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.

Previous Showing 11-15 of 15 results.

A364789 Initial digit of (n^n)^n (A002489(n)).

Original entry on oeis.org

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

Views

Author

Marco Ripà, Aug 08 2023

Keywords

Comments

a(0) = 1 is from (0^0)^0 = 1 per A002489.
The author conjectures that this sequence obeys the well-known Benford's law.

Examples

			a(4) = 4, since (4^4)^4 = 4^(4^2) = 4294967296.
		

Crossrefs

Cf. A229522 (final digit).

Programs

Formula

a(n) = floor(((n^n)^n)/10^floor(log_10((n^n)^n))).
a(n) = A000030(A002489(n)).

A362004 Initial digit of the decimal expansion of the tetration 2^^n (in Don Knuth's up-arrow notation).

Original entry on oeis.org

1, 2, 4, 1, 6, 2, 2
Offset: 0

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Author

Marco Ripà, Apr 02 2023

Keywords

Comments

The most significant digit of the base-10 representation of 2^(2^(2^...)) n times is given by floor(2^^n/10^(len(2^^n)-1)), where len(2^^n) indicates the number of digits of the argument.
Although it is known that 2^^6 starts with the digit 2 (see A241291), a(7) is not currently known (for details, see Googology link, section "First digits in tetration").

Examples

			a(4) = 6, since 2^^4 = 65536.
		

Crossrefs

Formula

a(n) = floor(2^^n/10^floor(log(2^^n))).

A246564 The n-th least-significant decimal digit of n^^n (in Don Knuth's up-arrow notation).

Original entry on oeis.org

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

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Author

Robert G. Wilson v, Aug 30 2014

Keywords

Comments

This sequence was inspired by the 41st Wohascum County problem.
The distribution of the first 500 terms beginning with 0: 101, 43, 40, 42, 29, 49, 43, 53, 58, 42.
The distribution does not conform to Benford's / Zipf's law, but seems to be evenly distributed once multiples of ten are excluded.

References

  • George T. Gilbert, Mark I. Krusemeyer and Loren C. Larson, The Wohascum County Problem Book, The Mathematical Association of America, Dolciani Mathematical Expositions No. 14, 1993, problem 41 "What is the fifth digit from the end (the ten thousands digit) of the number 5^5^5^5^5?", page 11 and solution on page 76.
  • Ilan Vardi, "Computational Recreations in Mathematica," Addison-Wesley Publishing Co., Redwood City, CA, 1991, pages 226-229.

Crossrefs

Programs

  • Mathematica
    (* first load "SuperPowerMod" from Vardi, see link above, and then *) f[n_] := Quotient[ SuperPowerMod[ n, n, 10^n], 10^(n - 1)]; Array[f, 105]

Formula

if n (mod 10) == 0 then a(n) = 0.

A361100 Decimal expansion of 2^(2^(2^(2^2))) = 2^^5.

Original entry on oeis.org

2, 0, 0, 3, 5, 2, 9, 9, 3, 0, 4, 0, 6, 8, 4, 6, 4, 6, 4, 9, 7, 9, 0, 7, 2, 3, 5, 1, 5, 6, 0, 2, 5, 5, 7, 5, 0, 4, 4, 7, 8, 2, 5, 4, 7, 5, 5, 6, 9, 7, 5, 1, 4, 1, 9, 2, 6, 5, 0, 1, 6, 9, 7, 3, 7, 1, 0, 8, 9, 4, 0, 5, 9, 5, 5, 6, 3, 1, 1, 4, 5, 3, 0, 8, 9, 5, 0
Offset: 19730

Views

Author

Marco Ripà, Mar 03 2023

Keywords

Comments

2^0 = 1, 2^1 = 2, 2^2 = 4, 2^2^2 = 2^^3 = (2^2)^2 = 16,
2^2^2^2 = 2^^4 = (((2^2)^2)^2)^2 = 65536
so that 2^2^2^2^2 = 2^^5 = 2^(2^(2^(2^2))) = 2^65536 = 20035299304068464649790723515602557504478254755697514192650169737108940595563...

Examples

			2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348
(...19529 digits omitted...)
5775699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156736.
The above example line shows the first one hundred decimal digits and the last one hundred digits with the number of unrepresented digits in parentheses.
		

Crossrefs

Programs

  • Mathematica
    nbrdgt = 100; f[base_, exp_] := RealDigits[ 10^FractionalPart[ N[ exp*Log10[ base], nbrdgt + Floor[ Log10[ exp]] + 2]], 10, nbrdgt][[1]]; f[2, 2^2^2^2]
    IntegerDigits[2^65536][[;;100]] (* Paolo Xausa, Jan 31 2024 *)
  • Python
    def A361100(n): return (1<<(1<<(1<<(1<<(1<<1)))))//10**(39458-n)%10 # Chai Wah Wu, Apr 03 2023

Formula

Equals 2^2^2^2^2 = 2^^5 = (((((((((((((((2^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2.

A363746 Initial digit of the decimal expansion of the tetration n^^n (in Don Knuth's up-arrow notation).

Original entry on oeis.org

1, 1, 4, 7, 2
Offset: 0

Views

Author

Marco Ripà, Jun 19 2023

Keywords

Comments

a(5), the most significant digit of the tetration 5^^5, has been estimated to be equal to 1 (and this is also consistent with Benford's law), but there is not any strict proof at the present time and computers are not powerful enough to calculate it without uncertainty.

Examples

			a(3) = 7 since 3^^3 = 7625597484987.
		

Crossrefs

Cf. A000030, A004231 (n^^n), A241293 (4^^4 digits).

Formula

a(n) = floor(n^^n/10^floor(log_10(n^^n))).
a(n) = A000030(A004231(n)).
Previous Showing 11-15 of 15 results.