A036536 -id:A036536 - cp's OEIS frontend

A037069 Smallest prime containing exactly n 8's.

2, 83, 881, 8887, 88883, 888887, 28888883, 88888883, 888888883, 48888888883, 288888888889, 888888888887, 48888888888883, 88888888888889, 888888888888883, 18888888888888883, 88888888888888889, 2888888888888888887, 8888888888888888881, 388888888888888888889

Offset: 0

Patrick De Geest, Jan 04 1999

The last digit of n cannot be 8, therefore a(n) must have at least n+1 digits. It is probable (using [] for floor) that none among [10^n/9]*80 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*80 + b with 1 <= b <= 9 and one of the (first) digits 8 replaced by a digit among {0, ..., 7}. - M. F. Hasler, Feb 22 2016

M. F. Hasler, Table of n, a(n) for n = 0..200

Cf. A065591, A037068, A034388, A036507-A036536.

Cf. A037053, A037055, A037057, A037059, A037061, A037063, A037065, A037067, A037071.

f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 8], {n, 1, 18}]

A037069(n)={my(p, t=10^(n+1)\9*80); forvec(v=[[-1, n], [-8, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)))} \\ ~

a(n) = prime(A037068(n)). - Amiram Eldar, Jul 21 2025

Corrected by Jud McCranie, Jan 04 2001
More terms from Erich Friedman, Jun 03 2001
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016

A269250 Number of times the digit 0 appears in the decimal expansion of n^3.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 1, 1, 0, 0, 0, 1, 1, 0, 3, 0, 2, 0, 0, 0, 0, 0, 0, 1, 3, 0, 0, 2, 0, 0, 0, 2, 0, 1, 3, 0, 0, 1, 1, 0, 0, 0, 0, 1, 3, 0, 0, 0, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 1, 0, 0, 0, 0, 1, 6

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036527, i.e., A048365(n) = A036527(n)^(1/3) is the index of the first occurrence of n.

			1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, ..., 9^3 = 729 all have a(1) = a(2) = ... = a(9) = 0 digits '0'.
0^3 = 0 has a(0) = 1 digit '0'.
10^3 = 1000 has a(10) = 3 digits '0'.

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

Cf. A048365 = A036527^(1/3); A036528 - A036536 and A048366 - A048374.
Analog for the other digits 1, ..., 9: A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086008 (digit 0), and A086009 - A086017 for digits 1 - 9.

seq(numboccur(0, convert(n^3,base,10)), n=0..100); # Robert Israel, Feb 21 2016

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

A269250(n)=!n+#select(t->!t,digits(n^3))

A269241 Number of times the digit 1 appears in the decimal expansion of n^3.

Original entry on oeis.org

0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 2, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 2, 0, 1, 2, 2, 1, 2, 1, 1, 1, 1, 2, 2, 3, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 0, 1, 2, 1, 1, 0, 2, 0, 0, 1, 0, 0, 1, 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 A036528, i.e., A036528(n)^(1/3) = A048366(n) is the index of the first occurrence of n.

			0^3 = 0 has a(0) = 0 digits '1'.
1^3 = 1 has a(1) = 1 digit '1'.
2^3 = 8 has a(2) = 0 digits '1'.
3^3 = 27 has a(3) = 0 digits '1'.
4^3 = 64 has a(4) = 0 digits '1'.
5^3 = 125 has a(5) = 1 digit '1'.
11^3 = 1331 is the smallest cube to have a(11) = 2 digits '1'.

Cf. A036528 and A036527 - A036536; A048366 and A048365 - A048374.
Analog for the other digits 0, 2, ..., 9: A269250, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086009 (digit 1), and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 1], {n, 0, 99}] (* Alonso del Arte, Feb 20 2016 *)

PARI
```
A269241(n)=#select(t->t==1,digits(n^3))
```

A269242 Number of times the digit 2 appears in the decimal expansion of n^3.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036529, i.e., A036529(n)^(1/3) = A048367(n) is the index of the first occurrence of n.

			0^3 = 0 has a(0) = 0 digits '2'.
1^3 = 1 has a(1) = 0 digits '2'.
2^3 = 8 has a(2) = 0 digits '2'.
3^3 = 27 has a(3) = 1 digits '2'.
4^3 = 64 has a(4) = 0 digits '2'.
5^3 = 125 has a(5) = 1 digit '2'.
28^3 = 21952 is the least cube which has a(28) = 2 digits '2'.

