A066895 -id:A066895 - cp's OEIS frontend

A067185 Numbers k such that prime(k+1) + prime(k) == 1 (mod k).

1, 7, 17, 109, 281, 1669, 6592663, 16899113, 16899145, 16899295, 749973611, 5067034877, 5067034925, 34487825911, 1626582230375

Offset: 1

Benoit Cloitre, Feb 19 2002

Cf. A066895 (== 0 instead of 1).

a = b = 0; Do[b = Prime[n + 1]; If[Mod[a + b, n] == 1, Print[n]]; a = b, {n, 1, 10^9}]

isok(k) = (prime(k+1) + prime(k)) % k == 1; \\ Michel Marcus, Feb 17 2021

def A067185(max):
    terms = []
    p = 2
    for n in range(1, max + 1):
        q = next_prime(p)
        if not (p + q - 1) % n:
            terms.append(n)
        p = q
    return terms
# Eric M. Schmidt, Feb 05 2013

Edited and extended by Robert G. Wilson v, Feb 19 2002
Extended by David W. Wilson, Feb 20 2002
a(11)-a(14) from Donovan Johnson, Mar 10 2010
First term from Eric M. Schmidt, Feb 05 2013

A225939 Numbers k that divide prime(k) + prime(k-1).

Original entry on oeis.org

4, 6, 13, 14, 15, 74, 190, 688, 690, 6456, 40082, 251735, 251736, 251738, 399916, 637325, 637326, 637342, 637343, 2582372, 2582434, 4124456, 4124458, 6592686, 10553425, 10553433, 10553818, 27067038, 27067053, 43435902, 69709872, 69709877, 69709945, 69709954, 179992917

Offset: 1

Alex Ratushnyak, May 21 2013

nonn

			prime(3) + prime(4) = 5+7 = 12, because 12 is divisible by 4, the latter is in the sequence.
prime(5) + prime(6) = 11+13 = 24, because 24 is divisible by 6, the latter is in the sequence.

Giovanni Resta, Table of n, a(n) for n = 1..58 (terms < 1.5*10^12)

Cf. A001043, A023143, A045924, A066895, A066896.

#include 
#define TOP (1ULL<<32)
int main() {
  unsigned long long i, j, n = 1, prev;
  char *c = (char*)malloc(TOP/2);
  memset(c, 0, TOP/2);
  for (prev = 2, i = 3; i < TOP; i += 2)
    if (c[i>>1]==0) {
      if ((i+prev) % ++n == 0)  printf("%llu, ", n);
      for (prev = i, j = i*i>>1; j < TOP/2; j += i)  c[j] = 1;
    }
  return 0;
}

Select[Range[2,10^4],Divisible[Prime@#+Prime[#-1],#]&] (* Giorgos Kalogeropoulos, Aug 20 2021 *)

def is_a(n): return (nth_prime(n) + nth_prime(n-1)) % n == 0
filter(is_a, (2..1000))  # Peter Luschny, May 22 2013

cp's OEIS Frontend

A067185 Numbers k such that prime(k+1) + prime(k) == 1 (mod k).

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