76391

Appears in sequences

• Let p = n-th irregular prime, A000928(n). Then a(n) = smallest value of m such that numerator(Bernoulli(2*m)/(2*m)) / numerator(Bernoulli(2*m)/(2*m*(2*m-1))) equals p.at n=20A092291
• Expansion of (1+x)*(1-x)/(1 - x + x^2 + x^3).at n=38A180735
• G.f. A(x) satisfies: A(x) = 1 + x * A(x/(1 - x)^3) / (1 - x).at n=8A351813
• Numbers k such that prime(k), prime(k+1), prime(k+2), prime(k+3) and prime(k+4) all have the same last digit.at n=16A371390