%I
%S 1,2,3,4,5,6,7,8,9,2,4,6,8,1,3,5,7,9,3,6,9,3,6,9,3,6,9,4,8,3,7,2,6,1,
%T 5,9,5,1,6,2,7,3,8,4,9,6,3,9,6,3,9,6,3,9,7,5,3,1,8,6,2,4,9,8,7,6,5,4,
%U 3,2,1,9,9,9,9,9,9,9,9,9,9,1,2,3,4,5,6
%N Infinite array of nine columns, read by rows: A(j,k) = digital root of j*k for j >= 1, 1 <= k <= 9.
%C a(n) = digital root of (1+floor(n/9))*(1+((n1) mod 9)).
%C Sequence is periodic, period length is 81. The 9 by 9 array of the first 81 terms is symmetric.
%C To determine the digital root of a product a*b by means of this sequence (or rather by means of the 9 by 9 array) reduce both a and b modulo 9  if a result is zero replace it by 9  to obtain c and d, then choose the dth element of the cth row, or alternatively the cth element of the dth row, i.e. A(c,d) or A(d,c); commutativity of multiplication is reflected in the symmetry of the array.
%e Array begins:
%e 1 2 3 4 5 6 7 8 9
%e 2 4 6 8 1 3 5 7 9
%e 3 6 9 3 6 9 3 6 9
%e 4 8 3 7 2 6 1 5 9
%e 5 1 6 2 7 3 8 4 9
%e 6 3 9 6 3 9 6 3 9
%e 7 5 3 1 8 6 4 2 9
%e 8 7 6 5 4 3 2 1 9
%e 9 9 9 9 9 9 9 9 9
%e Find the digital root of 197*799
%e (a) customarily: digital root of 197*799 = digital root of (digital root of 197)*(digital root of 799) = digital root of 8*7 = digital root of 56 = 2.
%e (b) using the array: 197 mod 9 = 8, 799 mod 9 = 7; A(8,7) = A(7,8) = 2.
%o (PARI) {digitalroot(n) = if(n<1, 0, (n1)%9+1)} {a(n) = digitalroot((1+floor(n/9))*(1+((n1)%9)))} {for(n=1, 105, print1(a(n), ","))} /* Klaus Brockhaus, Mar 28 2007 */
%Y Cf. A010888 (digital root of n).
%K nonn,tabf,base,easy
%O 1,2
%A Simon Alexander (simonalexander2005(AT)hotmail.com), Feb 07 2007
%E Edited by _Klaus Brockhaus_, Mar 28 2007
