A178716 a(n) = smallest prime whose sum of digits is A001651(n).

17, 19999999, 99899999999999, 299989999999999999999999999999999999999, 999999999998999999999999999999999999999999999999999999999, 9999999999999999999999999999999999999999999999999999999999999991999999999999999999999999999999999999999999999999

Offset: 1

Author

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Jun 07 2010

Keywords

• base
• nonn

Comments

As k runs through the numbers that are not multiples of 3, sequence gives smallest prime p whose sum of digits is k^3.

No primes exist whose sum of digits is a multiple of 3

k=4: 4^3 = 64 = 7 * 9 + 1, 8-digit prime:10^7+(10^7-1) = 2*10^7 - 1

k=5: 5^3 = 125 = 13 * 9 + 8, 14-digit prime: (10^2-1)*10^12+8*10^11+(10^11-1) = 10^14 - 10^11 - 1

k=7: 7^3 = 343 = 38 * 9 + 1: no such prime formed with 38 digits "9" and one digit "1" 343 = 37 * 9 + 2 + 8, gives as 4th term a 39-digit prime: 2*10^38+(10^3-1)*10^35+8*10^34+(10^34-1) = 3*10^38 - 10^34 - 1

k=8: 8^3 = 512 = 56 * 9 + 8, 57-digit prime: (10^11-1)*10^46+8*10^45+(10^45-1) = 10^57 - 10^45 - 1

k=10: 10^3 = 1000 = 111 * 9 + 1, 112-digit prime, 63 "leading" "9's", "1", 48 "ending" "9's": (10^63-1)*10^49+10^48+(10^48-1) = 10^112 - 8 * 10^48 - 1 = ((2*5)^(2^2^2))^7 - (2*((2*5)^(2^2^2)))^3 - 1

k=11: 11^3 = 1331 = 147 * 9 + 8: no such prime formed with 147 digits "9" and one digit "8" 1331 = 147 * 9 + 1 + 7, 149-digit prime: 10^148+(10^143-1)*10^5+7*10^4+(10^4-1) = 2*10^148 - 2*10^4 - 1

Crossrefs

Cf. A061105