cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A336843 Period of binary representation of 1/A003961(n): a(n) = A007733(A003961(n)).

Original entry on oeis.org

1, 2, 4, 6, 3, 4, 10, 18, 20, 6, 12, 12, 8, 10, 12, 54, 18, 20, 11, 6, 20, 12, 28, 36, 21, 8, 100, 30, 5, 12, 36, 162, 12, 18, 30, 60, 20, 22, 8, 18, 14, 20, 23, 12, 60, 28, 52, 108, 110, 42, 36, 24, 58, 100, 12, 90, 44, 10, 60, 12, 66, 36, 20, 486, 24, 12, 35, 18, 28, 30, 9, 180, 39, 20, 84, 66, 60, 8, 82, 54, 500
Offset: 1

Views

Author

Antti Karttunen, Aug 06 2020

Keywords

Crossrefs

Permutation of A002326.

Programs

  • PARI
    A336843(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); znorder(Mod(2,factorback(f))); };

Formula

a(n) = A007733(A003961(n)).
a(n) = A002326(A108228(n)) = A002326(A048673(n)-1).

A243050 Composite integers n such that n-1 divided by the binary period of 1/n (=A007733(n)) equals an integral power of 2.

Original entry on oeis.org

12801, 348161, 3225601, 104988673, 4294967297, 7816642561, 43796171521, 49413980161, 54745942917121, 51125767490519041, 18314818035992494081, 18446744073709551617
Offset: 1

Views

Author

Max Alekseyev, May 29 2014

Keywords

Comments

All terms are odd. If even n belongs to this sequence, then n-1 is odd and thus (n-1)/A007733(n) is also odd and thus must be equal to 1. On the other hand, for even n, A007733(n) < n/2 <= n-1, i.e., (n-1)/A007733(n) > 1, a contradiction.
Subsequence of A001567.
Contains all composite Fermat numbers A000215(k) = 2^(2^k)+1 (which are composite for 5<=k<=32 and conjecturally for any k>=5). In particular, a(5) = A000215(5), a(12) = A000215(6), and a(13) <= A000215(7) = 2^128+1.
Pseudoprimes n such that (n-1)/ord_{n}(2) = 2^k for some k, where ord_{n}(2) = A002326((n-1)/2) is the multiplicative order of 2 mod n. Composite numbers n such that Od(ord_{n}(2)) = Od(n-1), where ord_{n}(2) as above and Od(m) = A000265(m) is the odd part of m. Note that if Od(ord_{n}(2)) = Od(n-1), then ord_{n}(2)|(n-1). - Thomas Ordowski, Mar 13 2019

Examples

			n = a(6) = 7816642561 = 2^15 * 238545 + 1 is the first term, which is not Proth number (A080075). The binary period of 1/n is 954180 = (n-1)/2^13.
		

Crossrefs

Extensions

a(1)-a(3) from Lear Young; a(4)-a(5),a(9)-a(12) from Max Alekseyev; a(6),a(8) from Peter Kosinar; a(7) from Chris Boyd, May 29 2014.

A333746 Numbers k such that k, k+1 and k+2 have the same period of binary representation (A007733).

Original entry on oeis.org

23284, 77906, 509737, 717817, 996601, 1132177, 1550377, 3264241, 3896546, 4326962, 4491362, 4542457, 5978857, 7097161, 8981977, 9628921, 10140386, 11098201, 11472337, 12078217, 12699122, 13335457, 14079577, 16795417, 17796146, 17807017, 18832082, 20221106, 21096146
Offset: 1

Views

Author

Amiram Eldar, Apr 03 2020

Keywords

Comments

Numbers k such that A007733(k) = A007733(k+1) = A007733(k+2).

Examples

			23284 is a term since A007733(23284) = A007733(23285) = A007733(23286) = 388.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := MultiplicativeOrder[2, n/(2^IntegerExponent[n, 2])]; f1 = f[1]; f2 = f[2]; seq = {}; Do[f3 = f[n]; If[f1 == f2 && f2 == f3, AppendTo[seq, n-2]]; f1 = f2; f2 = f3, {n, 3, 10^6}]; seq

