A385560 Period of {binomial(N,9) mod n: N in Z}. Also, period of {binomial(N,10) mod n: N in Z}.
1, 16, 27, 32, 25, 432, 49, 64, 81, 400, 11, 864, 13, 784, 675, 128, 17, 1296, 19, 800, 1323, 176, 23, 1728, 125, 208, 243, 1568, 29, 10800, 31, 256, 297, 272, 1225, 2592, 37, 304, 351, 1600, 41, 21168, 43, 352, 2025, 368, 47, 3456, 343, 2000, 459, 416, 53, 3888, 275, 3136, 513, 464, 59, 21600
Offset: 1
Examples
For N == 0, 1, ..., 80 (mod 81), binomial(N,9) == {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 4, 4, 4, 1, 1, 1, 2, 2, 2, 8, 8, 8, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 7, 7, 7, 4, 4, 4, 5, 5, 5, 2, 2, 2, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 1, 1, 1, 7, 7, 7, 8, 8, 8, 5, 5, 5, 8, 8, 8} (mod 9), and binomial(N,10) == {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 7, 2, 6, 7, 8, 0, 2, 4, 6, 5, 4, 3, 5, 7, 0, 3, 6, 0, 3, 6, 0, 3, 6, 0, 4, 8, 3, 1, 8, 6, 1, 5, 0, 5, 1, 6, 8, 1, 3, 8, 4, 0, 6, 3, 0, 6, 3, 0, 6, 3, 0, 7, 5, 3, 4, 5, 6, 4, 2, 0, 8, 7, 6, 2, 7, 3, 2, 1} (mod 9).
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
A385560[n_] := If[n == 1, 1, n*Product[p^Floor[Log[p, 9]], {p, FactorInteger[n][[All, 1]]}]]; Array[A385560, 100] (* Paolo Xausa, Jul 07 2025 *) a[n_] := n * GCD[n, 6] * GCD[n, 210] * (2 - Mod[n, 2]); Array[a, 100] (* Amiram Eldar, Jul 07 2025 *)
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PARI
a(n, {choices=10}) = 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+3), a(3^e) = 3^(e+2), a(5^e) = 5^(e+1), a(7^e) = 7^(e+1), and a(p^e) = p^e for primes p >= 11.
From Amiram Eldar, Jul 07 2025: (Start)
a(n) = n * gcd(6, n) * gcd(210, n) * (2 - (n mod 2)).
Dirichlet g.f.: zeta(s-1) * (1 + 7/2*(s-1)) * (1 + 8/3*(s-1)) * (1 + 4/5*(s-1)) * (1 + 6/7*(s-1)).
Sum_{k=1..n} a(k) ~ (3861/140) * n^2. (End)