A154781 Sum of all numbers < n that appear as substring of n, written in decimal system.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 2, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 3, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 5, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 6, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 7, 15, 16, 8, 9, 10, 11, 12, 13, 14, 15, 8, 17

Offset: 0

Author

M. F. Hasler, Jan 16 2009

Keywords

• base
• easy
• nonn

Comments

The condition "< n" narrows the meaning of "substring" to the strict sense, i.e., excludes n itself.

Examples

			Since n=0,...,9 has a single digit, only n itself appears as substring in n, thus a(n)=0.
10 has { 0, 1, 10 } as substrings, thus a(10) = 0+1 = 1.
11 has { 1, 11 } as substrings, thus a(11) = 1.
12 has { 1, 2, 12 } as substrings, thus a(12) = 1+2 = 3.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A154770, A154771.

Programs

Mathematica

san[n_]:=Total[Union[FromDigits/@Flatten[Table[Partition[IntegerDigits[n],i,1],{i,IntegerLength[n]-1}],1]]]; Array[san,90,0] (* Harvey P. Dale, May 27 2017 *)

PARI

A154781(n) = { local(d=#Str(n)); n=vecsort(concat(vector(d,i,vector(d,j,n%10^j)+(d--&!n\=10))),,12);n*vector(#n,i,i>1)~ }

Python

def a(n):
    s = str(n)
    return sum(set(int(s[i:j]) for i in range(len(s)) for j in range(i+1, len(s)+1))) - n
print([a(n) for n in range(90)]) # Michael S. Branicky, Nov 08 2022

Formula

a(n) = A154771(n)-n. a(10^n) = A002275(n). a(n)>0 <=> n>9.