A363422 - OEIS

%I #23 Jul 27 2023 12:16:40

%S 351,621,886,5931,86673,97533,425322,430762,920781,3524751,4495491,

%T 4834872,5594151,5941971,6218001,6801381,6916671,8630841,32331001,

%U 44235301,57982563,67968432,68577483,69617484,71673981,88873491,89943354,119910901,338752611

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

%C k > reverse(k) for all terms, sometimes narrowly, see a(28) = 119910901.

%C This is easily shown: c=concat(a,b), c/a > (c-b)/a = 10^(#digits of b) > b; c > b*a.

%C Follows for triple or higher concatenations by induction.

%C Of the first 39 terms, 12 arise due to concatenations of only two numbers and are therefore also present in A281555.

%C No terms yet found with a product of more than five numbers.

%C 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 153 = 3*51.

%e 1395 = 5*9*31.

%e 1945944 = 44*9*54*91.

%e 1008126 = 6*21*8001.

%e 171548496 = 6*94*84*51*71.

%o (Python)

%o # Find numbers with a de-concatenation that multiplies to their reverse.

%o import math

%o def digits(x):

%o     y = []

%o     while x>0:

%o         y = [x%10] + y

%o         x//=10

%o     return y

%o def check(x):

%o     xx = digits(x)

%o     if xx[0] < xx[-1]:

%o         return

%o     for i in range(1,2**(len(xx)-1)):

%o         for dnum,digit in enumerate(xx):

%o             if dnum==0:

%o                 thisProd = [xx[0]]

%o             elif i&(2**(dnum-1)):

%o                 if digit==0:

%o                     break

%o                 thisProd += [digit]

%o             else:

%o                 thisProd[-1] = thisProd[-1]*10+digit

%o         answer = math.prod(thisProd)

%o         if not answer%10==xx[0]:

%o             continue

%o         if digits(answer)[-1::-1]==xx:

%o             print('\r'+str(thisProd).replace(', ','x')[1:-1])

%o             return

%o     return

%o i=0

%o while True:

%o     i += 1

%o     if not i%10000:

%o         print('\r'+str(i),end='')

%o     check(i)

%Y Cf. A267939, A281555, A265737.

%Y A267939 is contained in the intersection of this sequence and A281555.

%K nonn,base

%O 1,1

%A _David L. Reens_, Jun 01 2023