A268669 -id:A268669 - cp's OEIS frontend

A268670 a(n) = A006068(A268669(n)).

1, 3, 1, 7, 1, 3, 5, 15, 5, 3, 13, 7, 9, 11, 1, 31, 1, 11, 29, 7, 25, 27, 9, 15, 17, 19, 5, 23, 13, 3, 21, 63, 21, 3, 61, 23, 57, 59, 13, 15, 49, 51, 17, 55, 5, 19, 53, 31, 33, 35, 1, 39, 29, 11, 37, 47, 9, 27, 45, 7, 41, 43, 25, 127, 25, 43, 125, 7, 121, 123, 41, 47, 113, 115, 9, 119, 45, 27, 117, 31, 97, 99, 33, 103, 1

Offset: 1

Antti Karttunen, Feb 10 2016

All terms are odd, by definition.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A001317 (positions of ones).

Cf. A006068, A268669, A268671.

f[n_] := Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; g[n_] := If[OddQ@ #, n, g[#/2]] &@ f[n]; Table[f@ g@ n, {n, 85}]  (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

(define (A268670 n) (A006068 (A268669 n)))

a(n) = A006068(A268669(n)).

A276579 Squarefree numbers larger than one for which A268669(A048675(n)) is a power of two.

Original entry on oeis.org

2, 3, 5, 6, 7, 10, 11, 13, 15, 17, 19, 21, 22, 23, 29, 31, 35, 37, 39, 41, 43, 46, 47, 53, 55, 59, 61, 67, 71, 73, 77, 79, 83, 85, 87, 89, 91, 97, 101, 103, 107, 109, 113, 118, 127, 131, 133, 137, 139, 143, 149, 151, 155, 157, 163, 167, 173, 179, 181, 183, 187, 191, 193, 197, 199, 210, 211, 221, 223, 227, 229, 233

Offset: 1

Antti Karttunen, Sep 18 2016

nonn

Terms present in arrays A255483 & A276578, sorted into ascending order.

Antti Karttunen, Table of n, a(n) for n = 1..2000

Cf. A000120, A008683, A008966, A048675, A209229, A268669.
Cf. array A255483 (A276578).
Subsequence of A005117.
A000040 is a subsequence.

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

Original entry on oeis.org

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)

A268389 a(n) = greatest k such that polynomial (X+1)^k divides the polynomial (in polynomial ring GF(2)[X]) that is encoded in the binary expansion of n. (See the comments for details).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 4, 1, 0, 2, 0, 0, 1, 1, 0, 0, 2, 0, 1, 3, 0, 0, 1, 4, 0, 1, 0, 0, 2, 2, 0, 0, 1, 0, 3, 1, 0, 1, 0, 0, 5, 0, 1, 2, 0, 0, 2, 1, 0, 3, 0, 0, 1, 0, 2, 1, 0, 4, 0, 0, 1, 1, 0, 0, 3, 0, 1, 2, 0, 2, 0, 0, 1, 0, 6, 1, 0, 0, 1, 3, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 1, 5, 0, 0, 3, 1, 0, 2, 0, 0, 1, 0, 1, 2, 0, 1, 0, 0, 4, 3

Offset: 1

Antti Karttunen, Feb 10 2016

a(n) gives the number of iterations of map k -> A006068(k)/2 that are required (when starting from k = n) until k is an odd number.
A001317 gives the record positions and particularly, A001317(n) gives the first occurrence of n in this sequence.
When polynomials over GF(2) are encoded in the binary representation of n in a natural way where each polynomial b(n)*X^n+...+b(0)*X^0 over GF(2) is represented by the binary number b(n)*2^n+...+b(0)*2^0 in N (each coefficient b(k) is either 0 or 1), then a(n) = the number of times polynomial X+1 (encoded by 3, "11" in binary) divides the polynomial encoded by n.

			For n = 5 ("101" in binary) which encodes polynomial x^2 + 1, we see that it can be factored over GF(2) as (X+1)(X+1), and thus a(5) = 2.
For n = 8 ("1000" in binary) which encodes polynomial x^3, we see that it is not divisible in ring GF(2)[X] by polynomial X+1, thus a(8) = 0.
For n = 9 ("1001" in binary) which encodes polynomial x^3 + 1, we see that it can be factored over GF(2) as (X+1)(X^2 + X + 1), and thus a(9) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences operating on GF(2)[X]-polynomials

Cf. A001317, A006068, A007949, A048720, A048723, A048724, A057889, A268384, A235042, A268670.
Cf. A268669 (quotient left after (X+1)^a(n) has been divided out).
Cf. A268395 (partial sums).
Cf. A000069 (positions of zeros), A268679 (this sequence without zeros).
Cf. also A037096, A037097, A136386.

f[n_] := Which[n == 1, 0, OddQ@ #, 0, EvenQ@ #, 1 + f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Array[f, {120}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

a(n) = {my(b = binary(n), p = Pol(binary(n))*Mod(1,2), k = poldegree(p)); while (type(p/(x+1)^k*Mod(1,2)) != "t_POL", k--); k;} \\ Michel Marcus, Feb 12 2016

;; This employs the given recurrence and uses memoization-macro definec:
(definec (A268389 n) (if (odd? (A006068 n)) 0 (+ 1 (A268389 (/ (A006068 n) 2)))))
(define (A268389 n) (let loop ((n n) (s 0)) (let ((k (A006068 n))) (if (odd? k) s (loop (/ k 2) (+ 1 s)))))) ;; Computed in a loop, no memoization.

