A374559 Numbers k such that the binomial coefficient C(2*k, k) == 1 (mod 7).

16, 17, 23, 24, 57, 100, 101, 112, 119, 149, 150, 161, 168, 351, 393, 399, 688, 689, 700, 707, 784, 800, 801, 807, 808, 833, 849, 850, 856, 857, 1031, 1032, 1043, 1050, 1127, 1143, 1144, 1150, 1151, 1176, 1192, 1193, 1199, 1200, 2409, 2451, 2457, 2745, 2751, 2793, 2809, 2810, 2816, 2817, 2851

Offset

1,1

Comments

Let t(d,k) be the number of digits d in the base-7 representation of k. Then k is a term if and only if t(4,k) = t(5,k) = t(6,k) = 0, t(1,k) is divisible by 3 and t(2,k) + t(3,k) is even. This follows from Lucas's theorem on binomial coefficients.

If k is a term then so are 7*k, 49*k + 16, 49*k + 17, 49*k + 23, 49*k + 24 and 343*k + 57.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Wikipedia, Lucas's theorem

Example

a(3) = 23 is a term because C(46,23) = 8233430727600 and 8233430727600 == 1 (mod 7).

Maple

filter:= proc(n) local L;
   L:= convert(n, base, 7);
   max(L) <= 3 and numboccur(1, L) mod 3 = 0 and (numboccur(2, L) + numboccur(3, L))::even
end proc:
select(filter, [$1..10000]);

Mathematica

Select[Range[3000], Mod[Binomial[2#, #], 7]==1&] (* Stefano Spezia, Jul 11 2024 *)

Prog

(Python)
from itertools import count, islice
from gmpy2 import digits
def A374559_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n: max(d:=digits(n, 7))<='3' and not (d.count('1')%3 or (d.count('2')^d.count('3'))&1), count(max(startvalue, 1)))
A374559_list = list(islice(A374559_gen(), 10)) # Chai Wah Wu, Jul 13 2024

Crossrefs

Cf. A000984.

Sequence in context: A261344 A007636 A241751 * A138598 A244431 A151977

Adjacent sequences: A374556 A374557 A374558 * A374560 A374561 A374562

Keyword

nonn,look

Author

Robert Israel, Jul 11 2024

Status

approved