A270438 a(n) is the number of entries == 1 mod 4 in row n of Pascal's triangle.

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 8, 2, 4, 4, 4, 4, 8, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 16, 4, 8, 8

Offset: 0

Author

Robert Israel, Jul 12 2016

Keywords

• nonn
• easy

Comments

All entries are powers of 2.

Examples

			Row 3 of Pascal's triangle is (1,3,3,1) and has two entries == 1 (mod 4), so a(3) = 2.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.
• A. Granville, Zaphod Beeblebrox's Brain and the Fifty-ninth Row of Pascal's Triangle, The American Mathematical Monthly, 99(4) (1992), 318-331.

Crossrefs

Cf. A034931, A163000, A000120, A007318, A014081.

Programs

Maple

f:= proc(n) local L,m;
  L:= convert(n,base,2);
  m:= convert(L,`+`);
  if has(L[1..-2]+L[2..-1],2) then 2^(m-1) else 2^m fi
end proc:
map(f, [$0..1000]);

Mathematica

Count[#, 1] & /@ Table[Mod[Binomial[n, k], 4], {n, 0, 120}, {k, 0, n}] (* Michael De Vlieger, Feb 26 2017 *)

PARI

a(n) = 2^(hammingweight(n) - min(hammingweight(bitand(n, n>>1)),1)) \\ Charles R Greathouse IV, Jul 13 2016

Python

def A270438(n): return 1<>1)).bit_count() # Chai Wah Wu, Apr 24 2025

Formula

a(n) = 2^(A000120(n) - min(1, A014081(n))). [Davis & Webb]