OFFSET
1,4
COMMENTS
The algorithm: Take n consecutive primes starting with the a(n)-th prime: {p(a(n)), ..., p(a(n)+n-1)}. Calculate the absolute differences abs(p(i)-p(j)) for all relevant pairs i,j. The number of distinct entries of this n X n difference matrix equals binomial(n,2) = n(n-1)/2, the maximum possible. Also a(n) is the earliest index such that this diversity of differences reaches the maximum possible, binomial(n,2).
The diagonal [{p(i)-p(i)}] of difference matrices consists only of zeros and can be disregarded.
The complete diversity of k-1 consecutive prime differences [as in A079007] is a necessary but not sufficient condition for providing binomial(k,2) distinct entries in the corresponding k X k difference matrix of k consecutive primes. Consecutive prime differences are para-diagonal entries in the difference matrix. So the conditions here are stronger than in A079007.
FORMULA
a(n) = PrimePi(A099640(n)). - Amiram Eldar, Feb 23 2025
EXAMPLE
At n = 1, 2, 3, the maxima are binomial(n,2) = 0, 1, 3 reached at a(n) = 1, 1, 1 respectively.
For n = 7: a(7) = 374, primes = {p(374), ..., p(80)} = {2551, 2557, 2579, 2591, 2593, 2609, 2617}. Building the 7 X 7 matrix of values of abs(p(i)-p(j)), the number of its distinct positive entries equals binomial(7,2)=21, namely: {2, 6, 8, 12, 14, 16, 18, 22, 24, 26, 28, 30, 34, 36, 38, 40, 42, 52, 58, 60, 66}.
For n = 12: a(12) = 2332720, list of 12 primes = {p(n), ..., p(n+11)} = {38238461, ..., 38238737}. 12 X 12 matrix = {abs((p(i)-p(j))}, number of distinct entries = binomial(12,2) = 66, that of {2, 6, 8, ..., 266, 274, 276}.
MATHEMATICA
a[n_] := Module[{k = 1, v = Prime[Range[n]]}, While[CountDistinct@ Flatten@ Abs@ Outer[Plus, v, -v] - 1 != Binomial[n, 2], k++; v = Join[v[[2 ;; -1]], {NextPrime[v[[-1]]]}]]; k]; Array[a, 10] (* Amiram Eldar, Feb 23 2025 *)
PROG
(PARI) s(v) = {my(d = List()); for(i = 1, #v, for(j = 1, i-1, listput(d, abs(v[i] - v[j])))); #Set(d); }
a(n) = {my(k = 1, v = primes(n), t = n*(n-1)/2, j = 1); while(s(v) != t, k++; v[j] = nextprime(1 + if(j==1, v[n], v[j-1])); j++; if(j > n, j -= n)); k; } \\ Amiram Eldar, Feb 23 2025
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Labos Elemer, Oct 05 2004
EXTENSIONS
Edited by Jon E. Schoenfield, Oct 27 2019
a(14)-a(15) from Amiram Eldar, Feb 23 2025
STATUS
approved