A249732 Number of (not necessarily distinct) multiples of 4 on row n of Pascal's triangle.

0, 0, 0, 0, 2, 0, 1, 0, 6, 4, 3, 0, 7, 2, 3, 0, 14, 12, 11, 8, 13, 6, 7, 0, 19, 14, 11, 4, 17, 6, 7, 0, 30, 28, 27, 24, 29, 22, 23, 16, 33, 26, 23, 12, 29, 14, 15, 0, 43, 38, 35, 28, 37, 22, 23, 8, 45, 34, 27, 12, 37, 14, 15, 0, 62, 60, 59, 56, 61, 54, 55, 48, 65, 58, 55, 44, 61, 46, 47, 32

Offset

0,5

Comments

a(n) = Number of zeros on row n of A034931 (Pascal's triangle reduced modulo 4).

This should have a formula (see A048967).

Links

Antti Karttunen, Table of n, a(n) for n = 0..8192

Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.

Formula

Other identities:

a(n) <= A048277(n) for all n.

a(n) <= A048967(n) for all n.

Example

Row 9 of Pascal's triangle is: {1,9,36,84,126,126,84,36,9,1}. The terms 36 and 84 are only multiples of four, and both of them occur two times on that row, thus a(9) = 2*2 = 4.
Row 10 of Pascal's triangle is: {1,10,45,120,210,252,210,120,45,10,1}. The terms 120 (= 4*30) and 252 (= 4*63) are only multiples of four, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

Prog

(PARI)
A249732(n) = { my(c=0); for(k=0, n\2, if(!(binomial(n, k)%4), c += (if(k<(n/2), 2, 1)))); return(c); } \\ Slow...
for(n=0, 8192, write("b249732.txt", n, " ", A249732(n)));
(Python)
def A249732(n): return n+1-(2+((n>>1)&~n).bit_count()<<n.bit_count()>>1) # Chai Wah Wu, Jul 24 2025

Crossrefs

Cf. A007318, A034931, A048277, A048645, A072823, A048967, A062296, A187059, A249733.

Sequence in context: A249731 A335125 A335022 * A076694 A375535 A095403

Adjacent sequences: A249729 A249730 A249731 * A249733 A249734 A249735

Keyword

nonn

Author

Antti Karttunen, Nov 04 2014

Status

approved