415215

Appears in sequences

• Numbers k such that 11^k == -1 (mod k-1).at n=20A055694
• Sum of a(n) terms of 1/k^(6/7) first exceeds n.at n=38A056183
• a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 4*a(n-4) + 4*a(n-5) - 4*a(n-6) + 4*a(n-7) - 4*a(n-8) + 4*a(n-9) - 3*a(n-10) + 2*a(n-11) - 3*a(n-12) + 2*a(n-13) for n >= 16, with initial values as shown.at n=31A288511