A341759
Primes p such that Euler(p, 1) is an integer multiple of Bernoulli(p + 1, 1).
Original entry on oeis.org
5, 17, 41, 53, 293, 881, 2393, 4373, 5333, 5417, 6173, 7937, 12641, 13121, 14153, 16001, 16253, 18521, 23813, 27701, 37337, 42461, 48761, 50273, 58481, 74897, 82781, 88493, 106433, 113777, 143261, 174761, 195281, 227597, 236681, 249317, 297233, 336041, 341333
Offset: 1
Euler(17, 1) = 29127*Bernoulli(18, 1).
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Select[Select[Range[200000], IntegerQ[(2*(-1 + 2^#1))/#1] &] - 1, PrimeQ] (* Vaclav Kotesovec, Mar 24 2021 *)
Select[Prime[Range[30000]],Divisible[EulerE[#,1],BernoulliB[#+1]]&]//Quiet (* Harvey P. Dale, Nov 06 2022 *)
A337498
a(0) = 0; for n>0, a(n) = the smallest positive integer m not yet in the sequence with property that none of the numbers in row m of A193829 have occurred previously in the sequence.
Original entry on oeis.org
0, 2, 5, 6, 11, 13, 17, 18, 23, 25, 29, 31, 37, 41, 42, 47, 53, 54, 59, 61, 65, 67, 71, 73, 79, 83, 85, 89, 95, 97, 101, 103, 107, 109, 113, 121, 125, 127, 131, 137, 139, 145, 149, 151, 155, 157, 162, 167, 169, 173, 179, 181, 185, 191, 193, 197, 199, 205, 209, 211, 215, 221, 223, 227, 229
Offset: 0
a(1) = 2 as the consecutive divisors of 2 are 1,2 the difference of which is 1, which has not occurred previously in the sequence.
a(2) = 5. The consecutive divisors of 3 are 1,3, the difference of which is 2, but a(1) = 2 so 3 cannot appear. The consecutive divisors of 4 are 1,2,4, the differences of which are 1,2, but a(1) = 2 so 4 cannot appear. The consecutive divisors of 5 are 1,5, the difference of which is 4 and as 4 has not occurred previously in the sequence a(2) = 5.
a(3) = 6 as the consecutive divisors of 6 are 1,2,3,6, the differences of which are 1,1,3, and as neither 1 or 3 has occurred previously a(3) = 6.
a(4) = 11. The divisors of 7 are 1,7 with a difference of 6, the divisors of 8 are 1,2,4,8 with differences 1,2,4, the divisors of 9 are 1,3,9 with differences 2,6, and the divisors of 10 are 1,2,5,10, with differences 1,3,5. All of 6,2,2,5 have all occurred previously in the sequence. The divisors of 11 are 1,11 with a difference of 10, which has not occurred previously so a(4) = 11.
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