# [seqfan] RE : 1/sqrt[uphi[x]]= 1/sqrt[uphi[x]]= sqrt[k]*(1/sqrt[x] - 1/sqrt[y])

koh zbi74583.boat at orange.zero.jp
Thu Nov 13 07:42:16 CET 2008

```    Sorry for many typo.
I corrected them.

1/sqrt[uphi[x]]= 1/sqrt[uphi[y]]= sqrt[k]*(1/sqrt[x] - 1/sqrt[y])

I was calculating it in one month and no smaller example existed.
I think that it is complete.

I named it "Reciprocal Square Root differential Square Root integer multiply Unitary Phi Amicable Number"
I am not sure if it is correct English.
Could anyone tell me a grammatically Mathematically correct name?

%I A000001
%S A000001 5575680, 13620639744, 21296209880678400
%N A000001 Numbers m,n such that 1/(UnitaryPhi(m))^(1/2)=1/(UnitaryPhi(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2)), n<m
Sequence gives m number.
%e A000001 Factorization :
2^10*5*3^2*11^2
2^11*23*89*3^2*19^2
2^24*241*3^6*5^2*17^2

%Y A000001 A000002, A000003
%K A000001 none
%O A000001 0,1
%A A000001 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp

%I A000002
%S A000002 4920320, 12681574400, 20926651432960000
%N A000002 Numbers m,n such that 1/(UnitaryPhi(m))^(1/2)=1/(UnitaryPhi(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2)), n<m
Sequence gives n number.
%e A000002 Factorization :
2^10*5*31^2
2^11*23*89*5^2*11^2
2^24*241*5^4*7^2*13^2
%Y A000002 A000001,A000003
%K A000002 none
%O A000002 0,1
%A A000002 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp

%I A000003
%S A000003 341, 1805, 13600
%N A000003 Numbers m,n such that 1/(UnitaryPhi(m))^(1/2)=1/(UnitaryPhi(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2)), n<m
Sequence gives k number.
%Y A000003 A000001,A000002
%K A000003 none
%O A000003 0,1
%A A000003 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp

Yasutoshi

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