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A373469
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Least odd k such that C(2k, k) == 1 (mod A007775(n)), or 0 if no such k exists.
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2
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1, 17, 13, 2383, 37, 3, 3391, 185, 129, 419, 95, 139, 7, 7373, 497, 21, 89, 27, 319, 7, 23, 191, 277, 25, 33635, 137, 1957, 347, 879, 889, 47, 57, 411, 263, 63, 57, 63, 143, 62561, 363, 1679, 861, 285735, 1017, 545, 2605, 913, 1873, 735, 206349, 817, 407, 485, 49, 7605, 179817
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OFFSET
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1,2
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COMMENTS
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A007775 lists the odd numbers not divisible by 3 or 5. It seemed that these are exactly the odd numbers not in A086748 (= odd m such that C(2k,k) == 1 (mod m) has no odd solution k), i.e., the numbers in A086748 would exactly be the odd multiples of 3 and 5, but so far there was no proof or disproof for that. The present sequence gives an explicit proof, if it exists, for each x in A007775, that x is not in A086748.
It is likely that a(267) = 0, since A007775(267) = 1001 = 7 * 11 * 13 and there are likely finitely many k with C(2k,k) coprime to 1001, let alone C(2k,k) == 1 (mod 1001). See A030979 for a similar problem. - Max Alekseyev, Jul 12 2024
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LINKS
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PROG
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(PARI) /* helper function: compute C(n, k) mod prime p */
LucasT(n, k, p)={if(n>=k, my(kp = digits(k, p), np = digits(n, p)[-#kp..-1]); prod(i=1, #kp, binomial(np[i], kp[i]), Mod(1, p)))}
is1(k, f)={for(i=1, matsize(f)[1], LucasT(2*k, k, f[i, 1])==1||return); vecmax(f[, 2])==1 || binomial(2*k, k)%factorback(f)==1}
apply( {A373469(n, m=A007775(n), f=factor(m))=!f || forstep(k=3, oo, 2, is1(k, f) && return(k))}, [1..50])
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CROSSREFS
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Cf. A007775 (odd numbers not divisible by 3 or 5), A086748 (odd m such that C(2k,k)==1 (mod m) has no odd solution k).
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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