A257297 a(n) = (initial digit of n) * (n with initial digit removed).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 0, 1, 2, 3

Offset: 0

Author

M. F. Hasler, May 10 2015

Keywords

• nonn
• base
• easy

Comments

The initial 100 terms match those of A035930 (maximal product of any two numbers whose concatenation is n) and also those of A171765 (product of digits of n, or 0 for n<10), and except for initial terms, also A007954 (product of decimal digits of n) and A115300 (greatest digit of n * least digit of n).

Iterations of this map always end in 0, since a(n) < n. Sequence A257299 lists numbers for which these iterations reach 0 in exactly 9 steps, with the additional constraint of having each time a different initial digit.

If "initial" is replaced by "last" in the definition (A257850), then we get the same values up to a(100), but (10, 20, 30, ...) for n = 101, 102, 103, ..., again different from each of the 4 other sequences mentioned in the first comment. - M. F. Hasler, Sep 01 2021

Examples

			For n<10, a(n) = n*0 = 0, since removing the initial and only digit leaves nothing, i.e., zero (by convention).
a(10) = 1*0 = 0, a(12) = 1*2 = 2, ..., a(20) = 2*0 = 0, a(21) = 2*1 = 2, a(22) = 2*2 = 4, ...
a(99) = 9*9 = 81, a(100) = 1*00 = 0, a(101) = 1*01 = 1, ..., a(123) = 1*23, ...
		

Crossrefs

Cf. A257850, A007954, A035930, A080464, A115300, A169669, A111707, A088117, A187844.

Programs

Maple

a:= n-> `if`(n<10, 0, (s-> parse(s[1])*parse(s[2..-1]))(""||n)):
seq(a(n), n=0..120);  # Alois P. Heinz, Feb 12 2024

Mathematica

Table[Times@@FromDigits/@TakeDrop[IntegerDigits@n,1],{n,0,103}] (* Giorgos Kalogeropoulos, Sep 03 2021 *)

PARI

apply( {A257297(n)=vecprod(divrem(n,10^logint(n+!n,10)))}, [0..111]) \\ Edited by M. F. Hasler, Sep 01 2021

Python

def a(n): s = str(n); return 0 if len(s) < 2 else int(s[0])*int(s[1:])
print([a(n) for n in range(104)]) # Michael S. Branicky, Sep 01 2021

Formula

For 1 <= m <= 9 and n < 10^k, a(m*10^k + n) = m*n.

Extensions

a(101..103) corrected by M. F. Hasler, Sep 01 2021