A387051 Number of entries in the n-th row of Pascal's triangle not divisible by 32.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 17, 34, 27, 36, 29, 38, 35, 40, 33, 42, 39, 44, 41, 46, 45, 48, 41, 50, 47, 52, 49, 54, 53, 56, 53, 58, 57, 60, 59, 62, 62, 64, 17, 34, 43, 68

Offset: 0

Author

Chai Wah Wu, Aug 15 2025

Keywords

• nonn

Crossrefs

Cf. A001316, A006047, A194459, A382720-A382725, A386952, A387050.

Programs

Python

def A387051(n):
    n1 = n>>1
    n2 = n1>>1
    n3 = n2>>1
    n4 = n3>>1
    np = ~n
    n10, n100, n110 = (k1:=n1&np).bit_count(), (k2:=(k1>>1)&np).bit_count(), (k3:=n2&k1).bit_count()
    n1100, n1000, n1010, n1110 = (k5:=n3&k2).bit_count(), (k4:=(k2>>1)&np).bit_count(), (k6:=(k1>>2)&k1).bit_count(), (k7:=n3&k3).bit_count()
    n10000, n11000, n10100, n11100 = ((k4>>1)&np).bit_count(), (n4&k4).bit_count(), ((k6>>1)&np).bit_count(), (n4&k5).bit_count()
    n10010, n11010, n10110, n11110 = ((k2>>2)&k1).bit_count(), (n4&k6).bit_count(), ((k1>>3)&k3).bit_count(), (n4&k7).bit_count()
    c = n10*(n10*(n10*(n10+2)+((n100<<2)+n110)*12+35)+((((((n1000<<2)+n1010+n1100<<1)+n100<<1)+n1110<<1)+n110)*12+154))//24
    c += n100*((n100<<1)+n110+1<<2)+(((n10000<<2)+n1000+n10010+n10100+n11000+1<<2)+n10110+n11010+n11100<<2)+n1110+n11110+(n110*(n110+5)>>1)
    return c<>4