A206428 -id:A206428 - cp's OEIS frontend

A227428 Number of twos in row n of triangle A083093.

0, 0, 1, 0, 0, 2, 1, 2, 4, 0, 0, 2, 0, 0, 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, 13, 26, 40

Offset: 0

Reinhard Zumkeller, Jul 11 2013

nonn

"The number of entries with value r in the n-th row of Pascal's triangle modulo k is found to be 2^{#r^k (n)}, where now #_r^k (n) gives the number of occurrences of the digit r in the base-k representation of the integer n." [Wolfram] - _R. J. Mathar, Jul 26 2017 [This is not correct: there are entries in the sequence that are not powers of 2. - Antti Karttunen, Jul 26 2017]

			Example of Wilson's formula: a(26) = 13 = 2^(0-1)*(3^3-1) = 26/2, where A062756(26)=0, A081603(26)=3, 26=(222)_3. - _R. J. Mathar_, Jul 26 2017

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.
S. Wolfram, Geometry of binomial coefficients, Am. Math. Monthly 91 (9) (1984) 566-571.

Cf. A006047, A062296, A062756, A083093, A081603, A206424, A206428.

a227428 = sum . map (flip div 2) . a083093_row

A227428 := proc(n)
    local a;
    a := 0 ;
    for k from 0 to n do
        if A083093(n,k) = 2 then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A227428(n),n=0..20) ; # R. J. Mathar, Jul 26 2017

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

A227428(n) = sum(k=0,n,2==(binomial(n,k)%3)); \\ (Naive implementation, from the description) Antti Karttunen, Jul 26 2017

from sympy import binomial
def a(n):
    return sum(1 for k in range(n + 1) if binomial(n, k) % 3 == 2)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 26 2017

from sympy.ntheory import digits
def A227428(n):
    s = digits(n,3)[1:]
    return 3**s.count(2)-1<>1 # Chai Wah Wu, Jul 24 2025

(define (A227428 n) (* (A000079 (- (A062756 n) 1)) (+ -1 (A000244 (A081603 n))))) ;; After Wilson's direct formula, Antti Karttunen, Jul 26 2017

a(n) = A006047(n) - A206424(n) = n + 1 - A062296(n) - A206424(n).
a(n) = 2^(N_1-1)*(3^N_2-1) where N_1 = A062756(n), N_2 = A081603(n). [Wilson, Theorem 2, Wells] - R. J. Mathar, Jul 26 2017
a(n) = A206424(n) * ((3^A081603(n))-1) / ((3^A081603(n))+1). - Antti Karttunen, Jul 27 2017
a(n) = (1/2)*Sum_{k = 0..n} mod(C(n,k)^2 - C(n,k), 3). - Peter Bala, Dec 17 2020

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

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).

cp's OEIS Frontend

A227428 Number of twos in row n of triangle A083093.

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