A000225 a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
Offset: 0

Views

Author

Keywords

Comments

This is the Gaussian binomial coefficient [n,1] for q=2.
Number of rank-1 matroids over S_n.
Numbers k such that the k-th central binomial coefficient is odd: A001405(k) mod 2 = 1. - Labos Elemer, Mar 12 2003
This gives the (zero-based) positions of odd terms in the following convolution sequences: A000108, A007460, A007461, A007463, A007464, A061922.
Also solutions (with minimum number of moves) for the problem of Benares Temple, i.e., three diamond needles with n discs ordered by decreasing size on the first needle to place in the same order on the third one, without ever moving more than one disc at a time and without ever placing one disc at the top of a smaller one. - Xavier Acloque, Oct 18 2003
a(0) = 0, a(1) = 1; a(n) = smallest number such that a(n)-a(m) == 0 (mod (n-m+1)), for all m. - Amarnath Murthy, Oct 23 2003
Binomial transform of [1, 1/2, 1/3, ...] = [1/1, 3/2, 7/3, ...]; (2^n - 1)/n, n=1,2,3, ... - Gary W. Adamson, Apr 28 2005
Numbers whose binary representation is 111...1. E.g., the 7th term is (2^7) - 1 = 127 = 1111111 (in base 2). - Alexandre Wajnberg, Jun 08 2005
Number of nonempty subsets of a set with n elements. - Michael Somos, Sep 03 2006
For n >= 2, a(n) is the least Fibonacci n-step number that is not a power of 2. - Rick L. Shepherd, Nov 19 2007
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which either x is a subset of y or y is a subset of x. - Ross La Haye, Jan 10 2008
A simpler way to state this is that it is the number of pairs (x,y) where at least one of x and y is the empty set. - Franklin T. Adams-Watters, Oct 28 2011
2^n-1 is the sum of the elements in a Pascal triangle of depth n. - Brian Lewis (bsl04(AT)uark.edu), Feb 26 2008
Sequence generalized: a(n) = (A^n -1)/(A-1), n >= 1, A integer >= 2. This sequence has A=2; A003462 has A=3; A002450 has A=4; A003463 has A=5; A003464 has A=6; A023000 has A=7; A023001 has A=8; A002452 has A=9; A002275 has A=10; A016123 has A=11; A016125 has A=12; A091030 has A=13; A135519 has A=14; A135518 has A=15; A131865 has A=16; A091045 has A=17; A064108 has A=20. - Ctibor O. Zizka, Mar 03 2008
a(n) is also a Mersenne prime A000668 when n is a prime number in A000043. - Omar E. Pol, Aug 31 2008
a(n) is also a Mersenne number A001348 when n is prime. - Omar E. Pol, Sep 05 2008
With offset 1, = row sums of triangle A144081; and INVERT transform of A009545 starting with offset 1; where A009545 = expansion of sin(x)*exp(x). - Gary W. Adamson, Sep 10 2008
Numbers n such that A000120(n)/A070939(n) = 1. - Ctibor O. Zizka, Oct 15 2008
For n > 0, sequence is equal to partial sums of A000079; a(n) = A000203(A000079(n-1)). - Lekraj Beedassy, May 02 2009
Starting with offset 1 = the Jacobsthal sequence, A001045, (1, 1, 3, 5, 11, 21, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Numbers n such that n=2*phi(n+1)-1. - Farideh Firoozbakht, Jul 23 2009
a(n) = (a(n-1)+1)-th odd numbers = A005408(a(n-1)) for n >= 1. - Jaroslav Krizek, Sep 11 2009
Partial sums of a(n) for n >= 0 are A000295(n+1). Partial sums of a(n) for n >= 1 are A000295(n+1) and A130103(n+1). a(n) = A006127(n) - (n+1). - Jaroslav Krizek, Oct 16 2009
If n is even a(n) mod 3 = 0. This follows from the congruences 2^(2k) - 1 ~ 2*2*...*2 - 1 ~ 4*4*...*4 - 1 ~ 1*1*...*1 - 1 ~ 0 (mod 3). (Note that 2*2*...*2 has an even number of terms.) - Washington Bomfim, Oct 31 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 26 2010
This is the sequence A(0,1;1,2;2) = A(0,1;3,-2;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
a(n) = S(n+1,2), a Stirling number of the second kind. See the example below. - Dennis P. Walsh, Mar 29 2011
Entries of row a(n) in Pascal's triangle are all odd, while entries of row a(n)-1 have alternating parities of the form odd, even, odd, even, ..., odd.
Define the bar operation as an operation on signed permutations that flips the sign of each entry. Then a(n+1) is the number of signed permutations of length 2n that are equal to the bar of their reverse-complements and avoid the set of patterns {(-2,-1), (-1,+2), (+2,+1)}. (See the Hardt and Troyka reference.) - Justin M. Troyka, Aug 13 2011
A159780(a(n)) = n and A159780(m) < n for m < a(n). - Reinhard Zumkeller, Oct 21 2011
This sequence is also the number of proper subsets of a set with n elements. - Mohammad K. Azarian, Oct 27 2011
a(n) is the number k such that the number of iterations of the map k -> (3k +1)/2 == 1 (mod 2) until reaching (3k +1)/2 == 0 (mod 2) equals n. (see the Collatz problem). - Michel Lagneau, Jan 18 2012
For integers a, b, denote by a<+>b the least c >= a such that Hd(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>1. Thus this sequence is the Hamming analog of nonnegative integers. - Vladimir Shevelev, Feb 13 2012
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... apparently A007733. - R. J. Mathar, Aug 10 2012
Start with n. Each n generates a sublist {n-1,n-2,...,1}. Each element of each sublist also generates a sublist. Take the sum of all. E.g., 3->{2,1} and 2->{1}, so a(3)=3+2+1+1=7. - Jon Perry, Sep 02 2012
This is the Lucas U(P=3,Q=2) sequence. - R. J. Mathar, Oct 24 2012
The Mersenne numbers >= 7 are all Brazilian numbers, as repunits in base two. See Proposition 1 & 5.2 in Links: "Les nombres brésiliens". - Bernard Schott, Dec 26 2012
