A366032 Difference d between the least odd integer that would disprove Gilbreath's conjecture and prime(n).

2, 4, 2, 10, 6, 10, 6, 6, 12, 16, 16, 16, 8, 12, 10, 30, 20, 26, 34, 20, 28, 18, 26, 30, 36, 24, 28, 26, 30, 88, 54, 68, 44, 64, 46, 46, 48, 40, 36, 52, 32, 64, 46, 66, 36, 66, 94, 72, 66, 76, 60, 54, 56, 70, 58, 66, 74, 72, 76, 56, 84, 80, 88, 70, 92, 104, 78, 86, 100, 84, 66, 86, 84, 86, 96

Offset: 3

Author

Giorgos Kalogeropoulos, Sep 27 2023

Keywords

• nonn

Comments

In Gilbreath's conjecture the leading row lists the primes. In this sequence we take as leading row the first n-1 primes joined with the least odd integer k that disproves Gilbreath's conjecture instead of prime(n).

The terms of the sequence are the difference of this hypothetical number k and prime(n).

k is always greater than prime(n-1). The first 1000 terms show that k is greater than prime(n).

Although the first 1000 terms are positive, in theory a term can be negative: prime(n-1) < k < prime(n).

If we find a term that is zero then k = prime(n) and that would disprove the conjecture.

Examples

			The first term of the sequence is a(3) = 2 (offset is 3)
We start with the first 2 primes and instead of the third prime, we choose k=7.
  2,3  -->  2,3,7  instead of  2,3,5
  1         1,4                1,2
            3                  1
.
k=7 is the least odd integer that disproves the conjecture. So, a(3) = k-prime(3) = 7 - 5 = 2.
.
  2,3,5,7,11  -->  2,3,5,7,11,23  instead of  2,3,5,7,11,13
  1,2,2,4          1,2,2,4,12                 1,2,2,4,2
  1,0,2            1,0,2,8                    1,0,2,2
  1,2              1,2,6                      1,2,0
  1                1,4                        1,2
                   3                          1
k=23 is the least odd integer that disproves the conjecture. So, a(6) = k-prime(6) = 23 - 13 = 10.
		

Crossrefs

Cf. A036262, A363003.

Programs

Mathematica

Table[(k=Prime@n;While[Nest[Abs@*Differences,Join[Prime@Range@n,{k}],n]=={1},k=k+2];k)-NextPrime@Prime@n,{n,2,100}]

PARI

isok(v) = my(nb=#v); for (i=1, nb-1, v = vector(#v-1, k, abs(v[k+1]-v[k]));); v[1] == 1;
a(n) = my(v = primes(n-1), k=prime(n)); while (isok(concat(v, k)), k+=2); k - prime(n); \\ Michel Marcus, Sep 28 2023