A342387 Integers x such that there exists y, x>y>=1, satisfying (y+1)/(x+1) = (x||(y+1))/(y||(x+1)) where z||t is the concatenation of z and t.

20, 175, 1500, 29600, 253075, 1124039, 2163720, 1620864179, 3120083460, 13857908224, 118481007099, 2337285022799, 19983094049524, 170849530073079, 28815607761506104, 127985053235771120, 246364903884373539, 1094234263598927875, 184554358010701244300, 1577885049278315692375

Offset

1,1

Links

Chai Wah Wu, Table of n, a(n) for n = 1..949

Josep M. Brunat and Joan-Carles Lario, A problem on concatenated integers, arXiv:2103.05306 [math.NT], 2021.

Example

x=20 is a term with y=6 because 7/21 = 207/621.

Prog

(PARI) isok(x) = {for (y=1, x-1, if ((y+1)/(x+1) == eval(Str(x, y+1))/eval(Str(y, x+1)), return (y)); ); }
for (x=1, 10000, if (isok(x), print1(x, ", ")));
(Python)
A342387_list, x, s1, s2, m = [], 1, '1', '2', 10
while x < 10**6:
    for y in range(1, x):
        if (x+1)*int(s1+str(y+1)) == (y+1)*(y*m+x+1):
            A342387_list.append(x)
            break
    x += 1
    s1, s2 = s2, str(x+1)
    m = 10**(len(s2)) # Chai Wah Wu, Mar 10 2021
(Python)
# based on formula in Brunat and Lario 2021
xlist, ylist, A342387_list, x, y = [4, 20, 39], [1, 6, 12], [20], 39, 12
while len(A342387_list) < 100:
    if len(str(x+1)) == len(str(y+1))+1:
        A342387_list.append(x)
    x, y = 19*xlist[-3]+60*ylist[-3]+39, 6*xlist[-3]+19*ylist[-3]+12
    xlist, ylist = xlist[1:] + [x], ylist[1:] + [y] # Chai Wah Wu, Mar 10 2021

Crossrefs

Cf. A342388.

Sequence in context: A027791 A047819 A163689 * A140044 A027332 A159538

Adjacent sequences: A342384 A342385 A342386 * A342388 A342389 A342390

Keyword

nonn,base

Author

Michel Marcus, Mar 10 2021

Extensions

More terms from Chai Wah Wu, Mar 10 2021

Status

approved