Number of line segments after n-th stage in the H tree. - Omar E. Pol, Feb 16 2013
Row sums of triangle in A162741. - Reinhard Zumkeller, Jul 16 2013
a(n) is the highest power of 2 such that 2^a(n) divides (2^n)!. - Ivan N. Ianakiev, Aug 17 2013
In computer programming, these are the only unsigned numbers such that k&(k+1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013
Minimal number of moves needed to interchange n frogs in the frogs problem (see for example the NRICH 1246 link or the Britton link below). - N. J. A. Sloane, Jan 04 2014
a(n) !== 4 (mod 5); a(n) !== 10 (mod 11); a(n) !== 2, 4, 5, 6 (mod 7). - Carmine Suriano, Apr 06 2014
After 0, antidiagonal sums of the array formed by partial sums of integers (1, 2, 3, 4, ...). - Luciano Ancora, Apr 24 2015
a(n+1) equals the number of ternary words of length n avoiding 01,02. - Milan Janjic, Dec 16 2015
With offset 0 and another initial 0, the n-th term of 0, 0, 1, 3, 7, 15, ... is the number of commas required in the fully-expanded von Neumann definition of the ordinal number n. For example, 4 := {0, 1, 2, 3} := {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}, which uses seven commas. Also, for n>0, a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n - 1, where a single symbol (as usual) is always used to represent the empty set and spaces are ignored. E.g., a(5) = 31, the total such symbols for the ordinal 4. - Rick L. Shepherd, May 07 2016
With the quantum integers defined by [n+1]A001045%20are%20given%20by%20q%20=%20i%20*%20sqrt(2)%20for%20i%5E2%20=%20-1.%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Mersenne numbers are a(n+1) = q^n [n+1]_q with q = sqrt(2), whereas the signed Jacobsthal numbers A001045 are given by q = i * sqrt(2) for i^2 = -1. Cf. A239473. - _Tom Copeland, Sep 05 2016
For n>1: numbers n such that n - 1 divides sigma(n + 1). - Juri-Stepan Gerasimov, Oct 08 2016
This is also the second column of the Stirling2 triangle A008277 (see also A048993). - Wolfdieter Lang, Feb 21 2017
Except for the initial terms, the decimal representation of the x-axis of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 659", "Rule 721" and "Rule 734", based on the 5-celled von Neumann neighborhood initialized with a single on cell. - Robert Price, Mar 14 2017
a(n), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving partial injective mappings on a set with n elements. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Also the number of independent vertex sets and vertex covers in the complete bipartite graph K_{n-1,n-1}. - Eric W. Weisstein, Sep 21 2017
Sum_{k=0..n} p^k is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*p + binomial(i+j-1, i), in this case p=2 (empirical observation). - Tony Foster III, May 11 2019
The rational numbers r(n) = a(n+1)/2^(n+1) = a(n+1)/A000079(n+1) appear also as root of the n-th iteration f^{[n]}(c; x) = 2^(n+1)*x - a(n+1)*c of f(c; x) = f^{[0]}(c; x) = 2*x - c as r(n)*c. This entry is motivated by a riddle of Johann Peter Hebel (1760 - 1826): Erstes Rechnungsexempel(Ein merkwürdiges Rechnungs-Exempel) from 1803, with c = 24 and n = 2, leading to the root r(2)*24 = 21 as solution. See the link and reference. For the second problem, also involving the present sequence, see a comment in A130330. - Wolfdieter Lang, Oct 28 2019
a(n) is the sum of the smallest elements of all subsets of {1,2,..,n} that contain n. For example, a(3)=7; the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 7. - Enrique Navarrete, Aug 21 2020
a(n-1) is the number of nonempty subsets of {1,2,..,n} which don't have an element that is the size of the set. For example, for n = 4, a(3) = 7 and the subsets are {2}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,2,4}. - Enrique Navarrete, Nov 21 2020
From Eric W. Weisstein, Sep 04 2021: (Start)
Also the number of dominating sets in the complete graph K_n.
Also the number of minimum dominating sets in the n-helm graph for n >= 3. (End)
Conjecture: except for a(2)=3, numbers m such that 2^(m+1) - 2^j - 2^k - 1 is composite for all 0 <= j < k <= m. - Chai Wah Wu, Sep 08 2021
a(n) is the number of three-in-a-rows passing through a corner cell in n-dimensional tic-tac-toe. - Ben Orlin, Mar 15 2022
From Vladimir Pletser, Jan 27 2023: (Start)
a(n) == 1 (mod 30) for n == 1 (mod 4);
a(n) == 7 (mod 120) for n == 3 (mod 4);
(a(n) - 1)/30 = (a(n+2) - 7)/120 for n odd;
(a(n) - 1)/30 = (a(n+2) - 7)/120 = A131865(m) for n == 1 (mod 4) and m >= 0 with A131865(0) = 0. (End)
a(n) is the number of n-digit numbers whose smallest decimal digit is 8. - Stefano Spezia, Nov 15 2023
Also, number of nodes in a perfect binary tree of height n-1, or: number of squares (or triangles) after the n-th step of the construction of a Pythagorean tree: Start with a segment. At each step, construct squares having the most recent segment(s) as base, and isosceles right triangles having the opposite side of the squares as hypotenuse ("on top" of each square). The legs of these triangles will serve as the segments which are the bases of the squares in the next step. - M. F. Hasler, Mar 11 2024
a(n) is the length of the longest path in the n-dimensional hypercube. - Christian Barrientos, Apr 13 2024
a(n) is the diameter of the n-Hanoi graph. Equivalently, a(n) is the largest minimum number of moves between any two states of the Towers of Hanoi problem (aka problem of Benares Temple described above). - Allan Bickle, Aug 09 2024

