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A376336
Least positive integer m such that tau(m) is divisible by n, where the Ramanujan tau function is given by A000594.
1
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
1,2
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.
EXAMPLE
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.
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]
(* Alternative: *)
a[n_] := SelectFirst[Range[1, 30000], Divisible[RamanujanTau[#], n] &]; Array[a, 1000] (* Peter Luschny, Dec 22 2024 *)
PROG
(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
(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
CROSSREFS
Cf. A000594.
Sequence in context: A130155 A350403 A113516 * A226525 A120642 A216624
KEYWORD
nonn,changed
AUTHOR
Zhi-Wei Sun, Dec 22 2024
STATUS
approved