cp's OEIS Frontend

A379591 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (|q-r|)^3.

%I A379591 #17 Jan 14 2025 17:14:38
%S A379591 1000,456533,474552,1000000,69426531,1000000000,1000000000000,
%T A379591 1000000000000000,1000000000000000000,1000000000000000000000,
%U A379591 1000000000000000000000000,1000000000000000000000000000,1000000000000000000000000000000,1000000000000000000000000000000000
%N A379591 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (|q-r|)^3.
%e A379591 1000 = (|10-00|)^3.
%e A379591 456533 = (|456-533|)^3.
%e A379591 1000000 = (|100-0000|)^3.
%p A379591 filter:= proc(x) local m;
%p A379591 for m from 1 to ilog10(x) do
%p A379591   if x = abs((x mod 10^m) - floor(x/10^m))^3 then return true fi
%p A379591 od;
%p A379591 false
%p A379591 end proc:
%p A379591 select(filter, [seq(i^3,i=1..10^7)]); # _Robert Israel_, Jan 01 2025
%o A379591 (PARI) lista(nn) = my(d, m, q, s, t=1, v=List([])); while(t<nn, t*=10; fordivfactored(t+1, f, if(f[1]>1&&gcd(f[1], (t+1)/f[1])==1, d=[x[1]^x[2]|x<-f[2]~]; forvec(x=vector(#d, i, [0, 1]), s=lift(chinese(chinese(vector(#d, i, Mod((-1)^x[i], d[i]))), Mod(0, (t+1)/f[1]))); q=(s^3-s)/(t+1); if(q>0&&s+q<t, listput(v, s^3))); if(vecmax(f[2][, 1]%4)==1, forvec(x=vector(#d, i, [0, 1]), s=lift(chinese(chinese(vector(#d, i, m=Mod(lift(Mod(-1, f[2][i, 1])^(1/2)), d[i]); for(j=2, f[2][i, 2], m=(m^2-1)/(2*m)); (-1)^x[i]*m)), Mod(0, (t+1)/f[1]))); q=(s^3+s)/(t+1); if(q>=s&&q-s<t, listput(v, s^3))))))); Set(select(x->x<=nn, v)); \\ _Jinyuan Wang_, Jan 03 2025
%Y A379591 Cf. A102766, A228103.
%K A379591 nonn,base
%O A379591 1,1
%A A379591 _Travis Vasquez_, Dec 26 2024
%E A379591 a(11)-a(12) from _Michael S. Branicky_, Jan 01 2025
%E A379591 a(13)-a(14) from _Jinyuan Wang_, Jan 03 2025