Examples

			For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely,
  {a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - _Dennis P. Walsh_, Mar 29 2011
From _Justin M. Troyka_, Aug 13 2011: (Start)
Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are:
  (+1,+2,-3,-4),
  (+1,+3,-2,-4),
  (+1,-3,+2,-4),
  (+2,+4,-1,-3),
  (+3,+4,-1,-2),
  (-3,+1,-4,+2),
  (-3,-4,+1,+2). (End)
G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ...
For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3.
12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12  - _Allan Bickle_, Aug 07 2024
		

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
  • Johann Peter Hebel, Gesammelte Werke in sechs Bänden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 20-21, Loesung, S. 36-37. See also the link below.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 75-83.
  • 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, page 141.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, "Tower of Hanoi", Penguin Books, 1987, pp. 112-113.

Crossrefs

Cf. A000043 (Mersenne exponents).
Cf. A000668 (Mersenne primes).
Cf. A001348 (Mersenne numbers with n prime).
Cf. a(n)=A112492(n, 2). Rightmost column of A008969.
a(n) = A118654(n, 1) = A118654(n-1, 3), for n > 0.
Subsequence of A132781.
Smallest number whose base b sum of digits is n: this sequence (b=2), A062318 (b=3), A180516 (b=4), A181287 (b=5), A181288 (b=6), A181303 (b=7), A165804 (b=8), A140576 (b=9), A051885 (b=10).
Cf. A008277, A048993 (columns k=2), A000918, A130330.
Cf. A000225, A029858, A058809, A375256 (Hanoi graphs).

Programs

  • Haskell
    a000225 = (subtract 1) . (2 ^)
    a000225_list = iterate ((+ 1) . (* 2)) 0
    -- Reinhard Zumkeller, Mar 20 2012
    
  • Maple
    A000225 := n->2^n-1; [ seq(2^n-1,n=0..50) ];
    A000225:=1/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, sequence starting at a(1)
  • Mathematica
    a[n_] := 2^n - 1; Table[a[n], {n, 0, 30}] (* Stefan Steinerberger, Mar 30 2006 *)
    Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
    NestList[2 # + 1 &, 0, 32] (* Robert G. Wilson v, Feb 28 2011 *)
    2^Range[0, 20] - 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {1, 3}, 20] (* Eric W. Weisstein, Sep 21 2017 *)
    CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    A000225(n) = 2^n-1  \\ Michael B. Porter, Oct 27 2009
    
  • PARI
    concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015
    
  • Python
    def A000225(n): return (1<Chai Wah Wu, Jul 06 2022
  • SageMath
    def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019
    

