login
A363422
Numbers k which satisfy k = concat(a,b,...) and a*b*... = reverse(k), for some two or more a,b,...
0
351, 621, 886, 5931, 86673, 97533, 425322, 430762, 920781, 3524751, 4495491, 4834872, 5594151, 5941971, 6218001, 6801381, 6916671, 8630841, 32331001, 44235301, 57982563, 67968432, 68577483, 69617484, 71673981, 88873491, 89943354, 119910901, 338752611
OFFSET
1,1
COMMENTS
k > reverse(k) for all terms, sometimes narrowly, see a(28) = 119910901.
This is easily shown: c=concat(a,b), c/a > (c-b)/a = 10^(#digits of b) > b; c > b*a.
Follows for triple or higher concatenations by induction.
Of the first 39 terms, 12 arise due to concatenations of only two numbers and are therefore also present in A281555.
No terms yet found with a product of more than five numbers.
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.
EXAMPLE
153 = 3*51.
1395 = 5*9*31.
1945944 = 44*9*54*91.
1008126 = 6*21*8001.
171548496 = 6*94*84*51*71.
PROG
(Python)
# Find numbers with a de-concatenation that multiplies to their reverse.
import math
def digits(x):
y = []
while x>0:
y = [x%10] + y
x//=10
return y
def check(x):
xx = digits(x)
if xx[0] < xx[-1]:
return
for i in range(1, 2**(len(xx)-1)):
for dnum, digit in enumerate(xx):
if dnum==0:
thisProd = [xx[0]]
elif i&(2**(dnum-1)):
if digit==0:
break
thisProd += [digit]
else:
thisProd[-1] = thisProd[-1]*10+digit
answer = math.prod(thisProd)
if not answer%10==xx[0]:
continue
if digits(answer)[-1::-1]==xx:
print('\r'+str(thisProd).replace(', ', 'x')[1:-1])
return
return
i=0
while True:
i += 1
if not i%10000:
print('\r'+str(i), end='')
check(i)
CROSSREFS
A267939 is contained in the intersection of this sequence and A281555.
Sequence in context: A261640 A292990 A281555 * A267939 A355973 A092374
KEYWORD
nonn,base
AUTHOR
David L. Reens, Jun 01 2023
STATUS
approved