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.

Previous Showing 31-36 of 36 results.

A206427 Square array 2^(m-1)*(3^n+1), read by antidiagonals.

Original entry on oeis.org

1, 2, 2, 5, 4, 4, 14, 10, 8, 8, 41, 28, 20, 16, 16, 122, 82, 56, 40, 32, 32, 365, 244, 164, 112, 80, 64, 64, 1094, 730, 488, 328, 224, 160, 128, 128, 3281, 2188, 1460, 976, 656, 448, 320, 256, 256, 9842, 6562, 4376, 2920, 1952, 1312, 896, 640, 512, 512
Offset: 0

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Author

Marcus Jaiclin, Feb 07 2012

Keywords

Comments

Rectangular array giving the number of 1's in any row of Pascal's Triangle (mod 3) whose row number has exactly m 1's and n 2's in its ternary expansion (listed by antidiagonals).
a(m,n) is independent of the number of zeros in the ternary expansion of the row number.
a(m,n) gives a non-recursive formula for A206424.

Examples

			Initial 5 X 5 block of entries (upper corner is (m,n)=(0,0), m increases down, n increases across):
1    2    5   14   41
2    4   10   28   82
4    8   20   56  164
8   16   40  112  328
16  32   80  224  656
Pascal's Triangle (mod 3), row numbers in ternary:
1     <=  Row 0, m = 0, n = 0, 2^(-1)(3^0 + 1) = #1's = 1
1 1     <=  Row 1, m = 1, n = 0, 2^0(3^0 + 1) = #1's = 2
1 2 1     <=  Row 2, m = 0, n = 1, 2^(-1)(3^1 + 1) = #1's = 2
1 0 0 1     <=  Row 10, m = 1, n = 0, 2^0(3^0 + 1) = #1's = 2
1 1 0 1 1     <=  Row 11, m = 2, n = 0, 2^1(3^0 + 1) = #1's = 4
1 2 1 1 2 1     <=  Row 12, m = 1, n = 1, 2^0(3^1 + 1) = #1's = 4
1 0 0 2 0 0 1     <=  Row 20, m = 0, n = 1, 2^(-1)(3^1 + 1) = #1's = 2
1 1 0 2 2 0 1 1     <=  Row 21, m = 1, n = 1, 2^0(3^1 + 1) = #1's = 4
1 2 1 2 1 2 1 2 1     <=  Row 22, m = 0, n = 2, 2^(-1)(3^2 + 1) = #1's = 5
1 0 0 0 0 0 0 0 0 1     <=  Row 100, m = 1, n = 0, 2^0(3^0 + 1) = #1's = 2
		

Crossrefs

Formula

a(m, n) = 2^(m - 1)(3^n + 1).

A206428 Rectangular array, a(m,n) = 2^(m-1)*(3^n-1), read by antidiagonals.

Original entry on oeis.org

0, 1, 0, 4, 2, 0, 13, 8, 4, 0, 40, 26, 16, 8, 0, 121, 80, 52, 32, 16, 0, 364, 242, 160, 104, 64, 32, 0, 1093, 728, 484, 320, 208, 128, 64, 0, 3280, 2186, 1456, 968, 640, 416, 256, 128, 0, 9841, 6560, 4372, 2912, 1936, 1280, 832, 512, 256, 0
Offset: 0

Views

Author

Marcus Jaiclin, Feb 07 2012

Keywords

Comments

Number of 2's in any row of Pascal's triangle (mod 3) whose row number has exactly m 1's and n 2's in its ternary expansion.
a(m,n) is independent of the number of zeros in the ternary expansion of the row number.
a(m,n) gives a non-recursive formula for A227428.

Examples

			Initial 5 X 5 block of array (upper left corner is (0,0), row index m, column index n):
0    1    4   13   40
0    2    8   26   80
0    4   16   52  160
0    8   32  104  320
0   16   64  208  640
Pascal's Triangle (mod 3), row numbers in ternary:
1     <= Row 0, m=0, n=0, 2^(-1)(3^0-1) = #2's = 0
1 1     <= Row 1, m=1, n=0, 2^0(3^0-1) = #2's = 0
1 2 1     <= Row 2, m=0, n=1, 2^(-1)(3^1-1) = #2's = 1
1 0 0 1     <= Row 10, m=1, n=0, 2^0(3^0-1) = #2's = 0
1 1 0 1 1     <= Row 11, m=2, n=0, 2^1(3^0-1) = #2's = 0
1 2 1 1 2 1     <= Row 12, m=1, n=1, 2^0(3^1-1) = #2's = 2
1 0 0 2 0 0 1     <= Row 20, m=0, n=1, 2^(-1)(3^1-1) = #2's = 1
1 1 0 2 2 0 1 1     <= Row 21, m=1, n=1, 2^0(3^1-1) = #2's = 2
1 2 1 2 1 2 1 2 1     <= Row 22, m=0, n=2, 2^(-1)(3^2-1) = #2's = 4
1 0 0 0 0 0 0 0 0 1     <= Row 100, m=1, n=0, 2^0(3^0-1) = #2's = 0
		

Crossrefs

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

Views

Author

Alex Ratushnyak, Jun 06 2012

Keywords

Examples

			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;
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • Python
    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=',')

Extensions

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

Views

Author

Chai Wah Wu, Jul 21 2025

Keywords

Examples

			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;
  ...
		

Crossrefs

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).

Programs

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

Formula

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

Views

Author

Jonathan Vos Post, May 22 2010

Keywords

Comments

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 $SG_d$.

Examples

			1.127299070536616....
		

Crossrefs

A206425 Erroneous version of A227428.

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 1, 2, 4, 0, 0, 2, 0, 0, 4, 2, 4, 8, 1, 2, 4, 2, 4, 8, 1, 2, 4, 2, 4, 8, 4, 8, 13, 0, 0, 2, 0, 0, 4, 2, 4, 8, 0, 0, 4, 0, 0, 8, 4, 8, 16, 2, 4, 8, 4, 8, 16, 8, 16, 26, 1, 2, 4, 2, 4, 8, 4, 8, 13, 2, 4, 8, 4, 8, 16, 8, 16, 26, 4, 8, 13, 8, 16, 26
Offset: 0

Views

Author

Marcus Jaiclin, Feb 07 2012

Keywords

Comments

A006047(n) = A206424(n) + a(n).

Examples

			Example: Rows 0-8 of Pascal's Triangle (mod 3) are:
1                   So a(0) = 0
1 1                 So a(1) = 0
1 2 1               So a(2) = 1
1 0 0 1                 .
1 1 0 1 1               .
1 2 1 1 2 1             .
1 0 0 2 0 0 1
1 1 0 2 2 0 1 1
1 2 1 2 1 2 1 2 1
		

Crossrefs

Programs

  • Mathematica
    Table[Count[Mod[Binomial[n, Range[0, n]], 3], 2], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)
Previous Showing 31-36 of 36 results.