A269250 -id:A269250 - cp's OEIS frontend

A269248 Number of times the digit 8 appears in the decimal expansion of n^3.

0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 3, 1, 2, 1, 2, 0, 0, 0, 0

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036535, i.e., A036535(n)^(1/3) = A048373(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 3^3 = 27 and 4^3 = 64 all have a(0) = a(1) = a(3) = a(4) = 0 digits '8'.
2^3 = 8 has a(2) = 1 digit '8'.

Cf. A036535 and A036527 - A036536; A048373 and A048365 - A048374.
Analog for other digits 0, 1, ..., 7, 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269249.
Analog for squares: A086016, and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 8], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

PARI
```
A269248(n)=#select(t->t==8,digits(n^3))
```

A269249 Number of times the digit 9 appears in the decimal expansion of n^3.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 2, 0, 1, 1, 0, 0, 0, 0, 2, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 1, 2, 3, 0

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036535, i.e., A036535(n)^(1/3) = A048374(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, ... and 8^3 = 512 all have a(0) = a(1) = ... = a(8) = 0 digits '9'.
9^3 = 729 has a(9) = 1 digit '9'.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A036536 and A036527 - A036535; A048374 and A048365 - A048373.
Analog for the other digits 0, 1, ..., 8: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248.
Analog for squares: A086017 (digit 9) and A086008 - A086016 for digits 0 - 8.

DigitCount[(Range[0, 100])^3, 10, 9] (* G. C. Greubel, Dec 13 2016 *)

PARI
```
A269249(n)=#select(t->t==9,digits(n^3))
```

A333206 a(n) is the least decimal digit of n^3.

Original entry on oeis.org

0, 1, 8, 2, 4, 1, 1, 3, 1, 2, 0, 1, 1, 1, 2, 3, 0, 1, 2, 5, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 0, 1, 2, 3, 0, 2, 4, 0, 2, 1, 0, 1, 0, 0, 1, 1, 3, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 0, 1, 2, 2, 0, 1, 0, 0, 1, 2, 0, 0, 1, 3, 3, 2, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 6, 0, 0, 3, 3, 1, 1

Offset: 0

Robert Israel, Mar 12 2020

Dean Hickerson found an infinite sequence of n such that a(n) > 0 (see Guy, sec F24). Are there infinitely many such that a(n) > 1? If not, what is the greatest n with a(n)=k for each k > 1?

Heuristically, we should expect on the order of ((10-m)^3/100)^d terms n with d digits and a(n) >= m. Since 5^3/100 > 1 > 4^3/100 we should expect infinitely many terms with a(n) >= 5 but only finitely many terms with a(n) >= 6. See A291644 for a(n) = 5. There are only two n <= 10^6 with a(n) >= 6, namely a(2) = 8 and a(92) = 6.

			The least digit of 6^3=216 is 1, so a(6)=1.

R. Guy, Unsolved Problems in Number Theory (Third edition), Springer 2004.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A052044, A054054, A269250, A291639, A291640, A291641, A291642, A291643, A291644.

seq(min(convert(n^3,base,10)),n=0..200);

a(n) = A054054(n^3).

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A269248 Number of times the digit 8 appears in the decimal expansion of n^3.

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A269249 Number of times the digit 9 appears in the decimal expansion of n^3.

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A333206 a(n) is the least decimal digit of n^3.

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