# [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

```