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

0, 1, 10, 100, 147, 1000, 1099, 1407, 10000, 14007, 100000, 140007, 1000000, 1400007, 2124736, 10000000, 14000007, 100000000, 123456789, 140000007, 1000000000, 1234506789, 1400000007, 10000000000, 12345006789, 14000000007, 21247524736, 100000000000, 123450006789

Offset: 1

Author

Stefano Spezia, Nov 24 2023

Keywords

• nonn
• base

Comments

Among the terms of this sequence, there are:

the powers of 10 (cf. A011557);

the numbers of the form 14*10^h + 7 = 7*A199682(h) for h > 0;

the numbers of the form 12345*10^m + 6789 = A367650(m) for m > 3.

From Chai Wah Wu, Dec 01 2023: (Start)

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:

(a, b, c): (1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 4, 4), (1, 5, 5), (1, 7, 5),

(1, 9, 4), (1, 9, 5), (1, 9, 6), (1, 9, 7), (1, 9, 8), (1, 9, 9), (2, 6, 3).

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, ...

Similarly numbers of the form 21261 [2475]* 24738736 are terms of this sequence:

2126124738736, 21261247524738736, 212612475247524738736, ... (End)

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

Examples

			123456789 is a term since (987654321 - 1)/(123456789 + 1) = 8, which is an integer.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..44

Crossrefs

Cf. A004086, A011557, A199682, A367650.

Cf. A367727, A367728, A367740.

Programs

Mathematica

a={}; For[k=0, k<=10^10, k++,If[IntegerQ[(FromDigits[Reverse[IntegerDigits[k]]]-1)/(k+1)],AppendTo[a,k]]]; a
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 *)

PARI

isok(k) = denominator((fromdigits(Vecrev(digits(k))) - 1)/(k + 1)) == 1; \\ Michel Marcus, Nov 30 2023

Python

def digit_reversal(n):
    return int(str(n)[::-1])
def find_integers():
    result = []
    for k in range(0, 10**10):
        reversed_k = digit_reversal(k)
        if (reversed_k - 1) % (k + 1) == 0:
            result.append(k)
    return result
integers_list = find_integers()
print(integers_list)

Python

from itertools import product, count, islice
def A367593_gen(): # generator of terms
    yield from (0,1,10)
    for l in count(1):
        m = 10**(l+1)
        for d in product('0123456789',repeat=l):
            for a, b, c in ((1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 5, 5), (1, 7, 5)):
                k = a*m+int(s:=''.join(d))*10+b
                r = b*m+int(s[::-1])*10+a
                if c*(k+1)==r-1:
                    yield k
            a,b = 1,9
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            if not (r-1)%(k+1):
                yield k
        a,b,c=2,6,3
        for d in product('0123456789',repeat=l):
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            if c*(k+1)==r-1:
                yield k
A367593_list = list(islice(A367593_gen(),20)) # Chai Wah Wu, Dec 01 2023

Formula

A367728(a(n)) = 1.