[seqfan] Re: Unitary RMPN

zbi74583.boat at orange.zero.jp zbi74583.boat at orange.zero.jp
Tue Feb 4 04:41:44 CET 2014


    Maximilian,Hugo,Jack
    Thank you for comfirming and computing more terms

    >I don't know of an effective way to find these for the general case though,
    >> or to verify a value is the first possible, other than by crude
exhaustive
    >> search.
    >>

    >(That's what I did (*shame*). Took about a minute, thinking is slower...)

    I agree with you.
    But I like to think
    I compute them mentally, so somtimes I miss an easy example.

    >PS: you already found the smallest one in 2009 : see

    My God, I haven't remembered it.
    I might have a sickness of mind

    a(n) the smallest number such that UitarySigma(m)=(n+1)/n*m

    1  6
    2  2
    3  3
    4  4
    5  5
    6  2^3*3^3
    7  7
    8  8
    9  9
    10 2^5*3^3*5^3*7^2
    11 11
    12 2^6*3^3*5^3*7^2
    13 13
    14 2^10*5^4*7^2*41*79*157*313
    15 3^3*5^3*7^2
    16 16
    17 17
    18 2^9*3^5*31*61
    19 19
    20 2^10*5^4*41*79*15 *313
    21 22 2^15 * 3^11 * 5^5 * 7^2 * 17^2 * 29^3 * 67 * 271 * 331^2 * 521 * 661 *
1889
    22 -
    23 23
    24 552063590295800832


    b(n) secondly small number such that UitarySigma(m)=(n+1)/n*m

    1  2^2*3*5
    2  2^2*5
    3  3^2*5
    4  2^3*3^2
    5  2^3*3^3*5^3*7^2
    6  2^3*3^4*41
    7  5^2*7^2*13
    8  2^4*17
    9  2^9*3^5*19*31*61
    10 2^5*3^4*5^3*7^2*41
    11 2^5*11^2*31*61
    12 2^6*3^4*5^3*7^2*41


    15 3^4*5^3*7^2*41
    16 2^12*11^2*31*61*241

    18 2^9*3^7*23*137*547

    25 2^6*5^4*79*157*313

    27 3^4*41

    32 2^7*43



    If m=p~e then b(n)=0 Mod p^(e+1)
    I cocnjectured b(n)/a(n) is integer.

    Could you compute more term?



    Yasutoshi






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