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A117670
Triangle read by rows: partial sums of the Pascal triangle minus 1.
2
1, 2, 3, 3, 6, 7, 4, 10, 14, 15, 5, 15, 25, 30, 31, 6, 21, 41, 56, 62, 63, 7, 28, 63, 98, 119, 126, 127, 8, 36, 92, 162, 218, 246, 254, 255, 9, 45, 129, 255, 381, 465, 501, 510, 511, 10, 55, 175, 385, 637, 847, 967, 1012, 1022, 1023
OFFSET
1,2
COMMENTS
Imagine that you are in a building with floors starting at floor 1, the lowest floor and you have a large number of eggs. For each floor in the building, you want to know whether or not an egg dropped from that floor will break.
If an egg breaks when dropped from floor i, then all eggs are guaranteed to break when dropped from any floor j > i. Likewise, if an egg doesn't break when dropped from floor i, then all eggs are guaranteed to never break when dropped from any floor j <= i.
a(n,k) is the maximum number of floors where you can determine whether or not an egg will break when dropped from any floor, with the following restrictions: you may drop a maximum of n eggs (one at a time, from any floors of your choosing) and you may break a maximum of k eggs.
Each row of the triangle is the running sum of the corresponding row with the first 1 omitted of Pascal's triangle (A007318), see A008949, A054143, A193820.
The k-th entry in the n-th row is the number of possible combinations of on/off switches after k attempts to turn on a switch in a set of n distinguishable switches. An attempt to turn on the same switch twice does not result in a new combination. See example. - Sergei Viznyuk, Jun 24 2012
T(n,k) is the number of nonempty subsets of the n-set with at most k elements, see example. - Joerg Arndt, May 04 2014
LINKS
Google code.jam, Problem C. Egg Drop
Sergei Viznyuk, C-Program
Sergei Viznyuk, Local copy of C-Program
FORMULA
a(n,1) = n ; a(n,n) = 2^n-1; a(n+1,k+1) = 1 + a(n,k) + a(n,k-1), 0 < k < n.
a(n,k) = sum(binomial(n,m),m=1..k), 1 <= k <= n. (see the running sum comment above). - Wolfdieter Lang, Feb 07 2013
EXAMPLE
Triangle a(n,k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 ...
1: 1
2: 2 3
3: 3 6 7
4: 4 10 14 15
5: 5 15 25 30 31
6: 6 21 41 56 62 63
7: 7 28 63 98 119 126 127
8: 8 36 92 162 218 246 254 255
9: 9 45 129 255 381 465 501 510 511
10: 10 55 175 385 637 847 967 1012 1022 1023
... Reformatted and extended by Wolfdieter Lang, Feb 07 2013
From Sergei Viznyuk, Jun 24 2012: (Start)
For example, we have n=3 distinguishable switches A,B,C (third row above). We attempt k=2 times to turn on a switch at random. The possible resulting combinations are:
A=on, B=off, C=off (the same A switch was turned on 2 times)
A=off, B=on, C=off (the same B switch was turned on 2 times)
A=off, B=off, C=on (the same C switch was turned on 2 times)
A=on, B=on, C=off (switches A and B were turned on)
A=on, B=off, C=on (switches A and C were turned on)
A=off, B=on, C=on (switches B and C were turned on)
Thus, we have 6 different combinations, which is the number 6 at row n=3 column k=2 in the sequence above.
(End)
From Joerg Arndt, May 04 2014: (Start)
There are T(4,2) = 10 subsets of {0, 1, 2, 3}:
01: 1... { 0 }
02: 11.. { 0, 1 }
03: 111. { 0, 1, 2 }
04: 11.1 { 0, 1, 3 }
05: 1.1. { 0, 2 }
06: 1.11 { 0, 2, 3 }
07: 1..1 { 0, 3 }
08: .1.. { 1 }
09: .11. { 1, 2 }
10: .111 { 1, 2, 3 }
11: .1.1 { 1, 3 }
12: ..1. { 2 }
13: ..11 { 2, 3 }
14: ...1 { 3 }
(End)
MATHEMATICA
Table[Sum[Binomial[n, m], {m, k}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 25 2015 *)
PROG
(PARI) tabl(nrows) = {for (n=1, nrows, for (k=1, n, print1(sum(m=1, k, binomial(n, m)), ", "); ); print(); ); } \\ Michel Marcus, May 21 2013
CROSSREFS
Sequence in context: A251729 A187763 A187262 * A368253 A368260 A368262
KEYWORD
nonn,tabl
AUTHOR
Arie Bos, Jul 06 2008, Jul 08 2008
STATUS
approved