A152464 -id:A152464 - cp's OEIS frontend

A152054 Bouncy numbers (numbers whose digits are in neither increasing nor decreasing order).

101, 102, 103, 104, 105, 106, 107, 108, 109, 120, 121, 130, 131, 132, 140, 141, 142, 143, 150, 151, 152, 153, 154, 160, 161, 162, 163, 164, 165, 170, 171, 172, 173, 174, 175, 176, 180, 181, 182, 183, 184, 185, 186, 187, 190, 191, 192, 193, 194, 195, 196

Offset: 1

Jerome Abela (Jerome.Abela(AT)gmail.com), Nov 22 2008

Complement of union of A009994 and A009996. - Ray Chandler, Oct 25 2011

Number of n-digit bouncy numbers is 9*10^(n-1) - (n+18)*binomial(n+8, 8)/9 + 10. - Altug Alkan, Oct 02 2018

David F. Marrs, Table of n, a(n) for n = 1..10000
Project Euler, Non-bouncy numbers Problem 113

Cf. A152464. - Jon E. Schoenfield, Dec 06 2008

Cf. A009994, A009996, A204692.

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 *)

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

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
A152054_list = list(islice(A152054_gen(),30)) # Chai Wah Wu, Jul 28 2023

More terms from Jon E. Schoenfield, Dec 06 2008

A343462 Number of n-digit positive integers that undulate.

Original entry on oeis.org

9, 81, 525, 3105, 18939, 114381, 693129, 4195557, 25405586, 153820395, 931359050, 5639156409, 34143908573, 206733865761, 1251728824798, 7578945799704, 45888871327435, 277847147039527, 1682304127857000, 10185986079451152, 61673933253012813, 373422269794761171, 2260990733622821388

Offset: 1

David A. Corneth, Apr 16 2021

This is also the number of (2*n-1)-digit palindromes that undulate.

Classifying undulating numbers with n digits in initial digits and sign of first digit - second digit eases computation.

			a(2) = 81 as there are 90 2-digit positive integers (10, 11, ..., 99). Of those, 11, 22, ..., 99 do not undulate as there is a pair of consecutive digits that are equal. There are nine nonundulating 2-digit numbers, leaving 90-9 = 81 that do undulate.
134 does not undulate as there are two pairs of consecutive digits where the right one is in both cases either smaller or larger. (In this case 1 < 3 and 3 < 4.)
143 does undulate since 1 < 4 and 4 > 3.

Eric Weisstein's World of Mathematics, Undulating Number.
Index entries for linear recurrences with constant coefficients, signature (5,10,-20,-15,21,7,-8,-1,1).

Cf. A057332. Apart from the first 2 terms, the same as A152464.

first(n) = { my(res = vector(n), vup, vdown, nvup, nvdown); res[1] = 9; vup = vector(9, i, 1); vdown = vector(9, i, 1); for(i = 2, n, nvup = vector(9); nvdown = vector(9); nvdown[1] = vdown[9]; for(i = 2, 9, nvdown[i] = nvdown[i-1]+vup[i-1] ); for(i = 1, 8, nvup[i] = nvdown[9-i] ); vup = nvup; vdown = nvdown; res[i] = vecsum(vup)+vecsum(vdown)); res }

def aupton(terms):
  up, dn, alst = [0] + [1]*9, [0] + [1]*9, [9]
  for n in range(2, terms+1):
    up_next = [sum(dn[j] for j in range(i)) for i in range(10)]
    dn_next = [sum(up[j] for j in range(i+1, 10)) for i in range(10)]
    up, dn = up_next, dn_next
    alst.append(sum(up + dn))
  return alst
print(aupton(22)) # Michael S. Branicky, Apr 16 2021

# alternate program as a linear system
import numpy as np
from sympy import Matrix
def aupton(terms):
  x = Matrix([0] + [1]*9 + [0] + [1]*9)
  c = Matrix([[1]*20])
  z10 = np.zeros((10, 10), dtype=np.int64)
  o10 = np.ones((10, 10), dtype=np.int64)
  A = Matrix(np.block([[z10, np.tril(o10, -1)], [np.triu(o10, +1), z10]]))
  alst = [9]
  for n in range(2, terms+1):
    x = A*x
    alst.append((c*x)[0])
  return alst
print(aupton(22)) # Michael S. Branicky, Apr 16 2021

a(n) = 45*a(n-2) - 330*a(n-4) + 924*a(n-6) - 1287*a(n-8) + 1001*a(n-10) - 455*a(n-12) + 120*a(n-14) - 17*a(n-16) + a(n-18) for n >= 20. - Michael S. Branicky, Apr 17 2021
From Chai Wah Wu, Apr 24 2021: (Start)
a(n) = 5*a(n-1) + 10*a(n-2) - 20*a(n-3) - 15*a(n-4) + 21*a(n-5) + 7*a(n-6) - 8*a(n-7) - a(n-8) + a(n-9) for n > 10.
G.f.: x*(-8*x^9 + 7*x^8 + 63*x^7 - 45*x^6 - 162*x^5 + 81*x^4 + 150*x^3 - 30*x^2 - 36*x - 9)/(x^9 - x^8 - 8*x^7 + 7*x^6 + 21*x^5 - 15*x^4 - 20*x^3 + 10*x^2 + 5*x - 1). (End)

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A152054 Bouncy numbers (numbers whose digits are in neither increasing nor decreasing order).

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A343462 Number of n-digit positive integers that undulate.

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