## Sequences## Contains- Introduction
- A067957: 1, 1, 2, 2, 4, 5, 7, 7, 24, 22, 29, 39, 67, 55, 386, ...
- A093387: 1, 2, 6, 12, 29, 58, 130, 260, 562, ...
- A093388: 1, 6, 42, 312, 2394, 18756, 149136, ...
- A094062: 1, 2, 6, 22, 83, ...
- A098853: 0, 1, 2, 4, 6, 18, 36, 12, 30, 162, 18, 330, 136, 858, 1092, 198, ...
- New: 3, 4, 13, 14, 76, 40, 459, ...
## IntroductionAt april 17, 2004, during the NMC2004, Neil Sloane presented a lecture about sequences and his online integer sequences site. After the lecture I spoke to Neil, and I promised him to send some missing sequences. These sequences can be found below. If you would like to contact me about these sequences, you can send a message to matthijsssequences. Unfortunately I have not so much time to do research. I have ideas for a lot of new projects and I found a lot of unknown sequences.## A067957describes the number of permutations a_1,a_2,...,a_n of 1,2,...,n such that for all j=1,2,...,n: a_j|SUM_{i=1}^{i=j}a_i.NEW_2 : 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 6, 1, 11, 9, 15, 14, 14, 23, ... describes the number of the subsets of permutations starting with 2.
NEW_3 : 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 12, 20, 20, 1, 163, 55, ... describes the number of the subsets of permutations starting with 3.
NEW_4 : 0, 0, 0, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 6, 8, 3, 14, 12, 18, 13, 14, 6, 26, 13, 198, 152, 220, 118, 1033, 807,... describes the number of the subsets of permutations starting with 4.
## A0933870, 0, 1, 2, 6, 12, 29, 58, 130, 260, 562, 1124, 2380, 4760, 9949, 19898, 41226, 82452, 169766, 339532, 695860, 1391720, 2842226, 5684452, 11576916, 23153832, 47050564, 94101128, 190876696, 381753392, 773201629, 1546403258, 3128164186, 6256328372, 126423015342^(n-1)-binomial(n, floor(n/2),
statements.zip ## A0933881, 6, 42, 312, 2394, 18756, 149136, 1199232, 9729882, 79527084, 654089292, 5408896752, 44941609584, 375002110944, 3141107339328, 26402533581312, 222635989516122, 1882882811380284, 15967419789558804, 135752058036988848, 1156869080242393644(n+1)^2 a_{n+1} = (17n^2+17n+6) a_n - 72n^2 a_{n-1}, this result was written down in my (master)thesis, august 26, 1983, unfortunately unpublished.
## A094062A camel can carry one banana at a time on his back. It is on diet and therefore can only have one banana at a time in its stomach. As soon as it has eaten a banana it walks a mile and then needs a new banana (in order to be able to continue its iterary).Let there be a stock of N bananas at the border of the
desert. How far can the camel penetrate into the desert? (Of
course it can form new stocks with transported bananas.)
Question: How many bananas are needed to penetrate (at least) k miles? We get the sequence : 1, 2, 6, 22, 83, ...
camel.zip
I think that $a(n) := ceil((3-sqrt(3))*4^(n-2)+1)$ is a better formula than your formula $ceil((3-sqrt(3))*4^(n-3))+1$. Another formula which looks better is $b(n) := ceil((3-sqrt(3))*4^(n-2-1/16^(n-3/2))+1)$. See Original text (in Dutch).
The problem can be reformulated in terms of a jeep and complete refills. See http://www.inf.fu-berlin.de/inst/ag-ti/publications/rote.en.html#AREA01 Optimal logistics for expeditions-the jeep problem with complete refilling. ## A098853At october 11, 2004 I submitted a sequence aboutEgyptian fractions.
Let the Egyptian algorithm applied on the quotient p/q be defined by
- substract the largest quotient 1/
*n*smaller or equal then*p*/*q*. - if not zero go to (1).
q such that on n/q the Egyptian algorithm can be applied n times.
Since the n-th term is of the form k*n+1, only k has been written down.
0, 1, 2, 4, 6, 18, 36, 12, 30, 162, 18, 330, 136, 858, 1092, 198, 1470 I would like to thank Greg Martin for his reply on this sequence.
Suppose you have to solve a crossword-puzzle of |