A260213 Numbers j such that j = (c_1 + k)*(c_2 + k)*...*(c_m + k) for some k > 0 where c_1, c_2, ..., c_m is the centesimal expansion of j.

114, 120, 147, 198, 264, 420, 500, 506, 513, 525, 533, 550, 558, 568, 581, 648, 1102, 1116, 1168, 1302, 1320, 1377, 1680, 1692, 1710, 1720, 1734, 1755, 1771, 1872, 2106, 2132, 2310, 2332, 2380, 2664, 2714, 2736, 2790, 2914, 2940, 3312

Offset: 1

Author

Pieter Post, Jul 19 2015

Keywords

• base
• nonn
• less

Comments

k cannot be larger than 99, because in that case the product of the terms c_i+k is larger than the number j itself. For j up to 10^12, the highest value for k is 71, for 910602 = (71+91)*(71+6)*(71+2).

All terms j < 10000 have the following property. j = c_1//c_2, so j = (c_1 + k)*(c_2 + k). Let kk = c_1 + c_2 + k then j = (kk - c_1)*(kk - c_2). For example, 513 = (5 + 14)*(13 + 14), kk = 5 + 13 + 14 = 32, so 513 = (32 - 5)*(32 - 13).

Examples

			114 = (1 + 5)*(14 + 5) and 114 = (20 - 1)*(20 - 14).
1710 = (17 + 28)*(10 + 28) and 1710 = (55 - 17)*(55 - 10).
		

Links

• Pieter Post and Giovanni Resta, Table of n, a(n) for n = 1..295 (terms < 10^12, first 141 terms from Pieter Post)

Crossrefs

Cf. A055482.

Programs

PARI

is(n)=my(d=digits(n,100),t); while((t=vecprod(d))99 \\ Charles R Greathouse IV, Aug 28 2015

Python

def pod(n,m,a):
    kk = 1
    while n > 0:
        kk= kk*(n%m+a)
        n =int(n//m)
    return kk
for c in range (1,10000):
    for a in range (1,100):
        if c==pod(c,100,a):
            print (c)