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A125959
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Infinite array of nine columns, read by rows: A(j,k) = digital root of j*k for j >= 1, 1 <= k <= 9.
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0
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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, 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, 3, 2, 1, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 1, 2, 3, 4, 5, 6
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OFFSET
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1,2
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COMMENTS
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a(n) = digital root of (1+floor(n/9))*(1+((n-1) mod 9)).
Sequence is periodic, period length is 81. The 9 by 9 array of the first 81 terms is symmetric.
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 d-th element of the c-th row, or alternatively the c-th element of the d-th row, i.e. A(c,d) or A(d,c); commutativity of multiplication is reflected in the symmetry of the array.
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LINKS
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EXAMPLE
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Array begins:
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 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 4 2 9
8 7 6 5 4 3 2 1 9
9 9 9 9 9 9 9 9 9
Find the digital root of 197*799
(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.
(b) using the array: 197 mod 9 = 8, 799 mod 9 = 7; A(8,7) = A(7,8) = 2.
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PROG
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(PARI) {digitalroot(n) = if(n<1, 0, (n-1)%9+1)} {a(n) = digitalroot((1+floor(n/9))*(1+((n-1)%9)))} {for(n=1, 105, print1(a(n), ", "))} /* Klaus Brockhaus, Mar 28 2007 */
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CROSSREFS
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KEYWORD
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nonn,tabf,base,easy
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AUTHOR
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Simon Alexander (simonalexander2005(AT)hotmail.com), Feb 07 2007
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EXTENSIONS
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STATUS
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approved
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