OFFSET
1,2
COMMENTS
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 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
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
LINKS
M. F. Hasler and Max Alekseyev, Table of n, a(n) for n = 1..266
PROG
(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}
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
M. F. Hasler, Jul 12 2024
EXTENSIONS
a(43)-a(56) from Max Alekseyev, Jul 12 2024
STATUS
approved