29030400

Appears in sequences

• Dwork-Kontsevich sequence evaluated at 2*n.at n=7A007757
• Denominators of coefficients in expansion of sqrt(sin(x)/x) (even powers only).at n=4A008992
• a(n) = n!/lcm{1,2,...,n} = (n-1)!/lcm{C(n-1,0), C(n-1,1), ..., C(n-1,n-1)}.at n=15A025527
• a(n) = n!/lcm{1,2,...,n} = (n-1)!/lcm{C(n-1,0), C(n-1,1), ..., C(n-1,n-1)}.at n=16A025527
• Numbers m such that uphi(sigma(m)) = 2m, where the unitary phi function (A047994) is defined by: if x = p1^r1*p2^r2*p3^r3*... then uphi(x) = (p1^r1 - 1)*(p2^r2 - 1)*(p3^r3 - 1)*...at n=27A030165
• E.g.f. x^3/(1-x)^2.at n=10A052571
• E.g.f. 1/(1-x-x^5).at n=10A052632
• a(n) = gcd(n!, n!*(1 + 1/2 + 1/3 + ... + 1/n)).at n=15A056612
• a(n) = gcd(n!, n!*(1 + 1/2 + 1/3 + ... + 1/n)).at n=16A056612
• Product of the composite numbers between n and 2n (both inclusive).at n=7A073840
• Denominator of the coefficient of x^n in log(-log(1-x)/x).at n=7A075267
• Denominator of Cotesian number C(n,1).at n=10A100644
• a(1) = 10, a(n) = a(n-1) times the number of digits in a(n-1).at n=12A110804
• The first four terms of the sequence are doubled, then those numbers are tripled and then those numbers are quadrupled, etc.at n=39A115425
• Amicable quadruples: the values of sigma corresponding to A036471-A036474.at n=9A116148
• Amicable quadruples: the values of sigma corresponding to A036471-A036474.at n=10A116148
• a(n) = A001783(n)/A038610(n).at n=16A128247
• For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(n).at n=15A131657
• For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(n).at n=16A131657
• For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum__{k=j+1..j*n} (1/k)), and let u(n) be the largest integer for which exp(B_n(z)/(u(n)*A_n(z))) has integral coefficients. The sequence is u(n).at n=15A131658