

A242388


Triangle read by rows: T(n,k) = n*2^(k1) + 1, 1 <= k <= n.


0



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

1,1


COMMENTS

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).


REFERENCES

H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner's Sons, New York, 1967, pp. 7, 228, #18.


LINKS

Table of n, a(n) for n=1..55.


FORMULA

T(n,1) = n + 1, T(n,k) = 2*T(n,k1)  1 for n >= 1 and 2 <= k <= n.


EXAMPLE

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
...


PROG

(PARI) T(n, k) = if(n > 0 && k > 0 && k <= n, n*2^(k1) + 1)


CROSSREFS

Cf. A036289.
Sequence in context: A028691 A246353 A194009 * A257985 A089557 A249328
Adjacent sequences: A242385 A242386 A242387 * A242389 A242390 A242391


KEYWORD

nonn,tabl


AUTHOR

Rick L. Shepherd, May 12 2014


STATUS

approved



