• Number Sequences

    From David Entwistle@qnivq.ragjvfgyr@ogvagrearg.pbz to rec.puzzles on Tue Oct 21 13:18:42 2025
    From Newsgroup: rec.puzzles

    October 10th, 2025, was Neil Sloane's 86th birthday. A very happy birthday
    to Neil. Among may other things, Neil founded The On-line Encyclopedia of Integer Sequences. I hadn't realized that he was the same chap appearing
    on the Numberphile videos.

    http://neilsloane.com/

    https://oeis.org

    https://www.youtube.com/playlist?list=PLt5AfwLFPxWJXQqPe_llzWmTHMPb9QvV2

    Here are a few sequences to see if you can identify and extend them. They
    may roughly get harder as you work through them. The first five (a - e)
    are listed on the OEIS.


    a) 1, 11, 21, 1211, 111221, ?

    b) 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, ?

    c) 5, 16, 24, 72, 48, 120, ?

    d) 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, ?

    e) 1, 8, 9, 1, 8, 9, ?

    The following three number sequences (f - h) are from the first GCHQ
    Puzzle Book. I don't know the answers to these, without working them out,
    or looking them up in the answers section. They will more cryptic and
    probably not be on the OEIS.

    153. Number sequence V

    f) 8, 7, 1, 3, 3, 12, 13, 17, 21, ?

    g) 9, 1, 9, 20, 7, 11, 15, 13, 17, ?

    h) 1, 2, 4, 5, 6, 8, 10, 40, 46, 60, 61, 64, 80, 84, ?

    Feel free to post your own.
    --
    David Entwistle
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  • From David Entwistle@qnivq.ragjvfgyr@ogvagrearg.pbz to rec.puzzles on Fri Oct 24 07:53:00 2025
    From Newsgroup: rec.puzzles

    On Tue, 21 Oct 2025 13:18:42 -0000 (UTC), David Entwistle wrote:

    a) 1, 11, 21, 1211, 111221, ?

    b) 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, ?

    c) 5, 16, 24, 72, 48, 120, ?

    d) 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, ?

    e) 1, 8, 9, 1, 8, 9, ?

    SPOILER.
    POILER.
    OILER.
    ILER.
    LER.
    ER.
    R.
    .

    Answers:

    a) 1, 11, 21, 1211, 111221, 312211
    A005150 - Look and Say sequence: describe the previous term! (method A - initial term is 1).

    b) 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3
    A005589 - Number of letters in the US English name of n, excluding spaces
    and hyphens.

    c) 5, 16, 24, 72, 48, 120, 72
    A069482 - a(n) = prime(n+1)^2 - prime(n)^2.

    d) 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4
    A056992 - Digital roots of square numbers

    e) 1, 8, 9, 1, 8, 9, 1
    A073636 - Period 3: repeat [1, 8, 9] ; Digital root of A000578(n) = n^3
    for n >= 1.


    I think c), d) and e) are particularly interesting. How the difference
    between the squares of adjacent primes evolves, that the digital root of
    any perfect square will be 1, 4, 7 or 9 and that the digital root of any perfect cube will be either 1, 8 or 9.
    --
    David Entwistle
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  • From Phil Carmody@pc+usenet@asdf.org to rec.puzzles on Thu Nov 6 11:55:37 2025
    From Newsgroup: rec.puzzles

    David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:
    Answers:

    a) 1, 11, 21, 1211, 111221, 312211
    A005150 - Look and Say sequence: describe the previous term! (method A - initial term is 1).

    b) 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3
    A005589 - Number of letters in the US English name of n, excluding spaces and hyphens.

    c) 5, 16, 24, 72, 48, 120, 72
    A069482 - a(n) = prime(n+1)^2 - prime(n)^2.

    d) 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4
    A056992 - Digital roots of square numbers

    e) 1, 8, 9, 1, 8, 9, 1
    A073636 - Period 3: repeat [1, 8, 9] ; Digital root of A000578(n) = n^3
    for n >= 1.


    I think c), d) and e) are particularly interesting. How the difference between the squares of adjacent primes evolves, that the digital root of
    any perfect square will be 1, 4, 7 or 9 and that the digital root of any perfect cube will be either 1, 8 or 9.

    I'm annoyed I didn't get (c) given my affinity for prime numbers, I'm
    not sure any amount of additional thinking would have got me there.
    However, adding "try summing them" to my toolbox might be helpful in the future, I normally don't go beyond taking their differences.

    The other two are hardly surprising:
    if x=3n, for n>0, then x^2=9n^2 so will have digital root 9
    if x=3n+/-1, then x^2=9n^2+/-6n+1, so will have digital root 1, 4, or 7

    if x=3n, for n>0, then x^3=9(3n^3), so will have digital root 9
    if x=3n+1, then x^3=9(3n^3+3n^2+n)+1 so will have digital root 1
    if x=3n+2, then x^3=9(3n^3+6n^2+4n)+8 so will have digital root 8

    Phil
    --
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