A362002 - OEIS

%I #28 Apr 23 2023 23:32:17

%S 0,1,5,10,25,50,100,250,500,1000,2500,5000,10000,10005,10625,25000,

%T 50000,100000,100005,100050,106250,250000,500000,1000000,1000005,

%U 1000050,1000500,1062500,2500000,5000000,10000000,10000005,10000025,10000050,10000500,10005000

%N Numbers k such that the digits of k^2 are a subsequence of the digits of k^3.

%C If n is a term, then so is 10*n.

%C 10^k + 5 is a term for all k >= 4.

%C 10^k + 25 is a term for all k >= 7.

%H Chai Wah Wu, <a href="/A362002/b362002.txt">Table of n, a(n) for n = 1..80</a>

%e a(5) = 25 is a term because the digits of 25^2 = 625 form a subsequence of those of 25^3 = 15625.

%e a(14) = 10005 is a term because the digits of 10005^2 = 100100025 form a subsequence of those of 10005^3 = 1001500750125.

%p filter:= proc(n) local L,nL,M,nM,i,j,k;

%p   L:= convert(n^2,base,10); nL:= nops(L);

%p   M:= convert(n^3,base,10); nM:= nops(M);

%p   j:= 1:

%p   for i from 1 to nL do

%p     if not member(L[i],M[j..nM],'k') then return false fi;

%p     j:= j+k;

%p   od;

%p   true

%p end proc:

%p select(filter, [$0..5*10^5]);

%t fQ[n_]:=Module[{len=IntegerLength[n^2],sq=ToString[n^2],qb=ToString[n^3]},

%t StringMatchQ[qb,StringInsert[sq,"**",Range[1,len+1]]]];

%t Select[Range[0,10^6],fQ[#]&] (* _Ivan N. Ianakiev_, Apr 06 2023 *)

%o (Python)

%o from itertools import count, islice, product

%o def ok(n):

%o     s, t = n**2, n**3

%o     while s and t:

%o         if t%10 == s%10:

%o             s //= 10

%o         t //= 10

%o     return s == 0

%o print([k for k in range(10**6+1) if ok(k)]) # _Michael S. Branicky_, Apr 02 2023

%o (Python)

%o from itertools import count, islice

%o def A362002_gen(startvalue=0): # generator of terms >= startvalue

%o     for k in count(max(startvalue,0)):

%o         c = iter(str((m:=k**2)*k))

%o         if all(map(lambda b:any(map(lambda a:a==b,c)),str(m))):

%o             yield k

%o A362002_list = list(islice(A362002_gen(),20)) # _Chai Wah Wu_, Apr 03 2023

%Y Cf. A046829, A362001.

%K nonn,base

%O 1,3

%A _Robert Israel_, Apr 02 2023

%E a(24) and beyond from _Michael S. Branicky_, Apr 02 2023