OFFSET
1,3
COMMENTS
Suggested by Gilbreath's conjecture (see A036262).
The sequence of primes considered in Gilbreath's conjecture is just one possibility out of all the sequences that can be generated starting from (2,3,...,prime(n)).
Conjecture: Prime(n+1) is always between prime(n)+2 and prime(n)+2*a(n).
LINKS
Eric Weisstein's World of Mathematics, Gilbreath's Conjecture
FORMULA
Given the sequence of primes S = (2,3,5,...,prime(n)) and the triangle of the forward absolute differences of S, let d(r) be the last rightmost absolute difference of row r, for r>1. Then a(n) = (1+ Sum_{r>1} d(r)) / 2.
EXAMPLE
a(1) = 1, because considering the sequence (2,k), k=3 is the only solution.
a(2) = 1, because considering the sequence (2,3,k), k=5 is the only solution.
To calculate a(3), consider the sequence (2,3,5) and calculate the triangle of absolute differences:
2,3,5
1,2
1
Consider the rightmost diagonal, which contains the absolute differences (2,1), then a(3) is given by a(3) = [(2+1) + 1]/2 = 2.
That is, a(3) is the result of summing the two differences contained in the rightmost diagonal (2,1). To this sum we add 1 and divide the total by 2.
So there are 2 integers k > prime(3) = 5 -- namely, 7,9 -- such that the forward absolute differences of (2,3,5,k) start with 1.
To be explicit, there are two triangles of absolute differences resulting from (2,3,5,k), namely
2 3 5 7
1 2 2
1 0
1
and
2 3 5 9
1 2 4
1 2
1
a(24) = 16 because there are 16 integers k > prime(25) = 97 -- namely, 99,101,...,129 -- such that all rows (after the first row) of the triangle of absolute differences of (2,3,5,7,11,13,17,...,97,k) start with 1.
PROG
(Python)
# run the function find_M(n), where n is the last prime number of your list of consecutive primes starting from 2
from sympy import isprime
def primes_less_N(n):
primes = []
for i in range(2, n+1):
if(isprime(i)==True):
primes.append(i)
return primes
def find_M(n):
primes = primes_less_N(n)
l = len(primes)
count = 1
if count == 1:
a = [abs(x - primes[i - 1]) for i, x in enumerate(primes)][1:]
count += 1
list_f = []
if count > 1:
while count <l+1:
list_f.append(a[len(a)-1])
b = [abs(x - a[i-1]) for i, x in enumerate(a)][1:]
a=b
count +=1
return (sum(list_f)+1)/2
(PARI) diffs(v) = {while (#v != 1, v = vector(#v-1, k, abs(v[k+1] - v[k])); ); v[1]; }
a(n) = {my(v = primes(n), m = vecmax(v)+1, nb = 0); if (!(m%2), m++); forstep (k=m, oo, 2, if (diffs(concat(v, k)) == 1, nb++, if (nb, break)); ); nb; } \\ Michel Marcus, Dec 03 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Fabio Mercurio, Nov 27 2019
STATUS
approved