A062802 a(1) = 2; a(n+1) = smallest prime > a(n) whose sum of digits is a(n).

2, 11, 29, 2999

Offset: 1

Author

G. L. Honaker, Jr., Jul 19 2001

Keywords

• nonn
• base

Comments

a(5) = 5*10^333 - 10^332 - 10^174 - 1 =

489999999999999999999999999999999999999999999999999999999999999999999999999\

99999999999999999999999999999999999999999999999999999999999999999999999\

99999999999998999999999999999999999999999999999999999999999999999999999\

99999999999999999999999999999999999999999999999999999999999999999999999\

9999999999999999999999999999999999999999999999

which is too large to include in the DATA field. - Don Reble, Sep 03 2006

Define b(n), n>=0, to be the smallest prime p such that applying the sum-of-digits function n successive times to p produces n distinct primes (excluding p itself). Is b(n) = a(n) for all n? The first four terms agree. - Felix Fröhlich, Aug 13 2015

It is very likely that this is the case, since although there are always larger "parent" primes with the same digital sum, they typically are at least twice as large (for p=2, these are 11, 101, ...; for p=11 these are 29, 47, 83, ...; for p=29 these are 2999, 3989, 4799, ...), and the number of *digits* of the next term is roughly proportional to this value, so even the "second best" choice would typically lead to a much larger "parent" prime. - M. F. Hasler, Aug 16 2015

Crossrefs

Different from A103830 after a(4).

Programs

Mathematica

a = {2}; k = 3; Do[While[Total@ IntegerDigits@ k != a[[n - 1]], k = NextPrime@ k]; AppendTo[a, k], {n, 2, 4}]; a (* Michael De Vlieger, Aug 20 2015 *)