Formula

G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. - Paul Barry, May 26 2003
a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. - Paul Barry, Jun 06 2003
Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n) = n + Sum_{i=0..n-1} a(i); a(0) = 0. - Rick L. Shepherd, Aug 04 2004
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). - Paul Barry, Aug 06 2004
a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). - Paul Barry, Aug 23 2004
Inverse binomial transform of A001047. Also U sequence of Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A099393(n-1) - A020522(n-1) for n > 0. - Reinhard Zumkeller, Feb 07 2006
a(n) = A119258(n,n-1) for n > 0. - Reinhard Zumkeller, May 11 2006
a(n) = 3*a(n-1) - 2*a(n-2); a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
Sum_{n>0} 1/a(n) = 1.606695152... = A065442, see A038631. - Philippe Deléham, Jun 27 2006
Stirling_2(n-k,2) starting from n=k+1. - Artur Jasinski, Nov 18 2006
a(n) = A125118(n,1) for n > 0. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1,2). - Ross La Haye, Jan 10 2008
a(n) = A024036(n)/A000051(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A024088(n)/A001576(n). -Reinhard Zumkeller, Feb 15 2009
a(2*n) = a(n)*A000051(n); a(n) = A173787(n,0). - Reinhard Zumkeller, Feb 28 2010
For n > 0: A179857(a(n)) = A024036(n) and A179857(m) < A024036(n) for m < a(n). - Reinhard Zumkeller, Jul 31 2010
From Enrique Pérez Herrero, Aug 21 2010: (Start)
a(n) = J_n(2), where J_n is the n-th Jordan Totient function: (A007434, is J_2).
a(n) = Sum_{d|2} d^n*mu(2/d). (End)
A036987(a(n)) = 1. - Reinhard Zumkeller, Mar 06 2012
a(n+1) = A044432(n) + A182028(n). - Reinhard Zumkeller, Apr 07 2012
a(n) = A007283(n)/3 - 1. - Martin Ettl, Nov 11 2012
a(n+1) = A001317(n) + A219843(n); A219843(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2012
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013
a(n) = A000203(2^(n-1)), n >= 1. - Ivan N. Ianakiev, Aug 17 2013
a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). - Mircea Merca, Dec 06 2013
a(0) = 0; a(n) = a(n-1) + 2^(n-1) for n >= 1. - Fred Daniel Kline, Feb 09 2014
a(n) = A125128(n) - A000325(n) + 1. - Miquel Cerda, Aug 07 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Binomial transform of A057427.
Sum_{n>=0} a(n)/n! = A090142. (End)
a(n) = A000918(n) + 1. - Miquel Cerda, Aug 09 2016
a(n+1) = (A095151(n+1) - A125128(n))/2. - Miquel Cerda, Aug 12 2016
a(n) = (A079583(n) - A000325(n+1))/2. - Miquel Cerda, Aug 15 2016
Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. - Anton Zakharov, Sep 05 2016
a(n) = (A083706(n-1) + A000325(n))/2. - Miquel Cerda, Sep 30 2016
a(n) = A005803(n) + A005408(n-1). - Miquel Cerda, Nov 25 2016
a(n) = A279396(n+2,2). - Wolfdieter Lang, Jan 10 2017
a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for A000918(n+1) with an interpretation. - Wolfdieter Lang, Jun 14 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n+m) = a(n)*a(m) + a(n) + a(m). - Yuchun Ji, Jul 27 2018
a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). - Taras Goy, Dec 23 2018
a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). - Tony Foster III, May 11 2019
From Peter Bala, Jun 27 2025: (Start)
For n >= 1, a(3*n)/a(n) = A001576(n), a(4*n)/a(n) = A034496(n), a(5*n)/a(n) = A020514(n) a(6*n)/a(n) = A034665(n), a(7*n)/a(n) = A020516(n) and a(8*n)/a(n) = A034674(n).
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
Modulo differences in offsets, exp( Sum_{n >= 1} a(k*n)/a(n)*x^n/n ) is the o.g.f. of A006095 (k = 3), A006096 (k = 4), A006097 (k = 5), A006110 (k = 6), A022189 (k = 7), A022190 (k = 8), A022191 (k = 9) and A022192 (k = 10).
The following are all examples of telescoping series:
Sum_{n >= 1} 2^n/(a(n)*a(n+1)) = 1; Sum_{n >= 1} 2^n/(a(n)*a(n+1)*a(n+2)) = 1/9.
In general, for k >= 1, Sum_{n >= 1} 2^n/(a(n)*a(n+1)*...*a(n+k)) = 1/(a(1)*a(2)*...*a(k)*a(k)).
Sum_{n >= 1} 2^n/(a(n)*a(n+2)) = 4/9, since 2^n/(a(n)*a(n+2)) = b(n) - b(n+1), where b(n) = (2/3)*(3*2^(n-1) - 1)/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+2)) = -2/9, since (-2)^n/(a(n)*a(n+2)) = c(n) - c(n+1), where c(n) = (1/3)*(-2)^n/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} 2^n/(a(n)*a(n+4)) = 18/175, since 2^n/(a(n)*a(n+4)) = d(n) - d(n+1), where d(n) = (120*8^n - 140*4^n + 45*2^n - 4)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+4)) = -26/525, since (-2)^n/(a(n)*a(n+4)) = e(n) - e(n+1), where e(n) = (-1)^n*(40*8^n - 24*4^n + 5*2^n)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)). (End)

Extensions

Name partially edited by Eric W. Weisstein, Sep 04 2021

A000051 a(n) = 2^n + 1.

Original entry on oeis.org

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

Views

Author

Keywords

Comments

Same as Pisot sequence L(2,3).
Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003
See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004
From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)* charpoly(A,3). - Milan Janjic, Jan 27 2010
First differences of A006127. - Reinhard Zumkeller, Apr 14 2011
The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012
Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012
Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012
For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014
Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014
a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014
For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017
Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017
Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017
Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

References

  • Paul Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 244.
  • 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, page 141.

Crossrefs

Apart from the initial 1, identical to A094373.
See A008776 for definitions of Pisot sequences.
Column 2 of array A103438.
Cf. A007583 (a((n-1)/2)/3 for odd n).

