Smallest k such that both of the consecutive Woodall numbers A003261(k) and A003261(k+1) are divisible by A014662(n), the n-th prime p with even order of 2 mod p.
A287145
Smallest k such that both of the consecutive Woodall numbers A003261(k) and A003261(k+1) are divisible by A014662(n), the n-th prime p with even order of 2 mod p.
Terms
- a(0) =4a(1) =13a(2) =64a(3) =89a(4) =83a(5) =188a(6) =433a(7) =701a(8) =449a(9) =342a(10) =1429a(11) =1768a(12) =1889a(13) =2276a(14) =3484a(15) =2423a(16) =5149a(17) =5776a(18) =2069a(19) =1693a(20) =8644a(21) =4793a(22) =9728a(23) =11173a(24) =4237a(25) =13364a(26) =15049a(27) =16108a(28) =16469a(29) =9455
External references
- oeis: A287145