A385556 Period of {binomial(N,4) mod n: N in Z}.

1, 8, 9, 16, 5, 72, 7, 32, 27, 40, 11, 144, 13, 56, 45, 64, 17, 216, 19, 80, 63, 88, 23, 288, 25, 104, 81, 112, 29, 360, 31, 128, 99, 136, 35, 432, 37, 152, 117, 160, 41, 504, 43, 176, 135, 184, 47, 576, 49, 200, 153, 208, 53, 648, 55, 224, 171, 232, 59, 720

Offset: 1

Author

Jianing Song, Jul 03 2025

Keywords

• nonn
• easy
• mult

Examples

			For N == 0, 1, ..., 31 (mod 32), binomial(N,4) == {0, 0, 0, 0, 1, 5, 7, 3, 6, 6, 2, 2, 7, 3, 1, 5, 4, 4, 4, 4, 5, 1, 3, 7, 2, 2, 6, 6, 3, 7, 5, 1} (mod 8).
		

Links

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

Crossrefs

Row n = 4 of A349593. A022998, A385555, A385557, A385558, A385559, and A385560 are respectively rows n = 2, 3, 5-6, 7, 8, and 9-10.

Cf. A000034, A089128.

Programs

Mathematica

A385556[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 4]], {p, FactorInteger[n][[All, 1]]}]];
Array[A385556, 100] (* Paolo Xausa, Jul 07 2025 *)
a[n_] := n * GCD[n, 6] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)

PARI

a(n, {choices=4}) = my(r=1, f=factor(n)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(choices, p)+e)); return(r)

Formula

Multiplicative with a(2^e) = 2^(e+2), a(3^e) = 3^(e+1), and a(p^e) = p^e for primes p >= 5.

From Amiram Eldar, Jul 07 2025: (Start)

a(n) = n * gcd(6, n) * (2 - (n mod 2)) = n * A089128(n) * A000034(n-1).

Dirichlet g.f.: zeta(s-1) * (1 + 3/2*(s-1)) * (1 + 2/3*(s-1)).

Sum_{k=1..n} a(k) ~ (25/12) * n^2. (End)