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A367593 Nonnegative integers k such that (R(k) - 1)/(k + 1) is an integer, where R(k) is the digit reversal of k.

%I A367593 #55 Aug 11 2024 16:53:03
%S A367593 0,1,10,100,147,1000,1099,1407,10000,14007,100000,140007,1000000,
%T A367593 1400007,2124736,10000000,14000007,100000000,123456789,140000007,
%U A367593 1000000000,1234506789,1400000007,10000000000,12345006789,14000000007,21247524736,100000000000,123450006789
%N A367593 Nonnegative integers k such that (R(k) - 1)/(k + 1) is an integer, where R(k) is the digit reversal of k.
%C A367593 Among the terms of this sequence, there are:
%C A367593   the powers of 10 (cf. A011557);
%C A367593   the numbers of the form 14*10^h + 7 = 7*A199682(h) for h > 0;
%C A367593   the numbers of the form 12345*10^m + 6789 = A367650(m) for m > 3.
%C A367593 From _Chai Wah Wu_, Dec 01 2023: (Start)
%C A367593 First digit of positive terms must be either 1 or 2. If first digit is 1, then last digit must be 0,1,4,5,7 or 9. If first digit is 2, then last digit is 6. In particular, if a, b, c are the first digit, last digit and (R(k)-1)/(k+1) of a term k>0, then (a, b, c) must take on values from one of the following triples:
%C A367593   (a, b, c): (1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 4, 4), (1, 5, 5), (1, 7, 5),
%C A367593   (1, 9, 4), (1, 9, 5), (1, 9, 6), (1, 9, 7), (1, 9, 8), (1, 9, 9), (2, 6, 3).
%C A367593 Numbers of the form 21 [2475]* 24736 are terms of this sequence, where [2475]* denote a (possibly zero) repetition of the digits 2475. The first few terms of this form are: 2124736, 21247524736, 212475247524736, ...
%C A367593 Similarly numbers of the form 21261 [2475]* 24738736 are terms of this sequence:
%C A367593   2126124738736, 21261247524738736, 212612475247524738736, ... (End)
%C A367593 More generally, numbers of the form 21 [261]^k [2475]* 2473 [873]^k 6 are terms of this sequence, where [261]^k denote the digits '261' repeated k times with k>=0: e.g. 21261261247524752475247524738738736, ... It appears that all terms with first digit 2 and last digit 6 are of this form. - _Chai Wah Wu_, Dec 02 2023
%H A367593 Chai Wah Wu, <a href="/A367593/b367593.txt">Table of n, a(n) for n = 1..44</a>
%F A367593 A367728(a(n)) = 1.
%e A367593 123456789 is a term since (987654321 - 1)/(123456789 + 1) = 8, which is an integer.
%t A367593 a={}; For[k=0, k<=10^10, k++,If[IntegerQ[(FromDigits[Reverse[IntegerDigits[k]]]-1)/(k+1)],AppendTo[a,k]]]; a
%t A367593 Select[Range[0,10^6],IntegerQ[(IntegerReverse[#]-1)/(#+1)]&] (* The program generates the first 13 terms of the sequence. *) (* _Harvey P. Dale_, Aug 11 2024 *)
%o A367593 (Python)
%o A367593 def digit_reversal(n):
%o A367593     return int(str(n)[::-1])
%o A367593 def find_integers():
%o A367593     result = []
%o A367593     for k in range(0, 10**10):
%o A367593         reversed_k = digit_reversal(k)
%o A367593         if (reversed_k - 1) % (k + 1) == 0:
%o A367593             result.append(k)
%o A367593     return result
%o A367593 integers_list = find_integers()
%o A367593 print(integers_list)
%o A367593 (Python)
%o A367593 from itertools import product, count, islice
%o A367593 def A367593_gen(): # generator of terms
%o A367593     yield from (0,1,10)
%o A367593     for l in count(1):
%o A367593         m = 10**(l+1)
%o A367593         for d in product('0123456789',repeat=l):
%o A367593             for a, b, c in ((1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 5, 5), (1, 7, 5)):
%o A367593                 k = a*m+int(s:=''.join(d))*10+b
%o A367593                 r = b*m+int(s[::-1])*10+a
%o A367593                 if c*(k+1)==r-1:
%o A367593                     yield k
%o A367593             a,b = 1,9
%o A367593             k = a*m+int(s:=''.join(d))*10+b
%o A367593             r = b*m+int(s[::-1])*10+a
%o A367593             if not (r-1)%(k+1):
%o A367593                 yield k
%o A367593         a,b,c=2,6,3
%o A367593         for d in product('0123456789',repeat=l):
%o A367593             k = a*m+int(s:=''.join(d))*10+b
%o A367593             r = b*m+int(s[::-1])*10+a
%o A367593             if c*(k+1)==r-1:
%o A367593                 yield k
%o A367593 A367593_list = list(islice(A367593_gen(),20)) # _Chai Wah Wu_, Dec 01 2023
%o A367593 (PARI) isok(k) = denominator((fromdigits(Vecrev(digits(k))) - 1)/(k + 1)) == 1; \\ _Michel Marcus_, Nov 30 2023
%Y A367593 Cf. A004086, A011557, A199682, A367650.
%Y A367593 Cf. A367727, A367728, A367740.
%K A367593 nonn,base
%O A367593 1,3
%A A367593 _Stefano Spezia_, Nov 24 2023