cp's OEIS Frontend

A363422 Numbers k which satisfy k = concat(a,b,...) and a*b*... = reverse(k), for some two or more a,b,...

%I A363422 #23 Jul 27 2023 12:16:40
%S A363422 351,621,886,5931,86673,97533,425322,430762,920781,3524751,4495491,
%T A363422 4834872,5594151,5941971,6218001,6801381,6916671,8630841,32331001,
%U A363422 44235301,57982563,67968432,68577483,69617484,71673981,88873491,89943354,119910901,338752611
%N A363422 Numbers k which satisfy k = concat(a,b,...) and a*b*... = reverse(k), for some two or more a,b,...
%C A363422 k > reverse(k) for all terms, sometimes narrowly, see a(28) = 119910901.
%C A363422 This is easily shown: c=concat(a,b), c/a > (c-b)/a = 10^(#digits of b) > b; c > b*a.
%C A363422 Follows for triple or higher concatenations by induction.
%C A363422 Of the first 39 terms, 12 arise due to concatenations of only two numbers and are therefore also present in A281555.
%C A363422 No terms yet found with a product of more than five numbers.
%C A363422 Sometimes a term B relates to an earlier term A via a particular number N for which B=concat(A,N) and reverse(B)=reverse(A)*reverse(N). This is true of B=a(15), A=a(2), and N=8001 for example.
%e A363422 153 = 3*51.
%e A363422 1395 = 5*9*31.
%e A363422 1945944 = 44*9*54*91.
%e A363422 1008126 = 6*21*8001.
%e A363422 171548496 = 6*94*84*51*71.
%o A363422 (Python)
%o A363422 # Find numbers with a de-concatenation that multiplies to their reverse.
%o A363422 import math
%o A363422 def digits(x):
%o A363422     y = []
%o A363422     while x>0:
%o A363422         y = [x%10] + y
%o A363422         x//=10
%o A363422     return y
%o A363422 def check(x):
%o A363422     xx = digits(x)
%o A363422     if xx[0] < xx[-1]:
%o A363422         return
%o A363422     for i in range(1,2**(len(xx)-1)):
%o A363422         for dnum,digit in enumerate(xx):
%o A363422             if dnum==0:
%o A363422                 thisProd = [xx[0]]
%o A363422             elif i&(2**(dnum-1)):
%o A363422                 if digit==0:
%o A363422                     break
%o A363422                 thisProd += [digit]
%o A363422             else:
%o A363422                 thisProd[-1] = thisProd[-1]*10+digit
%o A363422         answer = math.prod(thisProd)
%o A363422         if not answer%10==xx[0]:
%o A363422             continue
%o A363422         if digits(answer)[-1::-1]==xx:
%o A363422             print('\r'+str(thisProd).replace(', ','x')[1:-1])
%o A363422             return
%o A363422     return
%o A363422 i=0
%o A363422 while True:
%o A363422     i += 1
%o A363422     if not i%10000:
%o A363422         print('\r'+str(i),end='')
%o A363422     check(i)
%Y A363422 Cf. A267939, A281555, A265737.
%Y A363422 A267939 is contained in the intersection of this sequence and A281555.
%K A363422 nonn,base
%O A363422 1,1
%A A363422 _David L. Reens_, Jun 01 2023