Programs

  • Haskell
    a000051 = (+ 1) . a000079
    a000051_list = iterate ((subtract 1) . (* 2)) 2
    -- Reinhard Zumkeller, May 03 2012
    
  • Magma
    [2^n+1: n in [0..40]]; // G. C. Greubel, Jan 18 2025
  • Maple
    A000051:=-(-2+3*z)/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
    a := n -> add(binomial(n,k)*bernoulli(n-k,1)*2^(k+1)/(k+1),k=0..n); # Peter Luschny, Apr 20 2009
  • Mathematica
    Table[2^n + 1, {n,0,40}]
    2^Range[0,40] + 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {2, 3}, 40] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    a(n)=2^n+1
    
  • PARI
    first(n) = Vec((2 - 3*x)/((1 - x)*(1 - 2*x)) + O(x^n)) \\ Iain Fox, Dec 31 2017
    
  • Python
    def A000051(n): return (1<Chai Wah Wu, Dec 21 2022
    

Formula

a(n) = 2*a(n-1) - 1 = 3*a(n-1) - 2*a(n-2).
G.f.: (2-3*x)/((1-x)*(1-2*x)).
First differences of A052944. - Emeric Deutsch, Mar 04 2004
a(0) = 1, then a(n) = (Sum_{i=0..n-1} a(i)) - (n-2). - Gerald McGarvey, Jul 10 2004
Inverse binomial transform of A007689. Also, V sequence in Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A127904(n+1) for n>0. - Reinhard Zumkeller, Feb 05 2007
Equals binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = A000079(n)+1. - Omar E. Pol, May 18 2008
E.g.f.: exp(x) + exp(2*x). - Mohammad K. Azarian, Jan 02 2009
a(n) = A024036(n)/A000225(n). - Reinhard Zumkeller, Feb 14 2009
From Peter Luschny, Apr 20 2009: (Start)
A weighted binomial sum of the Bernoulli numbers A027641/A027642 with A027641(1)=1 (which amounts to the definition B_{n} = B_{n}(1)).
a(n) = Sum_{k=0..n} C(n,k)*B_{n-k}*2^(k+1)/(k+1). (See also A052584.) (End)
a(n) is the a(n-1)-th odd number for n >= 1. - Jaroslav Krizek, Apr 25 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
a(n)*A000225(n) = A000225(2*n).
a(n) = A173786(n,0). (End)
If p[i]=Fibonacci(i-4) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise, then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
a(n+2) = a(n) + a(n+1) + A000225(n). - Ivan N. Ianakiev, Jun 24 2012
a(A006521(n)) mod A006521(n) = 0. - Reinhard Zumkeller, Jul 17 2014
a(n) = 3*A007583((n-1)/2) for n odd. - Eric W. Weisstein, Jul 17 2017
Sum_{n>=0} 1/a(n) = A323482. - Amiram Eldar, Nov 11 2020

A002326 Multiplicative order of 2 mod 2n+1.

Original entry on oeis.org

1, 2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 28, 5, 10, 12, 36, 12, 20, 14, 12, 23, 21, 8, 52, 20, 18, 58, 60, 6, 12, 66, 22, 35, 9, 20, 30, 39, 54, 82, 8, 28, 11, 12, 10, 36, 48, 30, 100, 51, 12, 106, 36, 36, 28, 44, 12, 24, 110, 20, 100, 7, 14, 130, 18, 36, 68, 138, 46, 60, 28
Offset: 0

Views

Author

Keywords

Comments

In other words, least m > 0 such that 2n+1 divides 2^m-1.
Number of riffle shuffles of 2n+2 cards required to return a deck to initial state. A riffle shuffle replaces a list s(1), s(2), ..., s(m) with s(1), s((i/2)+1), s(2), s((i/2)+2), ... a(1) = 2 because a riffle shuffle of [1, 2, 3, 4] requires 2 iterations [1, 2, 3, 4] -> [1, 3, 2, 4] -> [1, 2, 3, 4] to restore the original order.
Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.
It is not difficult to prove that if 2n+1 is a prime then 2n is a multiple of a(n). But the converse is not true. Indeed, one can prove that a(2^(2t-1))=4t. Thus if n=2^(2t-1), where, for any m > 0, t=2^(m-1) then 2n is a multiple of a(n) while 2n+1 is a Fermat number which, as is well known, is not always a prime. It is an interesting problem to describe all composite numbers for which 2n is divisible by a(n). - Vladimir Shevelev, May 09 2008
For an algorithm of calculation of a(n) see author's comment in A179680. - Vladimir Shevelev, Jul 21 2010
From V. Raman, Sep 18 2012, Dec 10 2012: (Start)
If 2n+1 is prime, then the polynomial (x^(2n+1)+1)/(x+1) factors into 2n/a(n) polynomials of the same degree a(n) over GF(2).
If (x^(2n+1)+1)/(x+1) is irreducible over GF(2), then 2n+1 is prime, and 2 is a primitive root (mod 2n+1) (cf. A001122).
For all n > 0, a(n) is the degree of the largest irreducible polynomial factor for the polynomial (x^(2n+1)+1)/(x+1) over GF(2). (End)
a(n) is a factor of phi(2n+1) (A000010(2n+1)). - Douglas Boffey, Oct 21 2013
Conjecture: if p is an odd prime then a((p^3-1)/2) = p * a((p^2-1)/2). Because otherwise a((p^3-1)/2) < p * a((p^2-1)/2) iff a((p^3-1)/2) = a((p-1)/2) for a prime p. Equivalently p^3 divides 2^(p-1)-1, but no such prime p is known. - Thomas Ordowski, Feb 10 2014
A generalization of the previous conjecture: For each k>=2, if p is an odd prime then a(((p^(k+1))-1)/2) = p * a((p^k-1)/2). Computer testing of this generalized conjecture shows that there is no counterexample for k and p both up to 1000. - Ahmad J. Masad, Oct 17 2020

