%I #33 Oct 22 2023 15:58:25
%S 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,
%T 24,28,26,30,88,54,68,44,64,46,46,48,40,36,52,32,64,46,66,36,66,94,72,
%U 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
%N Difference d between the least odd integer that would disprove Gilbreath's conjecture and prime(n).
%C 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).
%C The terms of the sequence are the difference of this hypothetical number k and prime(n).
%C k is always greater than prime(n-1). The first 1000 terms show that k is greater than prime(n).
%C Although the first 1000 terms are positive, in theory a term can be negative: prime(n-1) < k < prime(n).
%C If we find a term that is zero then k = prime(n) and that would disprove the conjecture.
%e The first term of the sequence is a(3) = 2 (offset is 3)
%e We start with the first 2 primes and instead of the third prime, we choose k=7.
%e 2,3 --> 2,3,7 instead of 2,3,5
%e 1 1,4 1,2
%e 3 1
%e .
%e k=7 is the least odd integer that disproves the conjecture. So, a(3) = k-prime(3) = 7 - 5 = 2.
%e .
%e 2,3,5,7,11 --> 2,3,5,7,11,23 instead of 2,3,5,7,11,13
%e 1,2,2,4 1,2,2,4,12 1,2,2,4,2
%e 1,0,2 1,0,2,8 1,0,2,2
%e 1,2 1,2,6 1,2,0
%e 1 1,4 1,2
%e 3 1
%e k=23 is the least odd integer that disproves the conjecture. So, a(6) = k-prime(6) = 23 - 13 = 10.
%t 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}]
%o (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;
%o 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
%Y Cf. A036262, A363003.
%K nonn
%O 3,1
%A _Giorgos Kalogeropoulos_, Sep 27 2023