%I #40 Sep 11 2024 23:00:40
%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 highly possible that a(n) = 0 for n with m = A007775(n) divisible by three or more distinct primes, in which case values of k such that C(2k,k) coprime to m, let alone C(2k,k) == 1 (mod m), are very sparse and possibly finite. See A030979 for a similar problem. - _Max Alekseyev_, Jul 14 2024
%C Examples for moduli that have 3 distinct prime factors >5: a(603) = 57 associated with modulus A007775(603) = 2261 = 7*17*19. a(4333) = 23 associated with modulus A007775(4333) = 16247 = 7*11*211. a(6621) = 1709 associated with 11*37*61. a(6797)=19999 assocated with 7*11*331. - _R. J. Mathar_, Aug 09 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