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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.
1
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
1,1
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).
LINKS
Pieter Post and Giovanni Resta, Table of n, a(n) for n = 1..295 (terms < 10^12, first 141 terms from Pieter Post)
EXAMPLE
114 = (1 + 5)*(14 + 5) and 114 = (20 - 1)*(20 - 14).
1710 = (17 + 28)*(10 + 28) and 1710 = (55 - 17)*(55 - 10).
PROG
(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)
(PARI)
is(n)=my(d=digits(n, 100), t); while((t=vecprod(d))<n, d+=vector(#d, i, 1)); t==n && n>99 \\ Charles R Greathouse IV, Aug 28 2015
CROSSREFS
Cf. A055482.
Sequence in context: A294309 A138693 A272303 * A228961 A057440 A113537
KEYWORD
base,nonn,less
AUTHOR
Pieter Post, Jul 19 2015
STATUS
approved