A282473 Multiples of 9 which cannot be expressed as the difference between a natural number k and its digit sum s(k).

90, 189, 288, 387, 486, 585, 684, 783, 882, 981, 990, 1089, 1188, 1287, 1386, 1485, 1584, 1683, 1782, 1881, 1980, 1989, 2088, 2187, 2286, 2385, 2484, 2583, 2682, 2781, 2880, 2979, 2988, 3087, 3186, 3285, 3384, 3483, 3582, 3681, 3780, 3879, 3978, 3987, 4086, 4185, 4284, 4383, 4482, 4581, 4680, 4779, 4878, 4977, 4986

Offset: 1

Author

Sharvil Kesarwani, Feb 16 2017

Keywords

• nonn
• base
• easy

Comments

Based on empirical observations, it can be noted that the difference between consecutive terms is usually 99. However, this breaks down when the hundreds digit is 9, in which case the difference between consecutive terms is 9. This changes back, however.

Examples

			If k=90, then k-s(k)=81. If k=100, then k-s(k)=99. This is an increasing function, so k-s(k)=90 is unachievable.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Brilliant.org, Those are some nifty numbers
• Australian Mathematics Olympiad, Question 2, 2015, p. 84.

Crossrefs

Cf. A003052, A176995.

Programs

Mathematica

okQ[n_] := Catch[ Do[ If[ x- Total@ IntegerDigits@ x == n, Throw@ False], {x, n, n+ 9 IntegerLength[n]}]; True]; Select[9 Range[1000], okQ[#] &] (* Giovanni Resta, Feb 27 2017 *)

PARI

is(n)=for(k=n,n+9*#Str(n)+9, if(k-sumdigits(k)==n, return(0))); n%9==0 \\ Charles R Greathouse IV, Feb 27 2017

Python

from math import ceil
def a(n): #Outputs all numbers less than n which are in the sequence
    def s(n):
        r = 0
        while n:
            r, n = r + n% 10, n//10
        return r
    mult9=[]
    if n%9==0:
        for x in range(1, ceil(n/9)+1):
            mult9.append(9*x)
    else:
        for x in range(1, ceil(n/9)):
            mult9.append(9*x)
    for y in range(1, ceil(n/10)+1):
        mult9.remove(10*y-s(10*y))
    return mult9