A354212 - OEIS

%I #18 May 31 2022 11:02:10

%S 11688,116688,126888,1166688,1266888,11666688,12446778,12666888,

%T 116666688,123456789,124466778,126666888

%N Numbers k such that A297330(k)*k and k have the same digits but in a different order.

%C Contains all numbers of the forms 116...688, 12446...6778, and 126...6888 (with at least one 6).

%C All terms are divisible by 3.

%e a(1) = 11688 is a term because A297330(11688) = 7 and 7*11688 = 81816 has the same digits as 11688 in a different order.

%p filter:= proc(n) local L,m;

%p   L:= convert(n,base,10);

%p   m:= convert(map(abs,L[2..-1]-L[1..-2]),`+`);

%p   if m = 1 then return false fi;

%p   sort(L) = sort(convert(m*n,base,10))

%p end proc:

%p select(filter, [seq(i,i=3..10^7,3)]);

%o (PARI) f(n) = my(d=digits(n)); sum(i=2, #d, if (d[i]<d[i-1], d[i-1]-d[i])) + sum(i=2, #d, if (d[i]>d[i-1], d[i]-d[i-1])); \\ A297330

%o isok(k) = my(d=digits(k), dd = digits(k*f(k))); (d != dd) && vecsort(d) == vecsort(dd); \\ _Michel Marcus_, May 19 2022

%o (Python)

%o from itertools import count, islice

%o def A354212_gen(startvalue=1): # generator of terms >= startvalue

%o     for n in count(max(startvalue,1)):

%o         s = str(n)

%o         t = str(n*sum(abs(int(s[i])-int(s[i+1])) for i in range(len(s)-1)))

%o         if s != t and sorted(s) == sorted(t):

%o             yield n

%o A354212_list = list(islice(A354212_gen(),5)) # _Chai Wah Wu_, May 31 2022

%Y Cf. A297330.

%K nonn,base,more

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, May 19 2022