OFFSET
1,1
COMMENTS
Number of n-digit bouncy numbers is 9*10^(n-1) - (n+18)*binomial(n+8, 8)/9 + 10. - Altug Alkan, Oct 02 2018
LINKS
David F. Marrs, Table of n, a(n) for n = 1..10000
Project Euler, Non-bouncy numbers Problem 113
MATHEMATICA
Select[Range[0, 200], !LessEqual@@IntegerDigits[#] && !GreaterEqual@@IntegerDigits[#]&] (* Ray Chandler, Oct 25 2011 *)
bnQ[n_]:=Module[{didn=Differences[IntegerDigits[n]]}, Count[didn, _?(#>0&)]>0 && Count[didn, _?(#<0&)]>0]; Select[Range[100, 200], bnQ] (* Harvey P. Dale, Jun 13 2020 *)
PROG
(Python)
a = 1
b = 100
while a != 51:
if str(b) != ''.join(sorted(str(b))) and str(b) != ''.join(sorted(str(b)))[::-1]:
print(b)
a += 1
b += 1
# David F. Marrs, Sep 25 2018
(Python)
from itertools import count, islice
def A152054_gen(startvalue=1): # generator of terms >= startvalue
for n in count(max(startvalue, 1)):
l = len(s:=tuple(int(d) for d in str(n)))
for i in range(1, l-1):
if (s[i-1]-s[i])*(s[i]-s[i+1]) < 0:
yield n
break
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Jerome Abela (Jerome.Abela(AT)gmail.com), Nov 22 2008
EXTENSIONS
More terms from Jon E. Schoenfield, Dec 06 2008
STATUS
approved