A034930 -id:A034930 - cp's OEIS frontend

A095145 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 12.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 3, 8, 3, 6, 1, 1, 7, 9, 11, 11, 9, 7, 1, 1, 8, 4, 8, 10, 8, 4, 8, 1, 1, 9, 0, 0, 6, 6, 0, 0, 9, 1, 1, 10, 9, 0, 6, 0, 6, 0, 9, 10, 1, 1, 11, 7, 9, 6, 6, 6, 6, 9, 7, 11, 1, 1, 0, 6, 4, 3, 0, 0, 0, 3, 4, 6, 0, 1, 1, 1, 6, 10, 7, 3, 0, 0, 3, 7, 10, 6, 1, 1

Offset: 0

Robert G. Wilson v, May 29 2004

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, A095144, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), (this sequence) (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 12]

# uses python code from A034931 and A083093
from sympy.ntheory.modular import crt
def A095145(n): return crt([4,3],[A034931(n),A083093(n)])[0] # Chai Wah Wu, Jul 19 2025

T(i, j) = binomial(i, j) mod 12.

A275198 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 14.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 7, 7, 7, 7, 7, 7, 1, 1, 8, 0, 0, 0, 0, 0, 8, 1, 1, 9, 8, 0, 0, 0, 0, 8, 9, 1, 1, 10, 3, 8, 0, 0, 0, 8, 3, 10, 1, 1, 11, 13, 11, 8, 0, 0, 8, 11, 13, 11, 1, 1, 12, 10, 10, 5, 8, 0, 8, 5, 10, 10, 12, 1, 1, 13, 8, 6, 1, 13, 8, 8, 13, 1, 6, 8, 13, 1, 1, 0, 7, 0, 7, 0, 7, 2, 7, 0, 7, 0, 7, 0, 1

Offset: 0

Ilya Gutkovskiy, Aug 11 2016

			Triangle begins:
                      1,
                    1,  1,
                  1,  2,  1,
                1,  3,  3,  1,
              1,  4,  6,  4,  1,
            1,  5, 10, 10,  5,  1,
          1,  6,  1,  6,  1,  6,  1,
        1,  7,  7,  7,  7,  7,  7,  1,
      1,  8,  0,  0,  0,  0,  0,  8,  1,
    1,  9,  8,  0,  0,  0,  0,  8,  9,  1,
  1, 10,  3,  8,  0,  0,  0,  8,  3, 10,  1,
  ...

Cf. A007318, A070696.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), (this sequence) (m = 14), A034932 (m = 16).

Mod[Flatten[Table[Binomial[n, k], {n, 0, 14}, {k, 0, n}]], 14]

from math import comb, isqrt
from sympy.ntheory.modular import crt
def A275198(n):
    w, c = n-((r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))*(r+1)>>1), 1
    d = int(not ~r & w)
    while True:
        r, a = divmod(r,7)
        w, b = divmod(w,7)
        c = c*comb(a,b)%7
        if r<7 and w<7:
            c = c*comb(r,w)%7
            break
    return crt([7,2],[c,d])[0] # Chai Wah Wu, May 01 2025

T(n, k) = binomial(n, k) mod 14.

a(n) = A070696(A007318(n)).

A385285 a(n) = Sum_{k=0..n} (binomial(n, k) mod 8).

Original entry on oeis.org

1, 2, 4, 8, 16, 16, 32, 32, 16, 32, 24, 48, 40, 64, 40, 64, 16, 32, 40, 64, 48, 48, 88, 96, 40, 80, 88, 112, 72, 112, 112, 128, 16, 32, 40, 64, 64, 96, 120, 160, 48, 96, 72, 144, 104, 176, 160, 192, 40, 80, 104, 176, 120, 176, 176, 224, 72, 144, 160, 224, 176

Offset: 0

Peter Luschny, Jun 26 2025

a(n) is a multiple of A001316(n). - Chai Wah Wu, Jun 26 2025

John Tyler Rascoe, Table of n, a(n) for n = 0..8192
James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 8), European Journal of Combinatorics 19:1 (1998), pp. 45-62.
Chai Wah Wu, Sum of rows of Pascal's triangle modulo 8, blog post.

