A095026 - OEIS

%I #9 Jun 19 2015 08:11:55

%S 1,3,1,5,2,2,8,2,4,2,9,4,4,4,4,15,2,6,5,6,2,13,6,6,6,6,6,6,20,4,8,4,

%T 12,4,8,4,21,6,6,12,6,6,12,6,6,27,4,12,4,12,9,12,4,12,4,21,10,10,10,

%U 10,10,10,10,10,10,10,40,4,8,10,16,4,20,4,16,10,8,4,25,12,12,12,12,12,12,12

%N Lower triangle T(j,k) read by rows, where T(j,k) is the number of occurrences of the digit k-1 as least significant digit in the base-j multiplication table.

%C Sum_{k=1..j} T(j,k) = j^2.

%C Assumes a suitable continuation of the representation of digits in bases 11, 12 (9,A,B,..)

%H David Book, <a href="http://pleacher.com/mp/probweek/p2001/a012201.html">The Multiplying Digits Problem.</a>

%e a(2)=T(2,1)=3 because 3 of the 4 possible combinations of last digits in the

%e product of two binary numbers produce 0 as last digit of the result. a(3)=T(2,2)=1 because only ...1 * ...1 gives a result with last digit=1.

%e T(10,k)={27,4,12,4,12,9,12,4,12,4} gives the probability in percent (j^2=100) to get {0,1,2,...,9} as last decimal digit in the decimal representation of the product of two arbitrary integers.

%Y The first column T(n, 1)=A018804(n).

%K nonn,tabl,base

%O 1,2

%A _Hugo Pfoertner_, Jun 02 2004

%E More terms from _David Wasserman_, Jun 03 2004