A094639 Partial sums of squares of Catalan numbers (A000108).

1, 2, 6, 31, 227, 1991, 19415, 203456, 2248356, 25887400, 307993016, 3763786812, 47032778956, 598933188956, 7751562502556, 101741582076581, 1351906409905481, 18159677984049581, 246298405721739581

Offset: 0

Author

André F. Labossière, May 27 2004

Keywords

• easy
• nonn

Comments

Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are A000108(2) = 2 and A000108(3) = 5. Franklin T. Adams-Watters showed that the only semiprime Catalan number is A000108(4) = 14. The subsequence of primes in the partial sum of squares of Catalan numbers begins: 2, 31, 227, 101741582076581. [Jonathan Vos Post, May 27 2010]

Conjecture: For any positive integer n, the polynomial P_n(x) = sum_{k = 0}^n(C_k)^2*x^k (with C_k = binomial(2k, k)/(k+1)) is irreducible over the field of rational numbers. [Zhi-Wei Sun, Mar 23 2013]

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..838
• Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
• Joel E. Cohen, Variance Functions of Asymptotically Exponentially Increasing Integer Sequences Go Beyond Taylor's Law, J. Int. Seq., Vol. 25 (2022), Article 22.9.3.

Crossrefs

Cf. A000108, A094638, A014137, A001246, A033536, A000984, A006134, A082894, A002897, A079727.

Programs

Mathematica

Accumulate[CatalanNumber[Range[0,20]]^2] (* Harvey P. Dale, May 01 2011 *)

Formula

a(n) = Sum_{k=0..n} ((2k)!/(k!)^2/(k+1))^2. - Alexander Adamchuk, Feb 16 2008

Sum_{i=1..n} [c(i)]^2 = Sum_{i=1..n} [C(2*i-2, i-1)/i]^2 = (1/(n-1)!)^2 * [ n^C(2*n-4, 1) + {2*C(n-1, 2)}*n^(2*n-5) + {C(n-2, 0) + 4*C(n-2, 1) + 13*C(n-2, 2) + 22*C(n-2, 3) + 12*C(n-2, 4)}*n^C(2*n-6, 1) + {12*C(n-3, 1) + 152*C(n-3, 2) + 458*C(n-3, 3) + 640*C(n-3, 4) + 440*C(n-3, 5) + 120*C(n-3, 6)}*n^(2*n-7) + {40*C(n-4, 0) + 313*C(n-4, 1) + 2332*C(n-4, 2) + 9536*C(n-4, 3) + 21409*C(n-4, 4) + 28068*C(n-4, 5) + 21700*C(n-4, 6) + 9240*C(n-4, 7) + 1680*C(n-4, 8) + ... + C(n-3, 0)*((n-1)!)^2 ].

Recurrence: (n+1)^2*a(n) = (17*n^2 - 14*n + 5)*a(n-1) - 4*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Jul 01 2016

a(n) ~ 2^(4*n+4) /(15*Pi*n^3). - Vaclav Kotesovec, Jul 01 2016