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%I #35 Jul 13 2024 09:13:15
%S 1,17,13,2383,37,3,3391,185,129,419,95,139,7,7373,497,21,89,27,319,7,
%T 23,191,277,25,33635,137,1957,347,879,889,47,57,411,263,63,57,63,143,
%U 62561,363,1679,861,285735,1017,545,2605,913,1873,735,206349,817,407,485,49,7605,179817
%N Least odd k such that C(2k, k) == 1 (mod A007775(n)), or 0 if no such k exists.
%C 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.
%C 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
%H M. F. Hasler and Max Alekseyev, <a href="/A373469/b373469.txt">Table of n, a(n) for n = 1..266</a>
%o (PARI) /* helper function: compute C(n,k) mod prime p */
%o 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)))}
%o 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}
%o apply( {A373469(n, m=A007775(n), f=factor(m))=!f || forstep(k=3, oo, 2, is1(k,f) && return(k))}, [1..50])
%Y 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).
%Y Cf. A030979.
%K nonn,hard
%O 1,2
%A _M. F. Hasler_, Jul 12 2024
%E a(43)-a(56) from _Max Alekseyev_, Jul 12 2024
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