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A055129
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Repunits in different bases: table by antidiagonals of numbers written in base k as a string of n 1's.
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15
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1, 1, 2, 1, 3, 3, 1, 4, 7, 4, 1, 5, 13, 15, 5, 1, 6, 21, 40, 31, 6, 1, 7, 31, 85, 121, 63, 7, 1, 8, 43, 156, 341, 364, 127, 8, 1, 9, 57, 259, 781, 1365, 1093, 255, 9, 1, 10, 73, 400, 1555, 3906, 5461, 3280, 511, 10, 1, 11, 91, 585, 2801, 9331, 19531, 21845, 9841, 1023, 11
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OFFSET
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1,3
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LINKS
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FORMULA
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T(n, k) = (k^n-1)/(k-1) [with T(n, 1) = n] = T(n-1, k)+k^(n-1) = (k+1)*T(n-1, k)-k*T(n-2, k) [with T(0, k) = 0 and T(1, k) = 1].
T(n,k) = 1 + k * T(n-1,k) for k > 0 and n > 1.
Sum_{m=2..n} T(m-1,k)/Product_{i=2..m} T(i,k) = (1 - 1/Product_{i=2..n} T(i,k))/k for k > 0 and n > 1.
Sum_{n > 1} T(n-1,k)/Product_{i=2..n} T(i,k) = 1/k for k > 0.
Sum_{i=1..n} k^(i-1) / (T(i,k) * T(i+1,k)) = T(n,k) / T(n+1,k) for k > 0 and n > 0. (End)
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EXAMPLE
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T(3,5)=31 because 111 base 5 represents 25+5+1=31.
1 1 1 1 1 1 1
2 3 4 5 6 7 8
3 7 13 21 31 43 57
4 15 40 85 156 259 400
5 31 121 341 781 1555 2801
6 63 364 1365 3906 9331 19608
7 127 1093 5461 19531 55987 137257
Starting with the second column, the q-th column list the numbers that are written as 11...1 in base q. - John Keith, Apr 12 2021
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MAPLE
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add(k^j, j=0..n-1) ;
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MATHEMATICA
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Table[FromDigits[ConstantArray[1, #], k] &[n - k + 1], {n, 11}, {k, n, 1, -1}] // Flatten (* or *)
Table[If[k == 1, n, (k^# - 1)/(k - 1) &[n - k + 1]], {n, 11}, {k, n, 1, -1}] // Flatten (* Michael De Vlieger, Dec 11 2016 *)
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CROSSREFS
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Rows include A000012, A000027, A002061, A053698, A053699, A053700. Columns (see recurrence) include A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002275, A016123, A016125. Diagonals include A023037, A031973. Numbers in the table (apart from the first column and first two rows) are ordered in A053696.
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KEYWORD
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AUTHOR
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STATUS
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approved
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