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A100452
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Triangle read by rows, based on array described below.
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6
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1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 19, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 39, 40, 42, 44, 45, 48, 49, 49, 50, 51, 52, 55, 60, 63, 64, 63, 64, 66, 68, 70, 72, 77, 80, 81, 79, 80, 81, 84, 85, 90, 91, 96, 99, 100, 91, 92, 93, 96, 100, 102, 105, 112, 117, 120, 121
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graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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The interesting property of this array is that the main diagonal gives A000960.
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LINKS
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H. Killingbergtro and C. U. Jensen, Problem 116, Nord. Mat. Tidskr. 5 (1957), 160-161.
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FORMULA
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Form an array a(m,n) (n >= 1, 1 <= m <= n) by: a(1,n) = n^2 for all n; a(m+1,n) = (n-m)*floor( (a(m,n)-1)/(n-m) ) for 1 <= m <= n-1.
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EXAMPLE
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Array begins:
1 4 9 16 25 36 49 64 81 100 ...
3 8 15 24 35 48 63 80 99 ...
7 14 21 32 45 60 77 96 ...
13 20 30 44 55 72 91 ...
19 28 42 52 70 90 ...
and triangle begins:
1
3 4
7 8 9
13 14 15 16
19 20 21 24 25
27 28 30 32 35 36
...
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MATHEMATICA
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max=11; a[1, n_]:= n^2;
a[m_, n_]/; 1<m<=n := a[m, n]= (n-m+1)*Floor@((a[m-1, n] -1)/(n-m+1));
a[_, _]=0;
t= Table[a[m, n], {m, max}, {n, m, max}];
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PROG
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(Magma)
if k eq 1 then return n^2;
else return (n-k+1)*Floor((t(n, k-1) -1)/(n-k+1));
end if;
end function;
[t(n, n-k+1): k in [1..n], n in [1..15]]; // G. C. Greubel, Apr 07 2023
(SageMath)
if (k==1): return n^2
else: return (n-k+1)*((t(n, k-1) -1)//(n-k+1))
flatten([[t(n, n-k+1) for k in range(1, n+1)] for n in range(1, 16)]) # G. C. Greubel, Apr 07 2023
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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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