prime with a(9133) = 789977, and 9643 is a prime with a(9643) = 1001401.

Examples

			a(5) = 5 since tau(5) = 4830 is divisible by 5, but none of tau(1) = 1, tau(2) = -24, tau(3) = 252 and tau(4) = -1472 is a multiple of 5.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000594.

Programs

Mathematica

f[n_]:=f[n]=RamanujanTau[n]; L={}; Do[m=1;Label[bb];If[Mod[f[m],n]==0,L=Append[L,m];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,100}];Print[L]
(* Or: *)
a[n_] := SelectFirst[Range[1, 30000], Divisible[RamanujanTau[#], n] &]; Array[a, 1000] (* Peter Luschny, Dec 22 2024 *)

PARI

first(n) = {
	my(todo = [1..n], res = vector(n, i, oo));
	for(i = 1, oo,
		c = ramanujantau(i);
		for(j = 1, #todo,
			if(res[todo[j]] > i && c % todo[j] == 0,
				res[todo[j]] = i;
				todo = setminus(todo, Set(todo[j]));
				if(#todo == 0,
					return(res)
				)
			)
		);
);} \\ David A. Corneth, Dec 23 2024

SageMath

from itertools import count
tau = delta_qexp(30000)  # adjust search length for n > 1000
a = lambda n: next((k for k in count(1) if n.divides(tau[k])))
print([a(n) for n in srange(1, 1001)])  # Peter Luschny, Dec 22 2024