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François Marques, <a href="/A061341/b061341.txt">Table of n, a(n) for n = 1..100</a>
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Conjecture: the last digit of a(n) is in {1, 2, 3, 4, 5, 6, 8} and if it's 3 then there is k>=2 such that a(n) = 3*10^k+3.
- 1 then there is k>=3 such that a(n) = 6*10^k+1,
- 2 then a(n) = 22 or a(n) = 2272 or there is k>=4 such that a(n) = 15*10^k+12,
- 3 then there is k>=2 such that a(n) = 3*10^k+3.
(PARI) can_split(v, b=10, p=3)=if(#v==0, 1, for(k=1, #v, if(ispower(fromdigits(v[1..k], b), 3p) && can_split(v[k+1..#v], b, p), return(1))); return(0));
isok(n, b=10, p=3)=if(n%10b==0, return(0)); my(v=digits(n^3p, b)); for(k=1, #v-1, if(ispower(fromdigits(v[1..k], b), 3p) && can_split(v[k+1..#v], b, p), return(1))); return(0); \\ François Marques, Jul 11 12 2021
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- 1 then exists there is k>=3 such that a(n) = 6*10^k+1,
- 2 then a(n) = 22 or a(n) = 2272 or exists there is k>=4 such that a(n) = 15*10^k+12,
- 3 then exists there is k>=2 such that a(n) = 3*10^k+3.
From François Marques, Jul 11 2021: (Start)
If a and b are integers such that 3*a^2*b and 3*a*b^2 can be splited obtained as concatenations of cubes then a*10^k+b is a term in the list for k greater than the maximum of the number of digits in b^3, 3*a^2*b and 3*a*b^2.
Conjecture: the last digit of a(n) is in {1, 2, 3, 4, 5, 6} and if it's
- 1 then exists k>=3 such that a(n) = 6*10^k+1,
- 2 then a(n) = 22 or a(n) = 2272 or exists k>=4 such that a(n) = 15*10^k+12,
- 3 then exists k>=2 such that a(n) = 3*10^k+3.
(End)
2272^3 = 11728027648 = 1_1728_0_27_64_8 (where the underscores indicate concatenation).
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