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

Cf. A036529 and A036527 - A036536; A048367 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086010 (digit 2), and A086008 - A086017 for digits 0 - 9.

[Multiplicity(Intseq(n^3),2): n in [0..100]]; // Marius A. Burtea, Jan 26 2020

seq(numboccur(2,convert(n^3,base,10)),n=0..100); # Robert Israel, Jan 26 2020

Table[DigitCount[n^3, 10, 2], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

A269242(n)=#select(t->t==2,digits(n^3))

A269243 Number of times the digit 3 appears in the decimal expansion of n^3.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036530, i.e., A036530(n)^(1/3) = A048368(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, 5^3 = 125 and 6^3 = 216 all have a(0) = a(1) = ... = a(6) = 0 digits '3'.
7^3 = 343 has a(7) = 2 digits '3'.

Cf. A036530 and A036527 - A036536; A048368 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086011 (digit 3), and A086008 - A086017 for digits 0 - 9.

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

PARI
```
A269243(n)=#select(t->t==3,digits(n^3))
```

A269244 Number of times the digit 4 appears in the decimal expansion of n^3.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036531, i.e., A036531(n)^(1/3) = A048369(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 2^3 = 8 and 3^3 = 27 all have a(0) = a(1) = a(2) = a(3) = 0 digits '4'.
4^3 = 64 has a(4) = 1 digit '4'.
14^3 = 2744 has a(14) = 2 digits '4'.

Cf. A036531 and A036527 - A036536; A048369 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086012 (digit 4), and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 4], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

PARI
```
A269244(n)=#select(t->t==4,digits(n^3))
```

A269245 Number of times the digit 5 appears in the decimal expansion of n^3.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036532, i.e., A036532(n)^(1/3) = A048370(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 2^3 = 8 and 3^3 = 27 all have a(0) = a(1) = a(2) = a(3) = 0 digits '5'.
5^3 = 125 has a(5) = 1 digit '5'.

Cf. A036532 and A036527 - A036536; A048370 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086013 (digit 5), and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 5], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

PARI
```
A269245(n)=#select(t->t==5,digits(n^3))
```

A269246 Number of times the digit 6 appears in the decimal expansion of n^3.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 3, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 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 A036533, i.e., A036533(n)^(1/3) = A048371(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 2^3 = 8 and 3^3 = 27 all have a(0) = a(1) = a(2) = a(3) = 0 digits '6'.
4^3 = 64 has a(4) = 1 digit '6'.

Cf. A036533 and A036527 - A036536; A048371 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086014 (digit 6), and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 6], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

PARI
```
A269246(n)=#select(t->t==6,digits(n^3))
```

A269247 Number of times the digit 7 appears in the decimal expansion of n^3.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 20 2016

The cubes corresponding to the first occurrence of 1, 2, 3, ... are listed in A036534, i.e., A036534(n)^(1/3) = A048372(n) is the index of the first occurrence of n.

			0^3 = 0, 1^3 = 1, 2^3 = 8 and 4^3 = 64 all have a(0) = a(1) = a(2) = a(4) = 0 digits '7'.
3^3 = 27 has a(3) = 1 digit '7'.

Cf. A036534 and A036527 - A036536; A048372 and A048365 - A048374.
Analog for the other digits 0, 1, ..., 9: A269250, A269241, A269242, A269243, A269244, A269245, A269246, A269247, A269248, A269249.
Analog for squares: A086015 (digit 7), and A086008 - A086017 for digits 0 - 9.

Table[DigitCount[n^3, 10, 7], {n, 0, 100}] (* Robert Price, Mar 21 2020 *)

PARI
```
A269247(n)=#select(t->t==7,digits(n^3))
```

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

Original entry on oeis.org

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))
```

cp's OEIS Frontend

A037069 Smallest prime containing exactly n 8's.

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A269250 Number of times the digit 0 appears in the decimal expansion of n^3.

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A269241 Number of times the digit 1 appears in the decimal expansion of n^3.

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A269242 Number of times the digit 2 appears in the decimal expansion of n^3.

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A269243 Number of times the digit 3 appears in the decimal expansion of n^3.

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A269244 Number of times the digit 4 appears in the decimal expansion of n^3.

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A269245 Number of times the digit 5 appears in the decimal expansion of n^3.

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A269246 Number of times the digit 6 appears in the decimal expansion of n^3.

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