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Search: id:A083942
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| 0, 1, 8, 625, 13402696, 19720133460129649, 126747521841153485025455279433135688, 15141471069096667541622192498608408980462133134430650704600552060872705905
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OFFSET
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0,3
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COMMENT
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Conjecture: a(n) = Sum [ CatalanNumber[ k ], {k, 1, 2^n-1} ] = Sum[ (2k)!/(k!(k+1)!), {k, 1, 2^n-1} ]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 10 2007
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LINKS
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Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 10 2007, Table of n, a(n) for n = 0..11
Eric Weisstein, Link to a section of The World of Mathematics, Catalan Number.
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FORMULA
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a(n) = A080300(A002542(n)). [Provided that 2^((2^n)-1)*((2^((2^n)-1))-1) is indeed the formula for A002542.]
Conjecture: a(n) = Sum[ Binomial[ 2k, k ]/(k+1), {k, 1, 2^n-1} ] = Sum [ CatalanNumber[ k ], {k, 1, 2^n-1} ] = Sum[ (2k)!/(k!(k+1)!), {k, 1, 2^n-1} ]. a(n) = A014138(2^n-2) for n>0. - Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 10 2007
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MATHEMATICA
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Table[ Sum[ Binomial[ 2k, k ]/(k+1), {k, 1, 2^n-1} ], {n, 0, 10} ] - Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 10 2007
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CROSSREFS
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a(n) = A057118(A084108(n)).
Cf. A014138 = Partial sums of Catalan numbers (starting 1, 2, 5, ..., cf. A000108). Cf. A000108 = Catalan Numbers.
Sequence in context: A015023 A090923 A080320 this_sequence A027877 A129927 A159621
Adjacent sequences: A083939 A083940 A083941 this_sequence A083943 A083944 A083945
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KEYWORD
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nonn
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AUTHOR
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Antti Karttunen (MyFirstname.MySurname(AT)iki.fi) May 13 2003
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