Marcus Jaiclin - cp's OEIS frontend

User: Marcus Jaiclin

Marcus Jaiclin has authored 4 sequences.

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

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

Marcus Jaiclin, Feb 07 2012

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.

			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

Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5

Cf. A206428, A206424, A227428, A083093, A077267, A062756, A081603.

Cf. A062296, A006047, A007318.

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

Marcus Jaiclin, Feb 07 2012

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.

			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

Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5

Cf. A206427, A206424, A227428, A083093, A077267, A062756, A081603.

Cf. A062296, A006047, A007318.

A206424 The number of 1's in row n of Pascal's Triangle (mod 3).

Original entry on oeis.org

1, 2, 2, 2, 4, 4, 2, 4, 5, 2, 4, 4, 4, 8, 8, 4, 8, 10, 2, 4, 5, 4, 8, 10, 5, 10, 14, 2, 4, 4, 4, 8, 8, 4, 8, 10, 4, 8, 8, 8, 16, 16, 8, 16, 20, 4, 8, 10, 8, 16, 20, 10, 20, 28, 2, 4, 5, 4, 8, 10, 5, 10, 14, 4, 8, 10, 8, 16, 20, 10, 20, 28, 5, 10, 14, 10, 20, 28

Offset: 0

Marcus Jaiclin, Feb 07 2012

A006047(n) = a(n) + A227428(n).

a(n) = n + 1 - A062296(n) - A227428(n); number of ones in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

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

Reinhard Zumkeller (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..19683
R. Garfield and H. S. Wilf, The distribution of the binomial coefficients modulo p, J. Numb. Theory 41 (1) (1992) 1-5
Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5
D. L. Wells, Residue counts modulo three for the fibonacci triangle, Appl. Fib. Numbers, Proc. 6th Int Conf Fib. Numbers, Pullman, 1994 (1996) 521-536.
Avery Wilson, Pascal's Triangle Modulo 3, Mathematics Spectrum, 47-2 - January 2015, pp. 72-75.

Cf. A083093, A062296, A006047, A062756, A081603, A206427, A227428.

a206424 = length . filter (== 1) . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

Table[Count[Mod[Binomial[n, Range[0, n]], 3], 1], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)

A206424(n) = sum(k=0,n,1==(binomial(n,k)%3)); \\ (naive way) Antti Karttunen, Jul 26 2017

from sympy.ntheory import digits
def A206424(n):
    s = digits(n,3)[1:]
    return (3**s.count(2)+1)<>1 # Chai Wah Wu, Jul 24 2025

(define (A206424 n) (* (+ (A000244 (A081603 n)) 1) (A000079 (- (A062756 n) 1)))) ;; (fast way) Antti Karttunen, Jul 27 2017

From Antti Karttunen, Jul 27 2017: (Start)
a(n) = (3^k + 1)*2^(y-1), where y = A062756(n) and k = A081603(n). [See e.g. Wells or Wilson references.]
a(n) = A006047(n) - A227428(n).
(End)
From David A. Corneth and Antti Karttunen, Jul 27 2017: (Start)
Based on the first formula above, we have following identities:
a(3n) = a(n).
a(3n+1) = 2*a(n).
a(9n+4) = 4*a(n).
(End)
a(n) = (1/2)*Sum_{k = 0..n} mod(C(n,k) + C(n,k)^2, 3). - Peter Bala, Dec 17 2020

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

Marcus Jaiclin, Feb 07 2012

dead

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

			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

Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5

Cf. A083093, A062296, A006047.

Table[Count[Mod[Binomial[n, Range[0, n]], 3], 2], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)

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User: Marcus Jaiclin

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

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

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A206424 The number of 1's in row n of Pascal's Triangle (mod 3).

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A206425 Erroneous version of A227428.

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