cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Chai Wah Wu, Aug 15 2025

Keywords

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

A387108 Number of entries in the n-th row of Pascal's triangle not divisible by 25.

Original entry on oeis.org

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, 6, 12, 18, 24, 30, 15, 20, 25, 30, 35, 24, 28, 32, 36, 40, 33, 36, 39, 42, 45, 42, 44, 46, 48, 50, 11, 22, 33, 44, 55, 24, 33, 42, 51, 60, 37, 44, 51, 58, 65, 50, 55, 60
Offset: 0

Views

Author

Chai Wah Wu, Aug 16 2025

Keywords

Links

Crossrefs

Cf. A001316, A006047, A194459, A382720-A382725, A386952, A387050, A387051.

Programs

  • Python
    import re
from gmpy2 import digits
def A387108(n):
    s = digits(n,5)
    n1, n2, n3, n4 = s.count('1'), s.count('2'), s.count('3'), s.count('4')
    n10, n12, n13, n42, n43, n11 = s.count('10'), s.count('12'), s.count('13'), s.count('42'), s.count('43'), len(re.findall('(?=11)',s))
    n20, n21, n23, n30, n22 = s.count('20'), s.count('21'), s.count('23'), s.count('30'), len(re.findall('(?=22)',s))
    n31, n32, n40, n41, n33 = s.count('31'), s.count('32'), s.count('40'), s.count('41'), len(re.findall('(?=33)',s))
    return ((1440*n10+540*n11+240*n12+90*n13+1920*n20+720*(n21+1)+320*n22+120*n23+2160*n30+810*n31+360*n32+135*n33+2304*n40+864*n41+384*n42+144*n43)*3**n2*5**n4<<(n1+(n3<<1)))//45>>4

A387109 Number of entries in the n-th row of Pascal's triangle not divisible by 27.

Original entry on oeis.org

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, 10, 20, 30, 19, 26, 33, 28, 32, 36, 25, 32, 39, 32, 37, 42, 39, 42, 45, 40, 44, 48, 45, 48, 51, 50, 52, 54, 19, 38, 57, 34, 47, 60, 49, 56, 63, 40, 53, 66, 51, 60
Offset: 0

Views

Author

Chai Wah Wu, Aug 16 2025

Keywords

Links

Crossrefs

Cf. A001316, A006047, A194459, A382720-A382725, A386952, A387050, A387051, A387108.

Programs

  • Python
    import re
from gmpy2 import digits
def A387109(n):
    s = digits(n,3)
    n1, n2, n10, n20, n21, n11 = s.count('1'), s.count('2'), s.count('10'), s.count('20'), s.count('21'), len(re.findall('(?=11)',s))
    n100, n110, n120, n101, n111, n121 = s.count('100'), s.count('110'), s.count('120'), len(re.findall('(?=101)',s)), len(re.findall('(?=111)',s)), len(re.findall('(?=121)',s))
    n200, n201, n210, n211, n220, n221 = s.count('200'), s.count('201'), s.count('210'), s.count('211'), s.count('220'), s.count('221')
    c = 144*n10+63*n11+128*(n20+n220)+80*n21+864*n100+216*(n101+n110)+54*n111+96*n120+24*n121+1152*n200+288*(n201+n210+1)+72*n211+32*n221
    c += (m:=4*n10+n11)*(96*n20+24*n21+9*m)+16*(4*n20+n21)**2
    return (c*3**n2<>5
Showing 1-3 of 3 results.

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