A206424 -id:A206424 - cp's OEIS frontend

A006047 Number of entries in n-th row of Pascal's triangle not divisible by 3.

1, 2, 3, 2, 4, 6, 3, 6, 9, 2, 4, 6, 4, 8, 12, 6, 12, 18, 3, 6, 9, 6, 12, 18, 9, 18, 27, 2, 4, 6, 4, 8, 12, 6, 12, 18, 4, 8, 12, 8, 16, 24, 12, 24, 36, 6, 12, 18, 12, 24, 36, 18, 36, 54, 3, 6, 9, 6, 12, 18, 9, 18, 27, 6, 12, 18, 12, 24, 36, 18, 36, 54, 9, 18, 27, 18, 36, 54, 27, 54

Offset: 0

Jeffrey Shallit

nonn

Fixed point of the morphism a -> a, 2a, 3a, starting from a(1) = 1. - Robert G. Wilson v, Jan 24 2006
This is a particular case of the number of entries in n-th row of Pascal's triangle not divisible by a prime p, which is given by a simple recursion using ⊗, the Kronecker (or tensor) product of vectors. Let v_0=(1,2,...,p). Then v_{n+1}=v_0 ⊗ v_n, where the vector v_n contains the values for the first p^n rows of Pascal's triangle (rows 0 through p^n-1). - William B. Everett (bill(AT)chgnet.ru), Mar 29 2008
a(n) = A206424(n) + A227428(n); number of nonzero terms in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

			15 in base 3 is 120, here r=1 and s=1 so a(15) = 3*2 = 6.
William B. Everett's comment with p=3, n=2: v_0 = (1,2,3), v_1 = (1,2,3) => v_2 = (1*1,1*2,1*3,2*1,2*2,2*3,3*1,3*2,3*3) = (1,2,3,2,4,6,3,6,9), the first 3^2 values of the present sequence. - _Wolfdieter Lang_, Mar 19 2014

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Antti Karttunen, Table of n, a(n) for n = 0..19683 (terms 0..1000 from Reinhard Zumkeller).
J.-P. Allouche and Jeffrey Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197.
Michael Gilleland, Some Self-Similar Integer Sequences
H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62.1 (1977), 19-22. (Annotated scanned copy)
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Periodic minimum in the count of binomial coefficients not divisible by a prime, arXiv:2408.06817 [math.NT], 2024.
Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
Index entries for sequences that are fixed points of mappings

Cf. A001316, A003586, A038148, A053735, A083093, A089898, A206424, A227428, A286586, A286587, A286633.

a006047 = sum . map signum . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

p:=proc(n) local ct, k: ct:=0: for k from 0 to n do if binomial(n,k) mod 3 = 0 then else ct:=ct+1 fi od: end: seq(p(n),n=0..82); # Emeric Deutsch
f:= proc(n) option remember; ((n mod 3)+1)*procname(ceil((n+1)/3)-1) end proc:
f(0):= 1: f(1):= 2:
seq(f(i), i=0..100); # Robert Israel, Oct 15 2015

Nest[Flatten[ # /. a_Integer -> {a, 2a, 3a}] &, {1}, 4] (* Robert G. Wilson v, Jan 24 2006 *)
Nest[ Join[#, 2#, 3#] &, {1}, 4] (* Robert G. Wilson v, Jul 27 2014 *)

b(n)=if(n<3,n,if(n%3==0,3*b(n/3),if(n%3==1,1*b((n+2)/3),2*b((n+1)/3)))) \\ Ralf Stephan

A006047(n) = b(1+n); \\ (The above PARI-program by Ralf Stephan is for offset-1-version of this sequence.) - Antti Karttunen, May 28 2017

A006047(n) = { my(m=1, d); while(n, d = (n%3); m *= (1+d); n \= 3); m; }; \\ Antti Karttunen, May 28 2017

a(n) = prod(i=1,#d=digits(n, 3), (1+d[i])) \\ David A. Corneth, May 28 2017

upto(n) = my(res = [1], v); while(#res < n, v = concat(2*res, 3*res); res = concat(res, v)); res \\ David A. Corneth, May 29 2017

from sympy.ntheory.factor_ import digits
from sympy import prod
def a(n):
    d=digits(n, 3)
    return n + 1 if n<3 else prod(1 + d[i] for i in range(1, len(d)))
print([a(n) for n in range(51)]) # Indranil Ghosh, Jun 06 2017

from sympy.ntheory import digits
def A006047(n): return 3**(s:=digits(n,3)).count(2)<Chai Wah Wu, Apr 24 2025

(define (A006047 n) (if (zero? n) 1 (let ((d (mod n 3))) (* (+ 1 d) (A006047 (/ (- n d) 3)))))) ;; For R6RS standard. Use modulo instead of mod in older Schemes like MIT/GNU Scheme. - Antti Karttunen, May 28 2017

