A126256 Number of distinct terms in rows 0 through n of Pascal's triangle.

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

Offset: 0

Author

Nick Hobson, Dec 24 2006

Keywords

• easy
• nonn

Comments

An easy upper bound is 1 + floor(n^2/4) = A033638(n).

First differences are in A126257.

Examples

			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.
		

Links

• Nick Hobson, Table of n, a(n) for n = 0..1000
• Nick Hobson, Python program for this sequence

Crossrefs

Cf. A007318, A027424, A061786, A126254, A126255, A126257.

Cf. A199425.

Cf. A258318.

Programs

Haskell

-- import Data.List.Ordered (insertSet)
a126256 n = a126256_list !! n
a126256_list = f a007318_tabl [] where
   f (xs:xss) zs = g xs zs where
     g []     ys = length ys : f xss ys
     g (x:xs) ys = g xs (insertSet x ys)
-- Reinhard Zumkeller, May 26 2015, Nov 09 2011

Maple

seq(nops(`union`(seq({seq(binomial(n,k),k=0..n)},n=0..m))),m=0..57); # Emeric Deutsch, Aug 26 2007

Mathematica

Table[Length[Union[Flatten[Table[Binomial[n,k],{n,0,x},{k,0,n}]]]],{x,0,60}] (* Harvey P. Dale, Sep 10 2022 *)

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, ", "))

Python

def A126256(n):
    s, c = (1,), {1}
    for i in range(n):
        s = (1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + (1,)
        c.update(set(s))
    return len(c) # Chai Wah Wu, Oct 17 2023