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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. 2
20, 175, 1500, 29600, 253075, 1124039, 2163720, 1620864179, 3120083460, 13857908224, 118481007099, 2337285022799, 19983094049524, 170849530073079, 28815607761506104, 127985053235771120, 246364903884373539, 1094234263598927875, 184554358010701244300, 1577885049278315692375 (list; graph; refs; listen; history; text; internal format)
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

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Last modified October 26 14:28 EDT 2021. Contains 348267 sequences. (Running on oeis4.)