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A285388
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a(n) = numerator of ((1/n) * Sum_{k=0..n^2-1} binomial(2k,k)/4^k).
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14
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1, 35, 36465, 300540195, 79006629023595, 331884405207627584403, 22292910726608249789889125025, 11975573020964041433067793888190275875, 411646257111422564507234009694940786177843149765, 56592821660064550728377610673427602421565368547133335525825
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OFFSET
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1,2
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COMMENTS
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Editorial comment: This sequence arose from Ralf Steiner's attempt to prove Legendre's conjecture that there is a prime between N^2 and (N+1)^2 for all N. - N. J. A. Sloane, May 01 2017
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LINKS
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FORMULA
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a(n) is numerator of n*binomial(2 n^2, n^2)/2^(2*n^2 - 1). - Ralf Steiner, Apr 26 2017
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MATHEMATICA
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Table[Numerator[Sum[Binomial[2k, k]/4^k, {k, 0, n^2-1}]/n], {n, 1, 10}]
Numerator[Table[2^(1-2 n^2) n Binomial[2 n^2, n^2], {n, 1, 10}]] (* Ralf Steiner, Apr 22 2017 *)
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PROG
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(PARI) a(n) = m=n*binomial(2*n^2, n^2); m>>valuation(m, 2) \\ David A. Corneth, Apr 27 2017
(Python)
from sympy import binomial, Integer
def a(n): return (Integer(2)**(1 - 2*n**2)*n*binomial(2*n**2, n**2)).numerator() # Indranil Ghosh, Apr 27 2017
(Magma) [Numerator( n*(n^2+1)*Catalan(n^2)/2^(2*n^2-1) ): n in [1..21]]; // G. C. Greubel, Dec 11 2021
(Sage) [numerator( n*(n^2+1)*catalan_number(n^2)/2^(2*n^2-1) ) for n in (1..20)] # G. C. Greubel, Dec 11 2021
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CROSSREFS
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Cf. A000079, A000265, A056220, A060757, A201555, A285389 (denominators), A285406, A280655 (similar), A190732 (2/sqrt(Pi)), A285738 (greatest prime factor), A285717, A285730, A285786, A286264, A000290 (n^2), A056220 (2*n^2 -1), A286127 (sum a(n-1)/a(n)).
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KEYWORD
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nonn,frac
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AUTHOR
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EXTENSIONS
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Edited (including the removal of the author's claim that this leads to a proof of the Legendre conjecture) by N. J. A. Sloane, May 01 2017
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STATUS
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approved
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