Cf. A034930 (row sums of), A001316 (mod 2), A384715 (mod 4).

A385285 := n -> local k; add(modp(binomial(n, k), 8), k = 0..n): seq(A385285(n), n = 0..60);

A385285[n_] := Sum[Mod[Binomial[n, k], 8], {k, 0, n}];
Array[A385285, 100, 0] (* Paolo Xausa, Jun 26 2025 *)

a(n) = sum(k=0, n, binomial(n, k) % 8); \\ Michel Marcus, Jun 26 2025

def A385285(n):
    if n==0: return 1
    s = '0'+bin(n)[2:]
    n1 = n>>1
    n2 = n1>>1
    n3 = n2>>1
    np = ~n
    n1111 = (n3&n2&n1&n).bit_count()
    n11 = (n1&n).bit_count()
    n101 = (n2&(~n1)&n).bit_count()
    n100 = (n2&(~n1)&np).bit_count()
    n110 = (n2&n1&np).bit_count()
    n10 = (n1&np).bit_count()
    n1100 = (n3&n2&(~n1)&np).bit_count()
    m11 = s.count('0110')
    m111 = s.endswith('0111')
    c = (n100<<2)+n110+(n10*(n10-1)>>1)
    if n11==0:
        c += n10+(n100<<1)
    elif n11==1:
        c += (n10-n1100<<1)+n110
    else:
        c += n10<<1
    if not (n1111 or n11 or n101):
        c += 1
    elif not (n1111 or n11) and n101:
        c += 3
    elif m111:
        c += 4
    elif not (n1111 or n101 or m11) and n11:
        c += 2
    else:
        c += 4
    return c<Chai Wah Wu, Jun 26 2025

A213126 Rows of triangle formed using Pascal's rule, except sums in the n-th row are modulo n: T(n,0) = T(n,n) = 1 and T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod n.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 4, 3, 1, 1, 4, 1, 2, 1, 4, 1, 1, 5, 5, 3, 3, 5, 5, 1, 1, 6, 2, 0, 6, 0, 2, 6, 1, 1, 7, 8, 2, 6, 6, 2, 8, 7, 1, 1, 8, 5, 0, 8, 2, 8, 0, 5, 8, 1, 1, 9, 2, 5, 8, 10, 10, 8, 5, 2, 9, 1, 1, 10, 11, 7, 1, 6, 8, 6

Offset: 0

Alex Ratushnyak, Jun 06 2012

			Triangle begins:
  1;
  1,  1;
  1,  0,  1;
  1,  1,  1,  1;
  1,  2,  2,  2,  1;
  1,  3,  4,  4,  3,  1;
  1,  4,  1,  2,  1,  4,  1;
  1,  5,  5,  3,  3,  5,  5,  1;
  1,  6,  2,  0,  6,  0,  2,  6,  1;
  1,  7,  8,  2,  6,  6,  2,  8,  7,  1;
  1,  8,  5,  0,  8,  2,  8,  0,  5,  8,  1;
  1,  9,  2,  5,  8, 10, 10,  8,  5,  2,  9,  1;

Cf. A007318 - Pascal's triangle read by rows.

Cf. A047999, A083093, A034931, A034930, A034932, A008975.

