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A242388
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Triangle read by rows: T(n,k) = n*2^(k-1) + 1, 1 <= k <= n.
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0
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2, 3, 5, 4, 7, 13, 5, 9, 17, 33, 6, 11, 21, 41, 81, 7, 13, 25, 49, 97, 193, 8, 15, 29, 57, 113, 225, 449, 9, 17, 33, 65, 129, 257, 513, 1025, 10, 19, 37, 73, 145, 289, 577, 1153, 2305, 11, 21, 41, 81, 161, 321, 641, 1281, 2561, 5121
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OFFSET
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1,1
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COMMENTS
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Sum of row n is n*2^n = A036289(n).
For n > 1, row n is the generalized solution to Dudeney's problem 18; he explicitly considered the case n = 9 only: The least (positive) initial distribution (of cents) amongst n people so that if each person gives each other person the same number (of cents) as the receiver currently has, then the final distribution results in each person having the same quantity (which turns out to be 2^n). If m > 0 (not necessarily integral), then m*Row n also results in an equal final distribution (with m*2^n each).
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REFERENCES
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H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner's Sons, New York, 1967, pp. 7, 228, #18.
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LINKS
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FORMULA
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T(n,1) = n + 1, T(n,k) = 2*T(n,k-1) - 1 for n >= 1 and 2 <= k <= n.
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EXAMPLE
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Row 3 of the triangle is 4, 7, 13. After the person with 13 gives 4 to the person with 4 and 7 to the person with 7, then 8, 14, 2 is the current distribution. Now, when the person with 14 gives away 8 and 2, respectively, 16, 4, 4 ensues. After the person with 16 then gives away 4 twice, 8, 8, 8 (= 2^3 each) is the final distribution.
Triangle starts:
2,
3, 5,
4, 7, 13,
5, 9, 17, 33,
6, 11, 21, 41, 81,
7, 13, 25, 49, 97, 193,
8, 15, 29, 57, 113, 225, 449,
9, 17, 33, 65, 129, 257, 513, 1025,
10, 19, 37, 73, 145, 289, 577, 1153, 2305,
11, 21, 41, 81, 161, 321, 641, 1281, 2561, 5121
...
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PROG
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(PARI) T(n, k) = if(n > 0 && k > 0 && k <= n, n*2^(k-1) + 1)
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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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