A173019 -id:A173019 - cp's OEIS frontend

A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, 12884901891, 21474836485, 64424509455, 73014444049, 219043332147, 365072220245, 1095216660735, 1103806595329, 3311419785987

Offset: 0

N. J. A. Sloane

The members are all palindromic in binary, i.e., a subset of A006995. - Ralf Stephan, Sep 28 2004
J. H. Conway writes (in Math Forum): at least the first 31 numbers give odd-sided constructible polygons. See also A047999. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005 [This observation was also made in 1982 by N. L. White (see letter). - N. J. A. Sloane, Jun 15 2015]
Decimal number generated by the binary bits of the n-th generation of the Rule 60 elementary cellular automaton. Thus: 1; 0, 1, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1, 1, 1; 0, 0, 0, 0, 1, 0, 0, 0, 1; ... . - Eric W. Weisstein, Apr 08 2006
Limit_{n->oo} log(a(n))/n = log(2). - Bret Mulvey, May 17 2008
Equals row sums of triangle A166548; e.g., 17 = (2 + 4 + 6 + 4 + 1). - Gary W. Adamson, Oct 16 2009
Equals row sums of triangle A166555. - Gary W. Adamson, Oct 17 2009
For n >= 1, all terms are in A001969. - Vladimir Shevelev, Oct 25 2010
Let n,m >= 0 be such that no carries occur when adding them. Then a(n+m) = a(n)*a(m). - Vladimir Shevelev, Nov 28 2010
Let phi_a(n) be the number of a(k) <= a(n) and respectively prime to a(n) (i.e., totient function over {a(n)}). Then, for n >= 1, phi_a(n) = 2^v(n), where v(n) is the number of 0's in the binary representation of n. - Vladimir Shevelev, Nov 29 2010
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960. - Vladimir Shevelev, Jan 02 2011
Converting the rows of the powers of the k-nomial (k = 2^e where e >= 1) term-wise to binary and reading the concatenation as binary number gives every (k-1)st term of this sequence. Similarly with powers p^k of any prime. It might be interesting to study how this fails for powers of composites. - Joerg Arndt, Jan 07 2011
This sequence appears in Pascal's triangle mod 2 in another way, too. If we write it as
  1111111...
  10101010...
  11001100...
  10001000...
we get (taking the period part in each row):
  .(1) (base 2) = 1
  .(10) = 2/3
  .(1100) = 12/15 = 4/5
  .(1000) = 8/15
The k-th row, treated as a binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 16-18 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct Fermat primes (thus there are 31 constructible odd-sided polygons, since a polygon has at least 3 sides). a(0)=1 (empty product) and a(1) to a(31) are those 31 non-products of distinct Fermat primes.
It can be proved by induction that all terms of this sequence are products of distinct Fermat numbers (A000215):
a(0)=1 (empty product) are products of distinct Fermat numbers in { };
a(2^n+k) = a(k) * (2^(2^n)+1) = a(k) * F_n, n >= 0, 0 <= k <= 2^n - 1.
Thus for n >= 1, 0 <= k <= 2^n - 1, and
  a(k) = Product_{i=0..n-1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
  a(2^n+k) = Product_{i=0..n-1} F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.)  (End)
The bits in the binary expansion of a(n) give the coefficients of the n-th power of polynomial (X+1) in ring GF(2)[X]. E.g., 3 ("11" in binary) stands for (X+1)^1, 5 ("101" in binary) stands for (X+1)^2 = (X^2 + 1), and so on. - Antti Karttunen, Feb 10 2016

			Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From _Daniel Forgues_, Jun 18 2011: (Start)
  a(0) = 1 (empty product);
  a(1) = 3 = 1 * F_0 = a(2^0+0) = a(0) * F_0;
  a(2) = 5 = 1 * F_1 = a(2^1+0) = a(0) * F_1;
  a(3) = 15 = 3 * 5 = F_0 * F_1 = a(2^1+1) = a(1) * F_1;
  a(4) = 17 = 1 * F_2 = a(2^2+0) = a(0) * F_2;
  a(5) = 51 = 3 * 17 = F_0 * F_2 = a(2^2+1) = a(1) * F_2;
  a(6) = 85 = 5 * 17 = F_1 * F_2 = a(2^2+2) = a(2) * F_2;
  a(7) = 255 = 3 * 5 * 17 = F_0 * F_1 * F_2 = a(2^2+3) = a(3) * F_2;
  ... (End)

Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
Henry Wadsworth Gould, Exponential Binomial Coefficient Series, Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Amiram Eldar, Table of n, a(n) for n = 0..3321 (terms 0..300 from T. D. Noe, terms 301..1023 from Tilman Piesk)
Gary W. Adamson, Letter to N. J. A. Sloane, Apr 29 1994, and MS "Algorithm for the Chinese Remainder Theorem".
Gary W. Adamson and N. J. A. Sloane, Correspondence, May 1994, including Adamson's MSS "Algorithm for Generating nth Row of Pascal's Triangle, mod 2, from n", and "The Tower of Hanoi Wheel".
Cristian Cobeli and Alexandru Zaharescu, Promenade around Pascal Triangle-Number Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104), No. 1 (2013), pp. 73-98; alternative link.
Richard K. Guy, The Second Strong Law of Small Numbers, Math. Mag, Vol. 63, No. 1 (1990), pp. 3-20. [Annotated scanned copy]
Richard K. Guy and N. J. A. Sloane, Correspondence, 1988.
Denton Hewgill, A relationship between Pascal's triangle and Fermat numbers, Fib. Quart., Vol. 15, No. 2 (1977), pp. 183-184.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata with even rule numbers, Fig. 4.
Dr. Math, Regular polygon formulas.
P. Mathonet, M. Rigo, M. Stipulanti and N. Zénaïdi, On digital sequences associated with Pascal's triangle, arXiv:2201.06636 [math.NT], 2022.
OEIS Wiki, Constructible odd-sided polygons and Sierpinski's triangle.
Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, J. Alg. Number Theory, Vol. 7, No. 1 (2012) pp. 11-29, preprint, arXiv:1011.6083 [math.NT], 2010-2012.
John Frederick Sweeney, Clifford Clock and the Moolakaprithi Cube, 2014. See p. 26. - _N. J. A. Sloane_, Mar 20 2014
Eric Weisstein's World of Mathematics, Rule 60 and Rule 102.
Neil L. White, Letter to N. J. A. Sloane, Apr 15 1982.
R. G. Wilson, V, Letter to N. J. A. Sloane, Aug. 1993.
Index entries for sequences related to cellular automata
Index entries for sequences operating on polynomials in ring GF(2)[X]

