A126257 Number of distinct new terms in row n of Pascal's triangle.

1, 0, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 6, 7, 5, 7, 8, 9, 9, 9, 8, 11, 11, 12, 12, 13, 13, 13, 14, 15, 15, 16, 16, 17, 16, 17, 18, 19, 19, 20, 20, 21, 21, 22, 21, 23, 23, 24, 24, 25, 25, 26, 26, 27, 26, 26, 28, 29, 29, 30, 30, 31, 31, 32, 32, 32, 33, 34, 34, 34, 35, 36, 36, 37, 37, 38

Offset

0,5

Comments

Partial sums are in A126256.

n occurs a(n) times in A265912. - Reinhard Zumkeller, Dec 18 2015

Links

Nick Hobson, Table of n, a(n) for n = 0..1000

Nick Hobson, Python program for this sequence

Example

Row 6 of Pascal's triangle is: 1, 6, 15, 20, 15, 6, 1. Of these terms, only 15 and 20 do not appear in rows 0-5. Hence a(6)=2.

Prog

(PARI) lim=77; z=listcreate(1+lim^2\4); print1(1, ", "); r=1; for(a=1, lim, for(b=1, a\2, s=Str(binomial(a, b)); f=setsearch(z, s, 1); if(f, listinsert(z, s, f))); print1(1+#z-r, ", "); r=1+#z)
(Haskell)
import Data.List.Ordered (minus, union)
a126257 n = a126257_list !! n
a126257_list = f [] a034868_tabf where
   f zs (xs:xss) = (length ys) : f (ys `union` zs) xss
                   where ys = xs `minus` zs
-- Reinhard Zumkeller, Dec 18 2015
(Python)
def A126257(n):
    if n:
        s, c = (1, ), {1}
        for i in range(n-1):
            c.update(set(s:=(1, )+tuple(s[j]+s[j+1] for j in range(len(s)-1))+(1, )))
        return len(set((1, )+tuple(s[j]+s[j+1] for j in range(len(s)-1))+(1, ))-c)
    return 1 # Chai Wah Wu, Oct 17 2023

Crossrefs

Cf. A007318, A062854, A126254, A126255, A126256.

Cf. A034868, A265912.

Sequence in context: A194165 A004524 A265409 * A025773 A308303 A285763

Adjacent sequences: A126254 A126255 A126256 * A126258 A126259 A126260

Keyword

easy,nonn

Author

Nick Hobson, Dec 24 2006

Status

approved