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A014137
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Partial sums of Catalan numbers (A000108).
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276
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1, 2, 4, 9, 23, 65, 197, 626, 2056, 6918, 23714, 82500, 290512, 1033412, 3707852, 13402697, 48760367, 178405157, 656043857, 2423307047, 8987427467, 33453694487, 124936258127, 467995871777, 1757900019101
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| This is also the result of applying the transformation on generating functions A -> 1/((1-x)*(1-x*A)) to the g.f. for the Catalan numbers.
p divides a(p)-3 for prime p=3 and p=7,13,19,31,37,43..=A002476 Primes of form 6n + 1. p^2 divides a(p^2)-3 for prime p>3. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006
Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 Primes congruent to {0, 2} mod 3; and A045309 Primes p such that x^3 = n (integer) has only one solution mod p. Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk (alex(AT)kolmogorov.com), Feb 23 2007
For p prime >=5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan (callan(AT)stat.wisc.edu), Nov 29 2007
Hankel transform is A010892(n+1). [From Paul Barry (pbarry(AT)wit.ie), Apr 24 2009]
Equals INVERTi transform of A000245: (1, 3, 9, 28,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 15 2009]
The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367, see A121852. [From Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 10 2010]
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REFERENCES
| N. S. S. Gu, N. Y. Li and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
I. Pak, Partition identities and geometric bijections. Proc. Amer. Math. Soc. 132 (2004), 3457-3462.
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..200
W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra Appl. (2011), in press
Guo-Niu Han, Enumeration of Standard Puzzles
Hao Pan and Zhi-Wei Sun, A combinatorial identity with application to Catalan numbers
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FORMULA
| G.f.: (1-(1-4*x)^(1/2))/(2*x*(1-x)).
Sum_{i=1..n} c(i) = Sum_{i=1..n} C(2*i-2, i-1)/i = 1/(n-1)! * [ n^(n-2) +C(n, 2)*n^C(n-3, 1) +{8*C(n-4, 0) +19*C(n-4, 1) +24*C(n-4, 2) +14*C(n-4, 3) +3*C(n-4, 4)}*n^(n-4) +{18*C(n-5, 0) +82*C(n-5, 1) +229*C(n-5, 2) +323*C(n-5, 3) +244*C(n-5, 4) +95*C(n-5, 5) +15*C(n-5, 6)}*n^(n-5) +... +C(n-3, 0)*(n-1)! ] (where c() = Catalan numbers A000108). - Andre F. Labossiere (boronali(AT)laposte.net), May 17 2004
a(n) = Sum[(2k)!/(k!)^2/(k+1),{k,0,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006
Conjecture: (n+1)*a(n)+(1-5n)*a(n-1) +2*(2n-1)*a(n-2)=0. - R. J. Mathar, Dec 14 2011
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MATHEMATICA
| Table[Sum[(2k)!/(k!)^2/(k+1), {k, 0, n}], {n, 1, 30}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006
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PROG
| (PARI) Vec((1-(1-4*x)^(1/2))/(2*x*(1-x))+O(x^99)) \\ Charles R Greathouse IV, Feb 11 2011
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CROSSREFS
| a(n) = A014138(n-1)+1.
Cf. A000108, A094638, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A002476, A045309, A128287.
Cf. A000245 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 15 2009]
Sequence in context: A000083 A092668 A164039 * A007476 A202552 A129698
Adjacent sequences: A014134 A014135 A014136 * A014138 A014139 A014140
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KEYWORD
| nonn,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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