Examples

			From _Vladimir Shevelev_, Oct 03 2017: (Start)
Our algorithm for the calculation of a(n) in the author's comment in A179680 (see also the Sage program below) could be represented in the form of a "finite continued fraction". For example let n = 8, 2*n+1 = 17. We have
    1 + 17
    ------- + 17
       2
    ------------- + 17
           2
    ------------------- + 17
              2
    -------------------------- = 1
                 32
Here the denominators are the A006519 of the numerators: A006519(1+17) = 2, A006519(9+17) = 2, A006519(13+17) = 2, A006519(15+17) = 32. Summing the exponents of these powers of 2, we obtain the required result: a(8) = 1 + 1 + 1 + 5 = 8. Indeed, we have (((1*32 - 17)*2 - 17)*2 - 17)*2 - 17 = 1. So 32*2*2*2 - 1 == 0 (mod 17), 2^8 - 1 == 0 (mod 17). In the general case, note that all "partial fractions" (which indeed are integers) are odd residues modulo 2*n+1 in the interval [1, 2*n-1]. It is easy to prove that the first 1 appears not later than in the n-th step. (End)
		

References

  • E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I.
  • T. Folger, "Shuffling Into Hyperspace," Discover, 1991 (vol 12, no 1), pages 66-67.
  • M. Gardner, "Card Shuffles," Mathematical Carnival chapter 10, pages 123-138. New York: Vintage Books, 1977.
  • L. Lunelli and M. Lunelli, Tavola di congruenza a^n == 1 mod K per a=2,5,10, Atti Sem. Mat. Fis. Univ. Modena 10 (1960/61), 219-236 (1961).
  • J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 146, Exer. 21.3
  • 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).

Crossrefs

Cf. A024222, A006694 (number of cyclotomic cosets).
Cf. A014664 (order of 2 mod n-th prime).
Cf. A001122 (primes for which 2 is a primitive root).
Cf. A216838 (primes for which 2 is not a primitive root).
Bisections give A274298, A274299.
Partial sums: A359147.

Programs

  • GAP
    List([0..100],n->OrderMod(2,2*n+1)); # Muniru A Asiru, Feb 01 2019
    
  • Haskell
    import Data.List (findIndex)
    import Data.Maybe (fromJust)
    a002326 n = (+ 1) $ fromJust $
                findIndex ((== 0) . (`mod` (2 * n + 1))) $ tail a000225_list
    -- Reinhard Zumkeller, Apr 22 2013
    
  • Magma
    [ 1 ] cat [ Modorder(2, 2*n+1): n in [1..72] ]; // Klaus Brockhaus, Dec 03 2008
    
  • Maple
    a := n -> `if`(n=0, 1, numtheory:-order(2, 2*n+1)):
    seq(a(n), n=0..72);
  • Mathematica
    Table[MultiplicativeOrder[2, 2*n + 1], {n, 0, 100}] (* Robert G. Wilson v, Apr 05 2011 *)
  • PARI
    a(n)=if(n<0,0,znorder(Mod(2,2*n+1))) /* Michael Somos, Mar 31 2005 */
    
  • Python
    from sympy import n_order
    [n_order(2, 2*n+1) for n in range(73)] # Hermann Stamm-Wilbrandt, Jul 27 2021
  • Sage
    # From Peter Luschny, Oct 06 2017: (Start)
    [Mod(2,n).multiplicative_order() for n in (0..145) if gcd(n,2) == 1]
    # Algorithm from Vladimir Shevelev as described in A179680 and presented in Example.
    def A002326VS(n):
        s, m, N = 0, 1, 2*n + 1
        while True:
            k = N + m
            v = valuation(k, 2)
            s += v
            m = k >> v
            if m == 1: break
        return s
    [A002326VS(n) for n in (0..72)] # (End)
    

Formula

a((3^n-1)/2) = A025192(n). - Vladimir Shevelev, May 09 2008
Bisection of A007733: a(n) = A007733(2*n+1). - Max Alekseyev, Jun 11 2009
a((b(n)-1)/2) = n for odd n and even n such that b(n/2) != b(n), where b(n) = A005420(n). - Thomas Ordowski, Jan 11 2014
Note that a(2^n-1) = n+1 and a(2^n) = 2*(n+1). - Thomas Ordowski, Jan 16 2014
a(n) = A056239(A292239(n)) = A048675(A292265(n)). - Antti Karttunen, Oct 04 2017

Extensions

More terms from David W. Wilson, Jan 13 2000
More terms from Benoit Cloitre, Apr 11 2003

A305801 Lexicographically earliest infinite sequence such that a(i) = a(j) => f(i) = f(j), where f(n) = 0 if n is an odd prime, with f(n) = n for all other n.