If A006068(n) is odd, then a(n) = 0, otherwise a(n) = 1 + a(A006068(n)/2).
Other identities. For all n >= 0:
a(A001317(n)) = n. [The sequence works as a left inverse of A001317.]
A048720(A268669(n),A048723(3,a(n))) = A048720(A268669(n),A001317(a(n))) = n.
A048724^a(n) (A268669(n)) = n. [The a(n)-th fold application (power) of A048724 when applied to A268669(n) gives n back.]
a(n) = A007949(A235042(n)).
a(A057889(n)) = a(n).

A280500 Square array for division in ring GF(2)[X]: A(r,c) = r/c, or 0 if c is not a divisor of r, where the binary expansion of each number defines the corresponding (0,1)-polynomial.

Original entry on oeis.org

1, 0, 2, 0, 1, 3, 0, 0, 0, 4, 0, 0, 1, 2, 5, 0, 0, 0, 0, 0, 6, 0, 0, 0, 1, 3, 3, 7, 0, 0, 0, 0, 0, 2, 0, 8, 0, 0, 0, 0, 1, 0, 0, 4, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 1, 0, 2, 7, 5, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 12, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 6, 13, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 4, 0, 14, 0, 0, 0, 0, 0, 0, 0, 1, 3, 3, 0, 3, 0, 7, 15

Offset: 1

Antti Karttunen, Jan 09 2017

The array A(row,col) is read by descending antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

			The top left 17 X 17 corner of the array:
col: 1  2   3  4  5  6  7  8  9 10 11 12 13 14 15 16 17
     --------------------------------------------------
     1, 0,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     2, 1,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     3, 0,  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     4, 2,  0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     5, 0,  3, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     6, 3,  2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     7, 0,  0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
     8, 4,  0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0
     9, 0,  7, 0, 0, 0, 3, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0
    10, 5,  6, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0
    11, 0,  0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0
    12, 6,  4, 3, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0
    13, 0,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0
    14, 7,  0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0
    15, 0,  5, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0
    16, 8,  0, 4, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0
    17, 0, 15, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 1
    ---------------------------------------------------
7 ("111" in binary) encodes polynomial X^2 + X + 1, which is irreducible over GF(2) (7 is in A014580), thus it is divisible only by itself and 1, and for any other values of c than 1 and 7, A(7,c) = 0.
9 ("1001" in binary) encodes polynomial X^3 + 1, which is factored over GF(2) as (X+1)(X^2 + X + 1), and thus A(9,3) = 7 and A(9,7) = 3 because the polynomial X + 1 is encoded by 3 ("11" in binary).

Cf. A001317, A014580, A048720, A091255, A268389, A268669.
Cf. A280499 for the lower triangular region (A280494 for its transpose).
Cf. also A106449, A280502, A280504, A280506.

up_to = 10440;
A280500sq(a,b) = { my(Pa=Pol(binary(a))*Mod(1, 2), Pb=Pol(binary(b))*Mod(1, 2)); if(0!=lift(Pa % Pb), 0, fromdigits(Vec(lift(Pa/Pb)),2)); };
A280500list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A280500sq(col,(a-(col-1))))); (v); };
v280500 = A280500list(up_to);
A280500(n) = v280500[n]; \\ Antti Karttunen, Jan 05 2025

(define (A280500 n) (A280500bi (A002260 n) (A004736 n)))
;; A very naive implementation:
(define (A280500bi row col) (let loop ((d row)) (cond ((zero? d) d) ((= (A048720bi d col) row) d) (else (loop (- d 1)))))) ;; A048720bi implements the carryless binary multiplication A048720.

A(row,col) = the unique d such that A048720(d,col) = row, provided that such d exists, otherwise zero.
Other identities. For all n >= 1:
A(n, A001317(A268389(n))) = A268669(n).

A268671 a(n) = (A268670(n)+1) / 2.

