A008975 -id:A008975 - cp's OEIS frontend

A095140 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 1, 4, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 1, 3, 3, 1, 1, 4, 1, 4, 1, 1, 4, 1, 4, 1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 2, 4, 2, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 2, 1, 1, 2, 0, 1, 3, 3, 1

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(5))/log(5) = log(15)/log(5) = 1.68260... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095141, A095142, A034930, A095143, A008975, 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), (this sequence) (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).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 5]

from math import isqrt, comb
def A095140(n):
    def f(m,k):
        if m<5 and k<5: return comb(m,k)%5
        c,a = divmod(m,5)
        d,b = divmod(k,5)
        return f(c,d)*f(a,b)%5
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 5.

A095142 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 7.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 3, 3, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 0, 0, 1, 3, 3, 1, 1, 4, 6, 4, 1, 0, 0, 1, 4, 6, 4, 1, 1, 5, 3, 3, 5, 1, 0, 1, 5, 3, 3, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 6, 1, 6, 1, 6, 1

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(7))/log(7) = log(28)/log(7) = 1.71241... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A034930, A095143, A008975, 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), (this sequence) (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 7]

from math import comb, isqrt
def A095142(n):
    def f(m,k):
        if m<7 and k<7: return comb(m,k)%7
        c,a = divmod(m,7)
        d,b = divmod(k,7)
        return f(c,d)*f(a,b)%7
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 7.

A095144 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 11.

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

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(11))/log(11) = log(66)/log(11) = 1.74722... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
Zubeyir Cinkir and Aysegul Ozturkalan, An extension of Lucas Theorem, arXiv:2309.00109 [math.NT], 2023. See Figures 3 and 4 p. 6.
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, 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), (this sequence) (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

R[0]:= 1:
for  n from 1 to 20 do
  R[n]:= op([R[n-1],0] + [0,R[n-1]] mod 11);
od:
for n from 0 to 20 do R[n] od; # Robert Israel, Jan 02 2019

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 11]

from math import isqrt, comb
def A095144(n):
    def f(m,k):
        if m<11 and k<11: return comb(m,k)%11
        c,a = divmod(m,11)
        d,b = divmod(k,11)
        return f(c,d)*f(a,b)%11
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 11.
From Robert Israel, Jan 02 2019: (Start)
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 11 with T(n,0) = 1.
T(n,k) = (Product_i binomial(n_i, k_i)) mod 11, where n_i and k_i are the base-11 digits of n and k. (End)

A095141 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 6.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 0, 4, 1, 1, 5, 4, 4, 5, 1, 1, 0, 3, 2, 3, 0, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 2, 4, 2, 4, 2, 4, 2, 1, 1, 3, 0, 0, 0, 0, 0, 0, 3, 1, 1, 4, 3, 0, 0, 0, 0, 0, 3, 4, 1, 1, 5, 1, 3, 0, 0, 0, 0, 3, 1, 5, 1, 1, 0, 0, 4, 3, 0, 0, 0, 3, 4, 0, 0, 1, 1, 1, 0, 4, 1, 3, 0, 0, 3, 1, 4, 0, 1, 1

Offset: 0

Robert G. Wilson v, May 29 2004

Bill Gosper, Pastel-colored illustration of triangle
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Ivan Korec, Definability of Pascal's Triangles Modulo 4 and 6 and Some Other Binary Operations from Their Associated Equivalence Relations, Acta Univ. M. Belii Ser. Math. 4 (1996), pp. 53-66.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095142, A034930, A095143, A008975, 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), (this sequence) (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).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 6]
Graphics[Table[{%[Mod[Binomial[n, k], 6]/5], RegularPolygon[{4√3 (k - n/2), -6 n}, {4,π/6}, 6]}, {n, 0, 105}, {k, 0, n}]] (* Mma code for illustration, Bill Gosper, Aug 05 2017 *)

from math import isqrt, comb
from sympy.ntheory.modular import crt
def A095141(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,3)
        w, b = divmod(w,3)
        c = c*comb(a,b)%3
        if r<3 and w<3:
            c = c*comb(r,w)%3
            break
    return crt([3,2],[c,d])[0] # Chai Wah Wu, May 01 2025

T(i, j) = binomial(i, j) mod 6.

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

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, 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)).

A059734 Carryless 11^n base 10; a(n) is carryless sum of 10*a(n-1) and a(n-1).

Original entry on oeis.org

1, 11, 121, 1331, 14641, 150051, 1650561, 17155171, 188606881, 1964664691, 10500200501, 115502205511, 1260524250621, 13865766756831, 141412323214141, 1555535555355551, 16000880008800061, 176008680086800671

Offset: 0

Henry Bottomley, Feb 20 2001

Subsequence of A002113. - Chai Wah Wu, Jul 30 2025

			a(7)=17155171 since a(6)=1650561 and digits of a(7) are sum mod 10 of 1, 6+1=7, 5+6=1, 0+5=5, 5+0=5, 6+5=1, 1+6=7 and 1.

Seiichi Manyama, Table of n, a(n) for n = 0..999
David Applegate, Marc LeBrun and N. J. A. Sloane, Carryless Arithmetic (I): The Mod 10 Version

Cf. A004520, A006940, A008975, A361351, A002113.

Table[Sum[Mod[Binomial[n, m], 10]*10^m, {m, 0, n}], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 14 2008 *)

a(n) = fromdigits(Vec(Pol(digits(11))^n)%10); \\ Seiichi Manyama, Mar 10 2023

from math import comb, prod
from sympy.ntheory.modular import crt
from gmpy2 import digits
def A059734(n):
    k, l = 0, len(s:=digits(n,5))
    for m in range(n+1):
        t = digits(m,5).zfill(l)
        k = 10*k+crt([5,2],[prod(comb(int(s[i]),int(t[i]))%5 for i in range(l))%5,int(not ~n & m)])[0]
    return k # Chai Wah Wu, Jul 30 2025

a(n)=Sum[Mod[Binomial[n, m], 10]*10^m, {m, 0, n}]. - Roger L. Bagula and Gary W. Adamson, Sep 14 2008

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.

cp's OEIS Frontend

A095140 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

A095142 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 7.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

A095144 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 11.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A095141 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 6.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

A059734 Carryless 11^n base 10; a(n) is carryless sum of 10*a(n-1) and a(n-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI