|
Search: id:A126256
|
|
|
| A126256 |
|
Number of distinct terms in rows 0 through n of Pascal's triangle. |
|
+0
2
|
|
| 1, 1, 2, 3, 5, 7, 9, 12, 16, 20, 24, 29, 35, 41, 48, 53, 60, 68, 77, 86, 95, 103, 114, 125, 137, 149, 162, 175, 188, 202, 217, 232, 248, 264, 281, 297, 314, 332, 351, 370, 390, 410, 431, 452, 474, 495, 518, 541, 565, 589, 614, 639, 665, 691, 718, 744, 770, 798
(list; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
An easy upper bound is 1 + floor(n^2/4) = A033638(n). First differences are in A126257.
|
|
LINKS
|
N. Hobson, Table of n, a(n) for n = 0..1000
N. Hobson, Home page (listed in lieu of email address)
Nick Hobson, Python program for this sequence
|
|
EXAMPLE
|
There are 9 distinct terms in rows 0 through 6 of Pascal's triangle (1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1); hence a(6)=9.
|
|
MAPLE
|
seq(nops(`union`(seq({seq(binomial(n, k), k=0..n)}, n=0..m))), m=0..57); - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 26 2007
|
|
PROGRAM
|
(PARI) lim=57; z=listcreate(1+lim^2\4); for(n = 0, lim, for(r=1, n\2, s=Str(binomial(n, r)); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); print1(1+#z, ", "))
|
|
CROSSREFS
|
Cf. A007318, A027424, A061786, A126254-A126257.
Sequence in context: A133231 A074752 A039825 this_sequence A062438 A102424 A080000
Adjacent sequences: A126253 A126254 A126255 this_sequence A126257 A126258 A126259
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Nick Hobson Dec 24 2006
|
|
|
Search completed in 0.003 seconds
|
