cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A175398 Sequence of resulting numbers after iterations of {((((D_1^D_2)^D_3)^D_4)^... )^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

a(n) = 1 - 9 for infinitely many n.
E.g., a(n) = b (b = 1, 2, ..., 9) for numbers n = b*10^k + A002275(k), where k >= 1.
a(n) = 1 for numbers n such that A055642(A133500(n)) = 1 for n >= 1, e.g., if the number n starts with a digit 1 or contains a digit 0 or for n >= 1.
Sequences after k steps of defined iteration (k >= 0):
0th step: A001477: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ...
1st step: A133500: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1, ...
2nd step: A175399: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1296, 1, 1073741824, 25, 1, 3, 9, 128, 8, 4096, 1628413597910449, 72057594037927936, 221073919720733357899776, 1, 1, 4, 1, 1296, 1073741824, 1, 1, 1, ...
3rd step: A175400: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 32, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
4th step: A175401: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
See A175402 and A175403.

Examples

			For n = 29: a(29) = 9 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}. Resulting number is 9.

A175399 Sequence of resulting numbers after 2nd step of iteration defined in A175398.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1296, 1, 1073741824, 25, 1, 3, 9, 128, 8, 4096, 1628413597910449, 72057594037927936, 221073919720733357899776, 1, 1, 4, 1, 1296, 1073741824, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

See A175398, A175400, A175401, A175402, A175403.

Examples

			For n = 29: a(29) = 25 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}, and the 2nd step of the iteration ends with 25.

A175400 Sequence of resulting numbers after 3rd step of iteration defined in A175398.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 32, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

See A175398, A175399, A175401, A175402, A175403.

Examples

			For n = 29: a(29) = 32 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}and 3rd step of iteration ending with number 32.

A175401 Sequence of resulting numbers after 4th step of iteration defined in A175398.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

See A175398, A175399, A175400, A175402, A175403, A175405. [Jaroslav Krizek, May 05 2010]

Examples

			For n = 29: a(29) = 9 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9} and 4th step of iteration ending with number 9.

A175402 a(n) is the number of iterations of {r -> (((D_1^D_2)^D_3)^...)^D_k, where D_k is the k-th decimal digit of r} needed to reach a one-digit number, starting at r = n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 4, 1, 1, 1, 3, 2, 3, 3, 3, 3, 2, 1, 1, 2, 3, 3, 2, 2, 2, 3, 3, 1, 1, 3, 2, 3, 3, 2, 3, 2, 2, 1, 1, 4, 4, 2, 3, 3, 3, 2, 2, 1, 1, 4, 4, 2, 2, 2, 3, 2, 2, 1, 1, 3, 4, 2, 3, 3, 2, 2, 2, 1, 1, 2, 3, 3, 2, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Comments

Conjecture: max(a(n)) = 4.
Assuming that A020665(2) = 86, A020665(3) = 68, A020665(5) = 58, and A020665(7) = 35, this conjecture is true, since in that case the largest power of a decimal digit that has no 0 is 7^35, and of those powers p none have a(p) > 3. The only n for which a(n) = 4 are those where one iteration goes to 6^2, 7^2, 6^3, 7^3, or 2^9.

Examples

			For n = 29: a(29) = 4 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}.

Crossrefs

Cf. A175398, A175399, A175400, A175401, A175403, A175405.

Programs

  • PARI
    iter(n)=my(v=eval(Vec(Str(n))));v[1]^prod(i=2,#v,v[i])
a(n)=my(k=0);while(n>9,k++;n=iter(n));k

Extensions

Corrected, extended, comment, and program from Charles R Greathouse IV, Aug 03 2010

A175405 Numbers m such that the resulting number h after iterations of {((((D_1^D_2)^D_3)^D_4)^...)^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n is equal to 1 (h = 1).

Original entry on oeis.org

1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 24, 26, 27, 28, 30, 33, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50
Offset: 1

Views

Author

Jaroslav Krizek, May 01 2010

Keywords

Examples

			For number a(14) = 26 we have iteration: {2^6 = 64}, {6^4 = 1296}, {((1^2)^9)^6 = 1}.

Crossrefs

Cf. A175398, A175399, A175400, A175401, A175402, A175403.
Showing 1-6 of 6 results.

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