T[n_,k_]:=If[k==0 || k==n, 1, Mod[T[n - 1, k - 1] + T[n- 1, k], n]]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] // Flatten (* Indranil Ghosh, Apr 29 2017 *)

src = [0]*1024
dst = [0]*1024
for n in range(19):
    dst[0] = dst[n] = 1
    for k in range(1, n):
        dst[k] = (src[k-1]+src[k]) % n
    for k in range(n+1):
        src[k] = dst[k]
        print(dst[k], end=',')

Offset corrected by Joerg Arndt, Dec 05 2016

A386441 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 27.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 8, 8, 21, 7, 1, 1, 8, 1, 2, 16, 2, 1, 8, 1, 1, 9, 9, 3, 18, 18, 3, 9, 9, 1, 1, 10, 18, 12, 21, 9, 21, 12, 18, 10, 1, 1, 11, 1, 3, 6, 3, 3, 6, 3, 1, 11, 1, 1, 12, 12, 4, 9, 9, 6, 9, 9, 4, 12, 12, 1, 1, 13, 24, 16, 13, 18, 15, 15, 18, 13, 16, 24, 13, 1

Offset: 0

Chai Wah Wu, Jul 21 2025

			Triangle begins:
               1;
             1,  1;
           1,  2,  1;
         1,  3,  3,  1;
       1,  4,  6,  4,  1;
     1,  5,  10,  10,  5,  1;
   1,  6,  15,  20,  15,  6,  1;
 1,  7,  21,  8,   8,  21,  7,  1;
  ...

Chai Wah Wu, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
Kenneth S. Davis and William A. Webb, Lucas' Theorem for Prime Powers, Europ. J. Combinatorics, Vol. 11, No. 3 (1990), 229-233.

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A008975, A095143, A095144, A095145, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

T[i_,j_]:=Mod[Binomial[i,j],27]; Table[T[n,k],{n,0,13},{k,0,n}]//Flatten (* Stefano Spezia, Jul 22 2025 *)

from math import isqrt, comb
from sympy import multiplicity
from gmpy2 import digits
def A386441(n):
    def g1(s,w,e):
        c, d = 1, 0
        if len(s) == 0: return c, d
        a, b = int(s,3), int(w,3)
        if a>=b:
            k = comb(a,b)%27
            j = multiplicity(3,k)
            d += j*e
            k = k//3**j
            c = c*pow(k,e,27)%27
        else:
            if int(s[0:1],3)4: return 0
    s = s.zfill(3)
    w = w.zfill(l:=len(s))
    c, d = g1(s[:3],w[:3],1)
    for i in range(1,l-2):
        c0, d0 = g1(s[i:i+3],w[i:i+3],1)
        c1, d1 = g1(s[i:i+2],w[i:i+2],-1)
        c = c*c0*c1%27
        d += d0+d1
    return c*3**d%27

T(i, j) = binomial(i, j) mod 27.

A178206 Decimal representation of asymptotic growth constant for the number of acyclic orientations on the two-dimensional Sierpinski gasket SG2(n) in the large n limit.

Original entry on oeis.org

1, 1, 2, 7, 2, 9, 9, 0, 7, 0, 5, 3, 6, 6, 1, 6

Offset: 1

Jonathan Vos Post, May 22 2010

Proposition III.1, p.10 of Chang. The paper also studies the number of acyclic orientations on the generalized two-dimensional Sierpinski gasket $SG_{2,b}(n)$ at stage $n$ with $b$ equal to two and three, and determines the asymptotic behaviors. It also derives upper bounds for the asymptotic growth constants for $SG_{2,b}$ and $d$-dimensional Sierpinski gasket $SG_d$.

			1.127299070536616....

Shu-Chiuan Chang, Acyclic orientations on the Sierpinski gasket, May 20, 2010.

Cf. A007318, A054431, A001317, A008292, A083093, A034931, A034930, A008975, A034932, A047999.

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A095145 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 12.

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A275198 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 14.

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A385285 a(n) = Sum_{k=0..n} (binomial(n, k) mod 8).

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A213126 Rows of triangle formed using Pascal's rule, except sums in the n-th row are modulo n: T(n,0) = T(n,n) = 1 and T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod n.

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A386441 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 27.

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A178206 Decimal representation of asymptotic growth constant for the number of acyclic orientations on the two-dimensional Sierpinski gasket SG2(n) in the large n limit.

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