A268304 Odd numbers n such that binomial(6*n, 2*n) == -1 (mod 8).

1, 5, 21, 73, 85, 273, 293, 297, 329, 341, 529, 545, 1041, 1057, 1089, 1093, 1105, 1173, 1189, 1193, 1297, 1317, 1321, 1353, 1365, 2065, 2081, 2113, 2117, 2129, 2177, 2181, 2209, 2577, 2593, 4113, 4129, 4161, 4165, 4177, 4225, 4229, 4257, 4353, 4357, 4373, 4417, 4421, 4433

Offset: 1

Author

Michel Marcus, Jan 31 2016

Keywords

• nonn

Comments

The primes p of this sequence are those that give the even semiprimes of A268303.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000
• Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672. See Table 2 p. 2669.

Crossrefs

Cf. A234839, A268303.

Programs

Mathematica

Select[Range[1, 5000, 2], Mod[Binomial[6 #, 2 #], 8] == 7 &] (* Michael De Vlieger, Feb 07 2016 *)

PARI

isok(n) = (n%2) && Mod(binomial(6*n, 2*n), 8) == Mod(-1, 8);

Python

from _future_ import division
A268304_list, b, m1, m2 = [], 15, [21941965946880, -54854914867200, 49244258396160, -19011472727040, 2933960577120, -126898662960, 771887070, 385943535, 385945560],  [10569646080, -25763512320, 22419210240, -8309145600, 1209116160, -46992960, 415800, 311850, 311850]
for n in range(10**3):
    if b % 8 == 7:
        A268304_list.append(2*n+1)
    b = b*m1[-1]//m2[-1]
    for i in range(8):
        m1[i+1] += m1[i]
        m2[i+1] += m2[i] # Chai Wah Wu, Feb 05 2016