Cf. A000079, A000215 (Fermat numbers), A047999, A048720, A054432, A173019, A177882, A177897, A177960, A193231, A230116, A247282, A249184, A268391.
Cf. A038183 (odd bisection, 1D Cellular Automata Rule 90).
Iterates of A048724 (starting from 1).
Row 3 of A048723.
Positions of records in A268389.
Positions of ones in A268669 and A268384 (characteristic function).
Not the same as A045544 nor as A053576.
Cf. A045544.

a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
-- Reinhard Zumkeller, Nov 24 2012
(Scheme, with memoization-macro definec, two variants)
(definec (A001317 n) (if (zero? n) 1 (A048724 (A001317 (- n 1)))))
(definec (A001317 n) (if (zero? n) 1 (A048720bi 3 (A001317 (- n 1))))) ;; Where A048720bi implements the dyadic function given in A048720.
;; Antti Karttunen, Feb 10 2016

[&+[(Binomial(n, i) mod 2)*2^i: i in [0..n]]: n in [0..41]]; // Vincenzo Librandi, Feb 12 2016

A001317 := proc(n) local k; add((binomial(n,k) mod 2)*2^k, k=0..n); end;

a[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)
NestList[BitXor[#,2#]&,1,50] (* Harvey P. Dale, Aug 02 2021 *)

PARI
```
a(n)=sum(i=0,n,(binomial(n,i)%2)*2^i)
```

a=1; for(n=0, 66, print1(a,", "); a=bitxor(a,a<<1) ); \\ Joerg Arndt, Mar 27 2013

A001317(n,a=1)={for(k=1,n,a=bitxor(a,a<<1));a} \\ M. F. Hasler, Jun 06 2016

a(n) = subst(lift(Mod(1+'x,2)^n), 'x, 2); \\ Gheorghe Coserea, Nov 09 2017

from sympy import binomial
def a(n): return sum([(binomial(n, i)%2)*2**i for i in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

def A001317(n): return int(''.join(str(int(not(~n&k))) for k in range(n+1)),2) # Chai Wah Wu, Feb 04 2022

a(n+1) = a(n) XOR 2*a(n), where XOR is binary exclusive OR operator. - Paul D. Hanna, Apr 27 2003
a(n) = Product_{e(j, n) = 1} (2^(2^j) + 1), where e(j, n) is the j-th least significant digit in the binary representation of n (Roberts: see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
a(2*n+1) = 3*a(2*n). Proof: Since a(n) = Product_{k in K} (1 + 2^(2^k)), where K is the set of integers such that n = Sum_{k in K} 2^k, clearly K(2*n+1) = K(2*n) union {0}, hence a(2*n+1) = (1+2^(2^0))*a(2*n) = 3*a(2*n). - Emmanuel Ferrand and Ralf Stephan, Sep 28 2004
a(32*n) = 3 ^ (32 * n * log(2) / log(3)) + 1. - Bret Mulvey, May 17 2008
For n >= 1, A000120(a(n)) = 2^A000120(n). - Vladimir Shevelev, Oct 25 2010
a(2^n) = A000215(n); a(2^n-1) = a(2^n)-2; for n >= 1, m >= 0,
a(2^(n-1)-1)*a(2^n*m + 2^(n-1)) = 3*a(2^(n-1))*a(2^n*m + 2^(n-1)-2). - Vladimir Shevelev, Nov 28 2010
Sum_{k>=0} 1/a(k) = Product_{n>=0} (1 + 1/F_n), where F_n=A000215(n);
Sum_{k>=0} (-1)^(m(k))/a(k) = 1/2, where {m(n)} is Thue-Morse sequence (A010060).
If F_n is defined by F_n(z) = z^(2^n) + 1 and a(n) by (1/2)*Sum_{i>=0}(1-(-1)^{binomial(n,i)})*z^i, then, for z > 1, the latter two identities hold as well with the replacement 1/2 in the right hand side of the 2nd one by 1-1/z. - Vladimir Shevelev, Nov 29 2010
G.f.: Product_{k>=0} ( 1 + z^(2^k) + (2*z)^(2^k) ). - conjectured by Shamil Shakirov, proved by Vladimir Shevelev
a(n) = A000225(n+1) - A219843(n). - Reinhard Zumkeller, Nov 30 2012
From Antti Karttunen, Feb 10 2016: (Start)
a(0) = 1, and for n > 1, a(n) = A048720(3, a(n-1)) = A048724(a(n-1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)

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

cp's OEIS Frontend

A227434 Duplicate of A173019.

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A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

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

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Programs

Haskell

Magma

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