Write n in base 3; if the representation contains r 1's and s 2's then a(n) = 3^s * 2^r. Also a(n) = Sum_{k=0..n} (C(n, k)^2 mod 3). - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001
a(n) = b(n+1), with b(1)=1, b(2)=2, b(3n)=3b(n), b(3n+1)=b(n+1), b(3n+2)=2b(n+1). - Ralf Stephan, Sep 15 2003
G.f.: Product_{n>=0} (1+2*x^(3^n)+3*x^(2*3^n)) (Northshield). - Johannes W. Meijer, Jun 05 2011
G.f. g(x) satisfies g(x) = (1 + 2*x + 3*x^2)*g(x^3). - Robert Israel, Oct 15 2015
From Tom Edgar, Oct 15 2015: (Start)
a(3^k) = 2 for k>=0;
a(2*3^k) = 3 for k>=0;
a(n) = Product_{b_j != 0} a(b_j*3^j) where n = Sum_{j>=0} b_j*3^j is the ternary representation of n. (End)
A056239(a(n)) = A053735(n). - Antti Karttunen, Jun 03 2017
a(n) = Sum_{k = 0..n} mod(C(n,k)^2, 3). - Peter Bala, Dec 17 2020

More terms from Ralf Stephan, Sep 15 2003

A083093 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 3.

Original entry on oeis.org

1, 1, 1, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1

Offset: 0

Benoit Cloitre, Apr 22 2003

Start with [1], repeatedly apply the map 0 -> [000/000/000], 1 -> [111/120/100], 2 -> [222/210/200]. - Philippe Deléham, Apr 16 2009

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(3))/log(3) = log(6)/log(3) = 1.63092... (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

			.            Rows 0 .. 3^3:
.    0:                             1
.    1:                            1 1
.    2:                           1 2 1
.    3:                          1 0 0 1
.    4:                         1 1 0 1 1
.    5:                        1 2 1 1 2 1
.    6:                       1 0 0 2 0 0 1
.    7:                      1 1 0 2 2 0 1 1
.    8:                     1 2 1 2 1 2 1 2 1
.    9:                    1 0 0 0 0 0 0 0 0 1
.   10:                   1 1 0 0 0 0 0 0 0 1 1
.   11:                  1 2 1 0 0 0 0 0 0 1 2 1
.   12:                 1 0 0 1 0 0 0 0 0 1 0 0 1
.   13:                1 1 0 1 1 0 0 0 0 1 1 0 1 1
.   14:               1 2 1 1 2 1 0 0 0 1 2 1 1 2 1
.   15:              1 0 0 2 0 0 1 0 0 1 0 0 2 0 0 1
.   16:             1 1 0 2 2 0 1 1 0 1 1 0 2 2 0 1 1
.   17:            1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1
.   18:           1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1
.   19:          1 1 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 1 1
.   20:         1 2 1 0 0 0 0 0 0 2 1 2 0 0 0 0 0 0 1 2 1
.   21:        1 0 0 1 0 0 0 0 0 2 0 0 2 0 0 0 0 0 1 0 0 1
.   22:       1 1 0 1 1 0 0 0 0 2 2 0 2 2 0 0 0 0 1 1 0 1 1
.   23:      1 2 1 1 2 1 0 0 0 2 1 2 2 1 2 0 0 0 1 2 1 1 2 1
.   24:     1 0 0 2 0 0 1 0 0 2 0 0 1 0 0 2 0 0 1 0 0 2 0 0 1
.   25:    1 1 0 2 2 0 1 1 0 2 2 0 1 1 0 2 2 0 1 1 0 2 2 0 1 1
.   26:   1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
.   27:  1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 .
- _Reinhard Zumkeller_, Jul 11 2013

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.

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
J.-P. Allouche, F. von Haeseler, H.-O. Peitgen, and G. Skordev, Linear cellular automata, finite automata and Pascal's triangle, Disc. Appl. Math. 66 (1996) 1-22.
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.
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Lin Jiu and Christophe Vignat, On Binomial Identities in Arbitrary Bases, arXiv:1602.04149 [math.CO], 2016.
Y. Moshe, The density of 0's in recurrence double sequences, J. Number Theory, 103 (2003), 109-121.
Y. Moshe, The distribution of elements in automatic double sequences, Discr. Math., 297 (2005), 91-103.
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, A051638 (row sums), A090044, A047999, A034931, A034930, A008975, A034932, A062296, A006047.
Cf. A006996 (central terms), A173019, A206424, A227428.
Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), (this sequence) (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).

a083093 n k = a083093_tabl !! n !! k
a083093_row n = a083093_tabl !! n
a083093_tabl = iterate
   (\ws -> zipWith (\u v -> mod (u + v) 3) ([0] ++ ws) (ws ++ [0])) [1]
-- Reinhard Zumkeller, Jul 11 2013

