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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.

User: Matthew E. Coppenbarger

Matthew E. Coppenbarger's wiki page.

Matthew E. Coppenbarger has authored 4 sequences.

A352752 a(n) is the smallest nonnegative number that requires n applications of the modified Sisyphus function of order 3, x -> A350709(x) to reach any number in the cycle {4031, 4112, 4220}.

Original entry on oeis.org

4031, 1001, 0, 10, 1000000000
Offset: 0

Views

Author

Matthew E. Coppenbarger, Apr 01 2022

Keywords

Comments

The next term, a(5), is 1 0^100 1^9 2^10, a number with 120 digits, is too large to display.

Examples

			0 -> 1100 -> 4220 reaches an element of the cycle {4031, 4112, 4220} in two iterates and must be the smallest, so a(2) = 0

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Cf. A350709.

A352751 Modified Sisyphus function of order 4: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 4)(number of digits of n congruent to 1 modulo 4)(number of digits of n congruent to 2 modulo 4)(number of digits of n congruent to 3 modulo 4).

Original entry on oeis.org

11000, 10100, 10010, 10001, 11000, 10100, 10010, 10001, 11000, 10100, 21100, 20200, 20110, 20101, 21100, 20200, 20110, 20101, 21100, 20200, 21010, 20110, 20020, 20011, 21010, 20110, 20020, 20011, 21010, 20110, 21001, 20101, 20011, 20002, 21001, 20101, 20011, 20002, 21001, 20101, 22000, 21100, 21010
Offset: 0

Views

Author

Matthew E. Coppenbarger, Apr 01 2022

Keywords

Comments

If we start with n and repeatedly apply the map i -> a(i), we eventually get one of three cycles: {51220}, {50410, 52111, 53200}, or {51301}

Examples

			11 has two digits, both congruent to 1 modulo 4, so a(11) = 20200.
a(20) = 21010.
a(30) = 21001.
a(1111123567) = 100622.

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002, A350709.

Programs

  • Python
    def a(n, order=4):
    d, m = list(map(int, str(n))), [0]*order
    for di in d: m[di%order] += 1
    return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Apr 01 2022

A350709 Modified Sisyphus function of order 3: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 3)(number of digits of n congruent to 1 modulo 3)(number of digits of n congruent to 2 modulo 3).

Original entry on oeis.org

1100, 1010, 1001, 1100, 1010, 1001, 1100, 1010, 1001, 1100, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2200, 2110, 2101, 2200, 2110, 2101, 2200
Offset: 0

Views

Author

Matthew E. Coppenbarger, Mar 28 2022

Keywords

Comments

If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}

Examples

			11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.

References

  • M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002.

Programs

  • Python
    def a(n):
    d, m = list(map(int, str(n))), [0, 0, 0]
    for di in d: m[di%3] += 1
    return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Mar 28 2022

A301408 Repeatedly map numbers to the number of letters in the English name (include spaces, but no hyphens and no "and"s); sequence gives smallest nonnegative integer that needs exactly n iterations to get to 4.

Original entry on oeis.org

4, 0, 3, 1, 11, 23, 123, 101323373373
Offset: 0

Views

Author

Matthew E. Coppenbarger, Mar 20 2018

Keywords

Examples

			Since "twenty three" has 12 letters (including the space), "twelve" has 6 letters, "three" has 5 letters, "four" has four letters, and no other smaller nonnegative integer maps to 4 in exactly 5 iterations, then a(5) = 23.

References

  • M. Ecker, Number play, calculators, and card tricks: Mathemagical black holes, in E. Berlekamp and T. Rodgers, eds., The Mathemagician and Pied Puzzler: A Collection in Tribute to Martin Gardner, A. K. Peters, 1999, pp. 41-52.

Crossrefs

Cf. A227290, A005589, A016037 (these do not count the spaces in the names of larger integers).

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