Original entry on oeis.org

1, 2, 1, 4, 1, 2, 3, 8, 3, 2, 7, 4, 5, 6, 1, 16, 1, 6, 15, 4, 13, 14, 5, 8, 9, 10, 3, 12, 7, 2, 11, 32, 11, 2, 31, 12, 29, 30, 7, 8, 25, 26, 9, 28, 3, 10, 27, 16, 17, 18, 1, 20, 15, 6, 19, 24, 5, 14, 23, 4, 21, 22, 13, 64, 13, 22, 63, 4, 61, 62, 21, 24, 57, 58, 5, 60, 23, 14, 59, 16, 49, 50, 17, 52, 1

Offset: 1

Antti Karttunen, Feb 10 2016

Antti Karttunen, Table of n, a(n) for n = 1..13107

Cf. A001317 (positions of ones).

Cf. A268669, A268670.

f[n_] := Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; g[n_] := If[OddQ@ #, n, g[#/2]] &@ f[n]; Table[(f@ g@ n + 1)/2, {n, 85}]  (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

(define (A268671 n) (/ (+ 1 (A268670 n)) 2))

a(n) = (A268670(n)+1) / 2.

A269165 If A269162(n) = 0, then a(n) = n, otherwise a(n) = a(A269162(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 1, 8, 9, 10, 11, 12, 3, 2, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 1, 6, 5, 4, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 15, 2, 3, 12, 11, 10, 55, 8, 57, 58, 59, 60, 61, 62, 9, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81

Offset: 0

Antti Karttunen, Feb 21 2016

nonn

a(n) is the earliest finite ancestor pattern n in Rule-30 or n itself if n has no finite predecessors.
Starting from k = a(n) with any n and iterating map k -> A269160(k) exactly A269166(n) times yields n back.
Apart from zero no terms of A269163 occur so all terms after zero are in A269164. Each term of A269164 occurs an infinitely many times.

Antti Karttunen, Table of n, a(n) for n = 0..32767
Eric Weisstein's World of Mathematics, Rule 30
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata

Cf. A269160, A269163, A269164, A269166 (for a distance in A269162-steps to the ancestor pattern).
Cf. A110240 (indices of ones in this sequence).
Cf. also A268669.

;; This implementation is based on given recurrence and utilitizes the memoization-macro definec:
(definec (A269165 n) (let ((p (A269162 n))) (if (zero? p) n (A269165 p))))
;; This one computes the same with tail-recursive iteration:
(define (A269165 n) (let loop ((n n) (p (A269162 n))) (if (zero? p) n (loop p (A269162 p)))))

If A269162(n) = 0, then a(n) = n, otherwise a(n) = a(A269162(n)).

A277812 a(n) = the first odious number encountered when starting from k = n and iterating the map k -> A003188(A006068(k)/2).

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 7, 8, 4, 2, 11, 1, 13, 14, 7, 16, 8, 4, 19, 2, 21, 22, 11, 1, 25, 26, 13, 28, 14, 7, 31, 32, 16, 8, 35, 4, 37, 38, 19, 2, 41, 42, 21, 44, 22, 11, 47, 1, 49, 50, 25, 52, 26, 13, 55, 56, 28, 14, 59, 7, 61, 62, 31, 64, 32, 16, 67, 8, 69, 70, 35, 4, 73, 74, 37, 76, 38, 19, 79, 2, 81, 82, 41, 84, 42, 21, 87, 88, 44, 22, 91, 11, 93, 94, 47, 1, 97, 98, 49, 100

Offset: 1

Antti Karttunen, Nov 03 2016

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to binary expansion of n

Cf. A000069, A001969, A000265, A003188, A006068, A010060, A268669, A277813.
Cf. A277808 (gives the number of such iterations needed to reach a(n) from n).
Cf. A003945 (the positions of 1's in this sequence).

If A010060(n) = 1 [when n is one of the odious numbers, A000069], then a(n) = n, otherwise a(n) = a(A003188(A006068(n)/2)).
Other identities:
a(n) = A000069(A277813(n)).
If A010060(n) = 0 [when n is one of the evil numbers, A001969], then a(n)= a(A000265(n)) [the trailing zeros in binary expansion of n do not affect the result].
For all n >= 1, a(A000069(n)) = A000069(n). [By definition].
For all n > 1, a(A001969(n)) < A001969(n).

cp's OEIS Frontend

A268670 a(n) = A006068(A268669(n)).

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A276579 Squarefree numbers larger than one for which A268669(A048675(n)) is a power of two.

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

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A268389 a(n) = greatest k such that polynomial (X+1)^k divides the polynomial (in polynomial ring GF(2)[X]) that is encoded in the binary expansion of n. (See the comments for details).

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A280500 Square array for division in ring GF(2)[X]: A(r,c) = r/c, or 0 if c is not a divisor of r, where the binary expansion of each number defines the corresponding (0,1)-polynomial.

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A268671 a(n) = (A268670(n)+1) / 2.

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A269165 If A269162(n) = 0, then a(n) = n, otherwise a(n) = a(A269162(n)).

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