OFFSET
1,1
COMMENTS
Consecutive previous primes of the selected prime are ignored even though the may also have an identical count of digits.
LINKS
David A. Corneth, PARI program
EXAMPLE
a(2) = 17, because the two primes in the sequence starting at — namely [17, 19] — each contain the same number of even digits, and no earlier prime sequence meets this criterion.
n In [a(n), ...]
1 In [2], each of the 1 number contain 1 even digit.
2 In [17, 19], each of the 2 numbers contains 0 even digit.
3 In [13, 17, 19], each of the 3 numbers contains 0 even digit.
4 In [11, 13, 17, 19], each of the 4 numbers contains 0 even digit.
5 In [7, 11, 13, 17, 19], each of the 5 numbers contains 0 even digit.
6 In [5, 7, 11, 13, 17, 19], each of the 6 numbers contains 0 even digit.
7 In [3, 5, 7, 11, 13, 17, 19], each of the 7 numbers contains 0 even digit.
8 In [491, 499, 503, 509, 521, 523, 541, 547], each of the 8 numbers contains 1 even digit.
9 In [14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each of the 9 numbers contains 2 even digits.
10 In [14293, 14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each of the 10 numbers contains 2 even digits.
11 In [157259, 157271, 157273, 157277, 157279, 157291, 157303, 157307, 157321, 157327, 157349], each of the 11 numbers contains 1 even digit.
12 In [157253, 157259, 157271, 157273, 157277, 157279, 157291, 157303, 157307, 157321, 157327, 157349], each of the 12 numbers contains 1 even digit.
PROG
(PARI) \\ See Corneth link
(PARI)
card(p)={my(c=0, u=digits(p), n=sum(i=1, #u, u[i]%2==0), even=n); while(even==n, p=nextprime(p+1); u=digits(p); even=sum(i=1, #u, u[i]%2==0); c++); c}
data(pp=10^9)={my(u=vector(30), r=0); forprime(p=2, pp, my(n=card(p)); if(u[n]==0, u[n]=p; if(n>r, r=n))); u[1..r]}
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Jean-Marc Rebert, Jul 23 2025
EXTENSIONS
a(19)-a(20) from David A. Corneth, Jul 23 2025
a(21)-a(26) from Jean-Marc Rebert, Jul 24 2025
a(27)-a(30) from Giovanni Resta, Jul 24 2025
STATUS
approved