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A366032
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Difference d between the least odd integer that would disprove Gilbreath's conjecture and prime(n).
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0
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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
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OFFSET
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3,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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
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k=7 is the least odd integer that disproves the conjecture. So, a(3) = k-prime(3) = 7 - 5 = 2.
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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.
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MATHEMATICA
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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}]
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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