/* As triangle: */ [[Binomial(n,k) mod 3: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 15 2016

A083093 := proc(n,k)
    modp(binomial(n,k),3) ;
end proc:
seq(seq(A083093(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Jul 26 2017

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 3] (* Robert G. Wilson v, Jan 19 2004 *)

from sympy import binomial
def T(n, k):
    return binomial(n, k) % 3
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Jul 26 2017

from math import comb, isqrt
def A083093(n):
    def f(m,k):
        if m<3 and k<3: return comb(m,k)%3
        c,a = divmod(m,3)
        d,b = divmod(k,3)
        return f(c,d)*f(a,b)%3
    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 3.
T(n+1,k) = (T(n,k) + T(n,k-1)) mod 3. - Reinhard Zumkeller, Jul 11 2013
T(n,k) = Product_{i>=0} binomial(n_i,k_i) mod 3, where n = Sum_{i>=0} n_i*3^i and k = Sum_{i>=0} k_i*3^i, 0<=n_i, k_i <=2 [Allouche et al.]. - R. J. Mathar, Jul 26 2017

A062296 a(n) = number of entries in n-th row of Pascal's triangle divisible by 3.

Original entry on oeis.org

0, 0, 0, 2, 1, 0, 4, 2, 0, 8, 7, 6, 9, 6, 3, 10, 5, 0, 16, 14, 12, 16, 11, 6, 16, 8, 0, 26, 25, 24, 27, 24, 21, 28, 23, 18, 33, 30, 27, 32, 25, 18, 31, 20, 9, 40, 35, 30, 37, 26, 15, 34, 17, 0, 52, 50, 48, 52, 47, 42, 52, 44, 36, 58, 53, 48, 55, 44, 33, 52, 35, 18, 64, 56, 48, 58, 41

Offset: 0

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 02 2001

Number of zeros in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

			When n=3 the row is 1,3,3,1 so a(3) = 2.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
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.

Cf. A006047, A083093, A206424, A227428.

Cf. A062756, A081603.

a062296 = sum . map ((1 -) . signum) . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

p:=proc(n) local ct, k: ct:=0: for k from 0 to n do if binomial(n,k) mod 3 = 0 then else ct:=ct+1 fi od: end: seq(n+1-p(n),n=0..83); # Emeric Deutsch

a[n_] := Count[(Binomial[n, #] & )[Range[0, n]], _?(Divisible[#, 3] & )];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 26 2018 *)
Table[n + 1 - 2^(DigitCount[n, 3, 1])*3^(DigitCount[n, 3, 2]), {n, 0, 76}] (* Shenghui Yang, Jan 08 2025 *)

from sympy.ntheory import digits
def A062296(n):
    s = digits(n,3)[1:]
    return n+1-(3**s.count(2)<Chai Wah Wu, Jul 24 2025

a(n) = n + 1 - A006047(n).
a(n) = n + 1 - A206424(n) - A227428(n). - Reinhard Zumkeller, Jul 11 2013
a(n) = n + 1 - 2^A062756(n)*3^A081603(n). - Shenghui Yang, Jan 08 2025

More terms from Emeric Deutsch, Feb 03 2005

A227428 Number of twos in row n of triangle A083093.

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, 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

A290094 Restricted growth sequence transform of A290093.

Original entry on oeis.org

1, 2, 3, 2, 4, 5, 3, 5, 6, 2, 7, 5, 4, 8, 9, 5, 10, 11, 3, 5, 12, 5, 9, 13, 6, 11, 14, 2, 7, 5, 7, 15, 10, 5, 10, 11, 4, 15, 9, 8, 16, 17, 9, 18, 19, 5, 10, 13, 10, 18, 20, 11, 21, 22, 3, 5, 12, 5, 9, 13, 12, 13, 23, 5, 10, 13, 9, 17, 24, 13, 20, 25, 6, 11, 23, 11, 19, 25, 14, 22, 26, 2, 7, 5, 7, 15, 10, 5, 10, 11, 7, 27, 10, 15, 28, 18, 10, 29, 21, 5, 10, 13

Offset: 0

Antti Karttunen, Jul 26 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 0..59049

For all i, j: a(i) = a(j) <=> A290093(n) = A290093(n), thus this matches to all the same base-3 (ternary) related sequences as A290093: A006047, A053735, A062756, A081603, A117942, A206424, A227428, A290091, A290092, A290079, and many others.

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).

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.

cp's OEIS Frontend

A006047 Number of entries in n-th row of Pascal's triangle not divisible by 3.

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A083093 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 3.

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A062296 a(n) = number of entries in n-th row of Pascal's triangle divisible by 3.

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A227428 Number of twos in row n of triangle A083093.

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A290094 Restricted growth sequence transform of A290093.

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