[seqfan] Re: Unitary RMPN

[Jack Brennen]

Filling in some of your gaps:

Smallest number with this unitary sigma ratio:

a(22) ... 19235716742537891017605454376709022095843377283072000000 ...
                 [2, 25; 3, 10; 5, 6; 7, 3; 11, 3; 13, 3; 37, 2; 43, 2;
                  79, 1; 137, 1; 157, 1; 197, 1; 251, 1; 601, 1;
                  1013, 1; 1181, 1; 4051, 1]

Second smallest number with these unitary sigma ratios:

b(13) ... 43599025600 ... [2, 6; 5, 2; 13, 3; 79, 1; 157, 1]

b(14) ... 3381389490561780096000 ... [2, 10; 3, 4; 5, 3; 7, 2; 29, 2;
                                       41, 2; 53, 1; 211, 1; 421, 1]

b(20) ... 69007948786975104000 ... [2, 10; 3, 4; 5, 3; 29, 2; 41, 2;
                                     53, 1; 211, 1; 421, 1]

b(21) ... 174679332322727913624625142203061760000000 ...
                             [2, 15; 3, 11; 5, 7; 7, 2; 17, 2; 29, 3;
                              67, 1; 271, 1; 331, 2; 449, 1; 661, 1;
                              1889, 1]


Also note that b(14)/a(14) is not an integer, so the conjecture
is disproven.






On 2/3/2014 7:41 PM, zbi74583.boat at orange.zero.jp wrote:
>      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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