A133900 - OEIS

A133900

a(n) = period of the sequence {b(m), m>=0}, defined by b(m):=binomial(m+n,n) mod n.

1, 4, 9, 16, 25, 72, 49, 64, 81, 400, 121, 864, 169, 784, 675, 256, 289, 2592, 361, 1600, 1323, 3872, 529, 3456, 625, 5408, 729, 3136, 841, 324000, 961, 1024, 9801, 18496, 6125, 31104, 1369, 23104, 13689, 32000, 1681, 254016, 1849, 15488, 30375, 33856

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OFFSET

1,2

COMMENTS

This is the analog of the sequence of Pisano periods (A001175) for binomial numbers.

n^2 always divides a(n).

A prime p is a factor of a(n) if and only if it is a factor of n (i.e., a(n) and n have the same prime factors).

If n has N distinct prime factors, then n^(N+2-2d(n)) <= a(n) <= n^(N+1), where d(n) is the sum of the reciprocals of the first n terms of Sylvester's sequence, A000058. This implies that there are distinct bands in the graph. - Hari Rajesh, Oct 12 2025

LINKS

Hari Rajesh, Table of n, a(n) for n = 1..10000 (first 111 terms from Hieronymus Fischer, apart from a(82) and a(84) that were corrected)

FORMULA

a(n) = n^2 if n is a prime or a power of a prime.

a(n) = Product_{i} p_i^(k_i+s_i), where the prime factorisation of n is Product_{i} p_i^s_i for distinct primes p_i, and k_i = floor(log_(p_i)(n)). - Hari Rajesh, Oct 12 2025

EXAMPLE

a(3)=9 since binomial(m+3,3) mod 3, m>=0, is periodic with period length 3^2=9 (see A133883).

a(6)=72 since binomial(m+6,6) mod 6, m>=0, is periodic with period length 4*6^2=72 (see A133886).

PROG

(PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k, 2] += logint(n, f[k, 1])); factorback(f); \\ Michel Marcus, Oct 13 2025

CROSSREFS

Cf. A000040, A001175, A133872-A133880, A133882-A133890, A133910.

Cf. A133620-A133625, A133630, A038509, A133634, A133635, A133636.

Cf. A133905.

Sequence in context: A279096 A374927 A299153 * A143480 A230365 A274963

Adjacent sequences: A133897 A133898 A133899 * A133901 A133902 A133903

KEYWORD

nonn,look

AUTHOR

Hieronymus Fischer, Oct 15 2007, Oct 20 2007

STATUS

approved