A258318 -id:A258318 - cp's OEIS frontend

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

Nick Hobson, Dec 24 2006

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

First differences are in A126257.

			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.

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

Cf. A007318, A027424, A061786, A126254, A126255, A126257.
Cf. A199425.
Cf. A258318.

-- 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

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

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

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, ", "))

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

A258197 Arithmetic derivative of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 4, 5, 4, 0, 0, 1, 7, 7, 1, 0, 0, 5, 8, 24, 8, 5, 0, 0, 1, 10, 12, 12, 10, 1, 0, 0, 12, 32, 92, 59, 92, 32, 12, 0, 0, 6, 60, 124, 165, 165, 124, 60, 6, 0, 0, 7, 39, 244, 247, 456, 247, 244, 39, 7, 0, 0, 1, 16, 103, 371, 493, 493, 371, 103, 16, 1, 0

Offset: 0

Reinhard Zumkeller, May 26 2015

A258318(n) = number of distinct terms up to row n.

			.  First 10 rows:                     |  Pascal's triangle:
.  0:  0                              |  1
.  1:  0,0                            |  1,1
.  2:  0,1,0                          |  1,2,1
.  3:  0,1,1,0                        |  1,3,3,1
.  4:  0,4,5,4,0                      |  1,4,6,4,1
.  5:  0,1,7,7,1,0                    |  1,5,10,10,5,1
.  6:  0,5,8,24,8,5,0                 |  1,6,15,20,15,6,1
.  7:  0,1,10,12,12,10,1,0            |  1,7,21,35,35,21,7,1
.  8:  0,12,32,92,59,92,32,12,0       |  1,8,28,56,70,56,28,8,1
.  9:  0,6,60,124,165,165,124,60,6,0  |  1,9,36,84,126,126,84,36,9,1 .

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A003415, A007318, A258290 (central terms), A258317 (row sums), A068312, A258318.

a258197 n k = a258197_tabl !! n !! k
a258197_row n = a258197_tabl !! n
a258197_tabl = map (map a003415) a007318_tabl

ad[n_] := n * Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); ad[0] = ad[1] = 0; Table[ad[Binomial[n, k]], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 13 2025 *)

T(n,k) = A003415(A007318(n,k)), 0 <= k <= n.
For n > 0: T(n,1) = A003415(n).
For n > 1: T(n,2) = A068312(n-1).

cp's OEIS Frontend

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

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