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A342439
Let S(n,k) denote the set of primes < 10^n which are the sum of k consecutive primes, and let K = maximum k >= 2 such that S(n,k) is nonempty; then a(n) = max S(n,K).
5
5, 41, 953, 9521, 92951, 997651, 9964597, 99819619, 999715711, 9999419621, 99987684473, 999973156643, 9999946325147, 99999863884699, 999999149973119, 9999994503821977, 99999999469565483, 999999988375776737, 9999999776402081701
OFFSET
1,1
COMMENTS
Inspired by the 50th problem of Project Euler (see link).
There must be at least two consecutive primes in the sum.
The corresponding number K of consecutive primes to get this largest prime is A342440(n) and the first prime of these A342440(n) consecutive primes is A342453(n).
It can happen that the sums of K = A342440(n) consecutive primes give two (or more) distinct n-digit primes. In that case, a(n) is the greatest of these primes. Martin Ehrenstein proved that there are only two such cases when 1 <= n <= 19, for n = 7 and n = 15 (see corresponding examples).
Solutions and Python program are proposed in Dreamshire and Archive.today links. - Daniel Suteu, Mar 12 2021
EXAMPLE
a(1) = 5 = 2+3.
a(2) = 41 = 2 + 3 + 5 + 7 + 11 + 13; note that 97 = 29 + 31 + 37 is prime, sum of 3 consecutive primes, but 41 is obtained by adding 6 consecutive primes, so, 97 is not a term.
A342440(7) = 1587, and there exist two 7-digit primes that are sum of 1587 consecutive primes; as 9951191 = 5+...+13399 < 9964597 = 7+...+13411 hence a(7) = 9964597.
A342440(15) = 10695879 , and there exist two 15-digit primes that are sum of 10695879 consecutive primes; as 999998764608469 = 7+...+192682309 < 999999149973119 = 13+...+192682337, hence a(15) = 999999149973119.
KEYWORD
nonn,base,more
AUTHOR
Bernard Schott, Mar 12 2021
EXTENSIONS
Name improved by N. J. A. Sloane, Mar 12 2021
a(4)-a(17) from Daniel Suteu, Mar 12 2021
a(18)-a(19) from Martin Ehrenstein, Mar 13 2021
a(7) and a(15) corrected by Martin Ehrenstein, Mar 15 2021
STATUS
approved