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Numbers k whose 10's complement mod 10 of their digits is equal to phi(k), the Euler totient function of k.
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%I #19 Mar 09 2024 18:42:33

%S 74,834,80940,809400,833334,7414114,7422694,7539694,8094000,80940000,

%T 809400000,8094000000,80940000000,83335786566,809400000000,

%U 7539682539694,8094000000000,80940000000000

%N Numbers k whose 10's complement mod 10 of their digits is equal to phi(k), the Euler totient function of k.

%C Any number of the form 8094*10^j, with j>0, is part of the sequence because its Euler totient function is 2016*10^j.

%C Contains subsequence 834, 833334, 833333333333334, ... formed by numbers (10^k/4 + 2)/3 for k in A296059. - _Max Alekseyev_, Mar 09 2024

%e phi(74) = 36 that is the 10's complement of the digits of 74.

%p with(numtheory): P:=proc(q) local a,b,k,n;

%p for n from 1 to q do a:=convert(phi(n),base,10);

%p for k from 1 to nops(a) do a[k]:=(10-a[k]) mod 10; od; b:=0;

%p for k from 1 to nops(a) do b:=b*10+a[nops(a)-k+1]; od;

%p if b=n then print(n); fi; od; end: P(10^9);

%o (PARI) isok(x) = {my(dx = digits(x), dy = vector(#dx, k, (10-dx[k]) % 10)); fromdigits(dy) == eulerphi(x); } \\ _Michel Marcus_, Mar 12 2018

%Y Cf. A000010, A055120, A296059.

%K nonn,base,more

%O 1,1

%A _Paolo P. Lava_, Mar 07 2018

%E a(11)-a(15) from _Giovanni Resta_, Mar 09 2018

%E a(16)-a(18) from _Max Alekseyev_, Mar 09 2024