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Number of even entries in row n of Pascal's triangle (A007318).
20

%I #39 Jul 27 2023 11:03:33

%S 0,0,1,0,3,2,3,0,7,6,7,4,9,6,7,0,15,14,15,12,17,14,15,8,21,18,19,12,

%T 21,14,15,0,31,30,31,28,33,30,31,24,37,34,35,28,37,30,31,16,45,42,43,

%U 36,45,38,39,24,49,42,43,28,45,30,31,0,63,62,63,60,65,62,63,56,69,66,67

%N Number of even entries in row n of Pascal's triangle (A007318).

%C In rows 2^k - 1 all entries are odd.

%C a(n) = 0 (all the entries in the row are odd) iff n = 2^m - 1 for some m >= 0 and then n belongs to sequence A000225. - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001

%C Also number of zeros in n-th row of SierpiƄski's triangle (cf. A047999): a(n) = A023416(A001317(n)). - _Reinhard Zumkeller_, Nov 24 2012

%H Seiichi Manyama, <a href="/A048967/b048967.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1000 from T. D. Noe)

%F a(n) = n+1 - A001316(n) = n+1 - 2^A000120(n) = n+1 - Sum_{k=0..n} (C(n, k) mod 2) = Sum_{k=0..n} ((1 - C(n, k)) mod 2).

%F a(2n) = a(n) + n, a(2n+1) = 2a(n). - _Ralf Stephan_, Oct 07 2003

%F a(n) = row sums in A219463 = A000120(A219843(n)). - _Reinhard Zumkeller_, Nov 30 2012

%F A249304(n+1) = a(n+1) + a(n). - _Reinhard Zumkeller_, Nov 14 2014

%F G.f.: 1/(1 - x)^2 - Product_{k>=0} (1 + 2*x^(2^k)). - _Ilya Gutkovskiy_, Jul 19 2019

%e Row 4 is 1 4 6 4 1 with 3 even entries so a(4)=3.

%t Table[n + 1 - Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]

%t a[n_] := n + 1 - 2^DigitCount[n, 2, 1]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 27 2023 *)

%o (PARI) a(n)=if(n<1,0,if(n%2==0,a(n/2)+n/2,2*a((n-1)/2)))

%o (Haskell)

%o import Data.List (transpose)

%o a048967 n = a048967_list !! n

%o a048967_list = 0 : xs where

%o xs = 0 : concat (transpose [zipWith (+) [1..] xs, map (* 2) xs])

%o -- _Reinhard Zumkeller_, Nov 14 2014, Nov 24 2012

%o (Python)

%o def A048967(n): return n+1-(1<<n.bit_count()) # _Chai Wah Wu_, May 03 2023

%Y Cf. A007318, A001316, A000120, A000225, A038573.

%Y Cf. A249304.

%K easy,nonn

%O 0,5

%A _Brian Galebach_