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A207826
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Upper right triangle: Fill columns with the smallest possible positive integers not occurring earlier and such that T[n+1,k] = |T[n,k-1]-T[n,k]| or T[n,k-1]+T[n,k]. Second version (see comment).
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4
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1, 2, 3, 4, 6, 9, 7, 11, 5, 14, 8, 15, 26, 21, 35, 10, 18, 33, 59, 38, 73, 13, 23, 41, 74, 133, 95, 22, 12, 25, 48, 89, 163, 30, 65, 43, 16, 28, 53, 101, 190, 27, 57, 122, 79, 20, 36, 64, 117, 218, 408, 381, 324, 202, 123, 19, 39, 75, 139, 256, 474, 66, 315, 639, 437, 314, 32, 51, 90, 165, 304, 560, 86, 152, 467, 172, 265, 49, 24, 56, 107, 17
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OFFSET
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1,2
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COMMENTS
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This "second version" is obtained by discarding a candidate for T[1,k] when the column cannot be filled in the "greedy way", without exploring all possibilities by tracing back earlier choices of |a-b| vs a+b, when one "gets stuck" somewhere down in the column (i.e., the sum as well as the absolute difference already occurred).
This differs from the "optimal" version A207831.
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LINKS
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EXAMPLE
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Start filling the columns of the triangle with 1, 2, 3=1+2 (because 2-1 already used), 4, 6=2+4 (because 4-2 already used), and 9=3+6 (because 6-3 already used):
1 2 4
. 3 6
. . 9
Then try T[1,4]=5, but this is not possible, since T[2,4] cannot be 4+5 nor 5-4 (both used). So try T[1,4]=7 (since 6 already used), which will allow us to fill the whole column (with 7+4=11 (since 7-4 already used), 11-6=5, 9+5=14 (since 9-5=4 already used).
See the Example in A207831 for the difference (occurring in the 25th column) with that triangle: since the greedy way of filling the column would not work with T[1,25]=A207829(25)=83, we have T[1,25]=A207827(25)=91 here.
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PROG
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(PARI) /* assuming that the vector A207827 with the first line of the triangle has already been computed */
{T=matrix( #A=A207827, #A); u=Set(T[1, ]=A); for(j=2, #T, for(i=2, j, setsearch( u, T[i, j]=abs(T[i-1, j-1]-T[i-1, j])) & T[i, j]=T[i-1, j-1]+T[i-1, j]; u=setunion( u, Set( T[i, j] ))))}
for(j=1, #T, for(i=1, j, print1(T[i, j]", ")))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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