A213018 Largest possible right-truncatable base n semiprime, written in decimal notation.
349859, 96614184696363331, 21453921664462866568480385, 5396625577204731352098054139, 1230847457959658263441326143300761, 95861957783594714393831931415189937897, 246968512564969427282294385793684699270364003, 2275670244821939317343219562642735197101789412250091, 452359410421075824795509870868069265597540337861667320077
Offset: 5
Examples
a(5)=349859=42143414 in base 5 = 89*3931 4214341 in base 5 = 69971 = 11*6361 421434 in base 5 = 13994 = 2*6997 42143 in base 5 = 2798 = 2*1399 4214 in base 5 = 559 = 13*43 421 in base 5 = 111 = 3*37 42 in base 5 = 22 = 2*11 4 in base 5 = 4 = 2*2 a(6)=4223145115415551545111 in base 6 a(7)=644324264233631242462662622646 in base 7 a(8)=4267773725372537135533515117773 in base 8 a(9)=43741424882428682844851886888222774 in base 9 a(10)=95861957783594714393831931415189937897 in base 10 a(11)=4567476a2738a828994aa851a116aa886a95686a231 in base 11 a(12)=43a2971ba155719171a2b1b97777775b779a732b755572b7 in base 12 a(13)=9114448462c6c46b3c9937446466b43686a24668666732c4356 in base 13
Programs
-
Python
from sympy import factorint def fromdigits(t, b): return sum(b**i*di for i, di in enumerate(t[::-1])) def semiprime(n): return sum(factorint(n).values()) == 2 def a(n): m, s = 0, [(i,) for i in range(n) if semiprime(fromdigits((i,), n))] while len(s) > 0: m = fromdigits(max(s), n) cands = set(t+(d,) for t in s for d in tuple(range(n))) s = [c for c in cands if semiprime(fromdigits(c, n))] return m print([a(n) for n in range(5, 8)]) # Michael S. Branicky, Aug 04 2022
Comments