A247282 -id:A247282 - 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)

A246674 Run Length Transform of A000225.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 3, 3, 3, 9, 3, 3, 9, 21, 7, 7, 7, 21, 15, 15, 31, 63, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 3, 3, 3, 9, 3, 3, 9, 21, 3, 3, 3, 9, 9, 9, 21, 45, 7, 7, 7, 21, 7, 7, 21, 49, 15, 15, 15, 45, 31, 31, 63, 127, 1

Offset: 0

Antti Karttunen, Sep 08 2014

a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus a(115) = A000225(2) * A000225(3) = ((2^2)-1) * ((2^3)-1) = 3*7 = 21.
The same result can be also obtained more directly, by realizing that '111' and '11' are the binary representations of 7 and 3, and 7*3 = 21.
From _Omar E. Pol_, Feb 15 2015: (Start)
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,3;
1,1,3,7;
1,1,1,3,3,3,7,15;
1,1,1,3,1,1,3,7,3,3,3,9,7,7,15,31;
1,1,1,3,1,1,3,7,1,1,1,3,3,3,7,15,3,3,3,9,3,3,9,21,7,7,7,21,15,15,31,63;
...
Right border gives 1 together with the positive terms of A000225.
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8192

Cf. A003714 (gives the positions of ones).
Cf. A000225, A051179.
A001316 is obtained when the same transformation is applied to A000079, the powers of two.
Run Length Transforms of other sequences: A071053, A227349, A246588, A246595, A246596, A246660, A246661, A246685, A247282.

f[n_] := 2^n - 1; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)

# uses RLT function in A278159
def A246674(n): return RLT(n,lambda m: 2**m-1) # Chai Wah Wu, Feb 04 2022

For all n >= 0, a(A051179(n)) = A247282(A051179(n)) = A051179(n).

A286575 Run-length transform of A001316.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 2, 2, 4, 4, 4, 4, 8, 4, 8, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 2, 4, 4, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 4, 4, 8, 8, 8, 8, 16, 8, 16, 4, 8, 8, 8, 8, 16, 4, 8, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4

Offset: 0

Antti Karttunen, May 28 2017

			For n = 0, there are no 1-runs, and thus a(0) = 1 as an empty product.
For n = 29, "11101" in binary, there are two 1-runs, of lengths 1 and 3, thus a(29) = A001316(1) * A001316(3) = 2*4 = 8.

Cf. A000079, A001316, A005940, A037445, A227349, A247282, A286574.

Table[Times @@ Map[Sum[Mod[#, 2] &@ Binomial[#, k], {k, 0, #}] &@ Length@ # &, DeleteCases[Split@ IntegerDigits[n, 2], ?(First@ # == 0 &)]], {n, 0, 108}] (* _Michael De Vlieger, May 29 2017 *)

from sympy import factorint, prime, log
import math
def wt(n): return bin(n).count("1")
def a037445(n):
    f=factorint(n)
    return 2**sum([wt(f[i]) for i in f])
def A(n): return n - 2**int(math.floor(log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a(n): return a037445(b(n)) # Indranil Ghosh, May 30 2017

# use RLT function from A278159
def A286575(n): return RLT(n,lambda m: 2**(bin(m).count('1'))) # Chai Wah Wu, Feb 04 2022

(define (A286575 n) (fold-left (lambda (a r) (* a (A001316 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z)))))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
(define (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2))))))

a(n) = A037445(A005940(1+n)).

a(n) = A000079(A286574(n)).

A246685 Run Length Transform of sequence 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 3, 3, 3, 9, 3, 3, 9, 15, 5, 5, 5, 15, 17, 17, 257, 65537, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257

Offset: 0

Antti Karttunen, Sep 22 2014

nonn

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to sequence b = 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers, with b(1) = 1, b(2) = 3, b(3) = 5, ..., b(n) = 2^(2^(n-2)) + 1 for n >= 2).

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus we multiply the second and the third elements of the sequence 1, 3, 5, 17, 257, 65537, ... to get a(115) = 3*5 = 15.

Antti Karttunen, Table of n, a(n) for n = 0..1024

Cf. A003714 (gives the positions of ones).
Cf. A000215.
A001316 is obtained when the same transformation is applied to A000079, the powers of two. Cf. also A001317.
Run Length Transforms of other sequences: A071053, A227349, A246588, A246595, A246596, A246660, A246661, A246674, A247282.

f[n_] := Switch[n, 0|1, 1, , 2^(2^(n-2))+1]; Table[Times @@ (f[Length[#]] &) /@ Select[s = Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 95}] (* _Jean-François Alcover, Jul 11 2017 *)

# use RLT function from A278159
def A246685(n): return RLT(n,lambda m: 1 if m <= 1 else 2**(2**(m-2))+1) # Chai Wah Wu, Feb 04 2022

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

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A246674 Run Length Transform of A000225.

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A286575 Run-length transform of A001316.

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A246685 Run Length Transform of sequence 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers).

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