A393426 Smallest k multiple of n such that the Pisano period A001175(k) divides k.

1, 24, 24, 24, 120, 24, 336, 24, 72, 120, 1320, 24, 4368, 336, 120, 48, 1224, 72, 1368, 120, 336, 1320, 1104, 24, 600, 4368, 216, 336, 9744, 120, 3720, 96, 1320, 1224, 1680, 72, 50616, 1368, 4368, 120, 4920, 336, 56760, 1320, 360, 1104, 4512, 48, 2352, 600, 1224, 4368, 11448

Offset

1,2

Comments

Let P = A001175.

a(n) exists for all n: we can prove that P(k) divides k for k = 4*lcm(1,2,...,n), n >= 3. Write k = 2^(e_1)*3^(e_2)*5^(e_3)*...*(p_i)^(e_i), then:

- For p_i != 2 or 5, P((p_i)^(e_i)) divides 2 * (p_i)^(e_i-1) * (p_i-Legendre(p_i,5)) = 4 * (p_i)^(e_i-1) * (p_i-Legendre(p_i,5))/2. Since both (p_i)^(e_i-1) and (p_i-Legendre(p_i,5))/2 divide lcm(1,2,...,n), we have that P((p_i)^(e_i)) divides k;

- For p_i = 5, P(5^(e_3)) = 4*5^(e_3) divides k;

- For p_i = 2, P(2^(e_1)) = 3*2^(e_1-1) divides k.

Note that if n | k and P(k) | k, then P(n) | P(k) | k, so lcm(n,P(n)) | k. As a result, to obtain a(n), we start with n and apply continuously the iteration n -> lcm(n,P(n)). Here the iteration must halt because n dividing a(n) implies lcm(n,P(n)) dividing lcm(a(n),P(a(n))) = a(n). We will finally reach some fixed point k0 dividing a(n) such that n divides k0 and P(k0) divides k0, and then that k0 must be a(n) since a(n) is the smallest number with the property that n divides k and P(k) divides k.

By comments in A317971, a(n) is divisible by 24 for n > 1, so a(n) = a(2*n) = a(3*n) = a(4*n) = a(6*n) = a(8*n) = a(12*n) = a(24*n) for n > 1.

Links

Jianing Song, Table of n, a(n) for n = 1..10000

Example

a(13) = 4368 is the smallest k divisible by 13 such that A001175(k) divides k. Indeed, we have A001175(4368) = 336.
a(73) = 3694968 is the smallest k divisible by 73 such that A001175(k) divides k. Indeed, we have A001175(3694968) = 50616.

Prog

(PARI) \\ Pisano period of k
P(k) = {
\\ find maximum possible order
my(r=1, f=factor(k));
for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
    if(p%5==1||p%5==4, r=lcm(r, (p-1)*p^(e-1)));
    if(p%5==2||p%5==3, r=lcm(r, 2*(p+1)*p^(e-1)));
    if(p==5, r=lcm(r, 4*5^e));
);
my(ld=divisors(r), d); for(i=1, #ld, d=ld[i]; if(Mod([1, 1; 1, 0], k)^d==[1, 0; 0, 1], return(d)));
}
a(n) = my(s=P(n)); while(n%s!=0, n=lcm(n, s); s=P(n)); n

Crossrefs

Cf. A001175, A235249 (fixed point of n->A001175(n)), A317971.

Sequence in context: A362116 A217140 A235249 * A115028 A099543 A188311

Adjacent sequences: A393423 A393424 A393425 * A393427 A393428 A393429

Keyword

nonn

Author

Jianing Song, Apr 26 2026

Status

approved