A051679 Total number of even entries in first n rows of Pascal's triangle (the zeroth and first rows being 1; 1,1).

0, 0, 1, 1, 4, 6, 9, 9, 16, 22, 29, 33, 42, 48, 55, 55, 70, 84, 99, 111, 128, 142, 157, 165, 186, 204, 223, 235, 256, 270, 285, 285, 316, 346, 377, 405, 438, 468, 499, 523, 560, 594, 629, 657, 694, 724, 755, 771, 816, 858, 901, 937, 982, 1020, 1059, 1083, 1132

Offset: 0

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)

Keywords

• easy
• nice
• nonn

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 6, 29.
• Eric Weisstein's World of Mathematics, Sierpinski Sieve
• Index entries for sequences generated by sieves

Crossrefs

Cf. A006046.

Partial sums of A048967. - Michel Marcus, Feb 16 2016

Programs

Mathematica

f[n_] := n + 1 - Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ]; Table[ Sum[ f[k], {k, 0, n} ], {n, 0, 100} ]
Accumulate[Count[#,?EvenQ]&/@Table[Binomial[n,k],{n,0,60},{k,0,n}]] (* _Harvey P. Dale, Nov 26 2014 *)

PARI

a(n)=if(n<2,0,if(n%2==0,3*a(n/2)+n/4*(n/2-1),2*a((n-1)/2)+a((n+1)/2)+((n-1)/4)*((n+1)/2)))

Formula

a(0)=a(1)=0, a(2n) = 3a(n)+n(n-1)/2, a(2n+1) = 2a(n)+a(n+1)+n(n+1)/2. - Ralf Stephan, Oct 10 2003

n(n+3)/2 - A074330(n). - Ralf Stephan, Oct 10 2003