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A137821
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Numbers k such that Sum_{j=1..2k} Catalan(j) == 0 (mod 3).
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7
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1, 4, 6, 13, 15, 18, 19, 40, 42, 45, 46, 54, 55, 58, 60, 121, 123, 126, 127, 135, 136, 139, 141, 162, 163, 166, 168, 175, 177, 180, 181, 364, 366, 369, 370, 378, 379, 382, 384, 405, 406, 409, 411, 418, 420, 423, 424, 486, 487, 490, 492, 499, 501, 504, 505
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OFFSET
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1,2
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COMMENTS
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It would be natural to prepend an initial term a(1)=0 (for which the sum is to be considered empty, thus zero), but we omit it to avoid confusion w.r.t. indices of A107755.
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LINKS
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FORMULA
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a(2^j) = 2 a(2^j-1) + 1 (resp. +2) for j even (resp. odd).
Sum_{k=1..2n} Catalan(k) = Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1), thus:
{ a(m) } = { n>0 | Sum_{k=1..n} Catalan(2k-1) * (10k-1)/(2k+1) == 0 (mod 3) }.
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PROG
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(PARI) n=0; A137821=vector(499, i, { if( bitand(i, i-1), while(n++ & s+=binomial(4*n-2, 2*n-1)/(2*n)*(10*n-1)/(2*n+1), ), s=Mod(0, 3); n=2*n+1+log(i+.5)\log(2)%2 ); n})
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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