Original entry on oeis.org

1, 2, 3, 4, 3, 5, 3, 6, 7, 8, 3, 9, 3, 10, 11, 12, 3, 13, 3, 14, 15, 16, 3, 17, 18, 19, 20, 21, 3, 22, 3, 23, 24, 25, 26, 27, 3, 28, 29, 30, 3, 31, 3, 32, 33, 34, 3, 35, 36, 37, 38, 39, 3, 40, 41, 42, 43, 44, 3, 45, 3, 46, 47, 48, 49, 50, 3, 51, 52, 53, 3, 54, 3, 55, 56, 57, 58, 59, 3, 60, 61, 62, 3, 63, 64, 65, 66, 67, 3, 68, 69, 70, 71, 72, 73, 74, 3, 75, 76, 77, 3, 78, 3, 79, 80
Offset: 1

Views

Author

Antti Karttunen, Jun 14 2018

Keywords

Comments

The original name was: "Filter sequence for a(odd prime) = constant sequences", which stemmed from the fact that for all i, j, a(i) = a(j) => b(i) = b(j) for any sequence b that obtains a constant value for all odd primes A065091.
For example, we have for all i, j:
a(i) = a(j) => A305800(i) = A305800(j),
a(i) = a(j) => A007814(i) = A007814(j),
a(i) = a(j) => A305891(i) = A305891(j) => A291761(i) = A291761(j).
There are several filter sequences "above" this one (meaning that they have finer equivalence class partitioning), for example, we have, for all i, j:
[where odd primes are further distinguished by]
A305900(i) = A305900(j) => a(i) = a(j), [whether p = 3 or > 3]
A319350(i) = A319350(j) => a(i) = a(j), [A007733(p)]
A319704(i) = A319704(j) => a(i) = a(j), [p mod 4]
A319705(i) = A319705(j) => a(i) = a(j), [A286622(p)]
A331304(i) = A331304(j) => a(i) = a(j), [parity of A000720(p)]
A336855(i) = A336855(j) => a(i) = a(j). [distance to the next larger prime]

Crossrefs

Cf. A305900, A319350, A319704, A319705, A331304, A336855 (sequences with finer equivalence class partitioning).
Cf. also A003602, A103391, A295300, A305795, A324400, A331300, A336460 (for similar constructions or similarly useful sequences).

Programs

  • Mathematica
    Array[If[# <= 2, #, If[PrimeQ[#], 3, 2 + # - PrimePi[#]]] &, 105] (* Michael De Vlieger, Oct 18 2021 *)
  • PARI
    A305801(n) = if(n<=2,n,if(isprime(n),3,2+n-primepi(n)));

Formula

a(1) = 1, a(2) = 2; for n > 2, a(n) = 3 for odd primes, and a(n) = 2+n-A000720(n) for composite n.
For n > 2, a(n) = 1 + A305800(n).

Extensions

Name changed and Comment section rewritten by Antti Karttunen, Oct 17 2021

A007732 Period of decimal representation of 1/n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 2, 1, 6, 6, 1, 1, 16, 1, 18, 1, 6, 2, 22, 1, 1, 6, 3, 6, 28, 1, 15, 1, 2, 16, 6, 1, 3, 18, 6, 1, 5, 6, 21, 2, 1, 22, 46, 1, 42, 1, 16, 6, 13, 3, 2, 6, 18, 28, 58, 1, 60, 15, 6, 1, 6, 2, 33, 16, 22, 6, 35, 1, 8, 3, 1, 18, 6, 6, 13, 1, 9, 5, 41, 6, 16, 21, 28, 2, 44, 1
Offset: 1

Views

Author

N. J. A. Sloane, Hal Sampson [ hals(AT)easynet.com ]

Keywords

Comments

Appears to be a divisor of A007733*A007736. - Henry Bottomley, Dec 20 2001
Primes p such that a(p) = p-1 are in A001913. - Dmitry Kamenetsky, Nov 13 2008
When 1/n has a finite decimal expansion (namely, when n = 2^a*5^b), a(n) = 1 while A051626(n) = 0. - M. F. Hasler, Dec 14 2015
a(n.n) >= a(n) where n.n is A020338(n). - Davide Rotondo, Jun 13 2024

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, pp. 159 etc.

Crossrefs

Programs

  • Maple
    A007732 := proc(n)
        a132740 := 1 ;
        for pe in ifactors(n)[2] do
            if not op(1,pe) in {2,5} then
                a132740 := a132740*op(1,pe)^op(2,pe) ;
            end if;
        end do:
        if a132740 = 1 then
            1 ;
        else
            numtheory[order](10,a132740) ;
        end if;
    end proc:
    seq(A007732(n),n=1..50) ; # R. J. Mathar, May 05 2023
  • Mathematica
    Table[r = n/2^IntegerExponent[n, 2]/5^IntegerExponent[n, 5]; MultiplicativeOrder[10, r], {n, 100}] (* T. D. Noe, Oct 17 2012 *)
  • PARI
    a(n)=znorder(Mod(10,n/2^valuation(n,2)/5^valuation(n,5))) \\ Charles R Greathouse IV, Jan 14 2013
    
  • Python
    from sympy import n_order, multiplicity
    def A007732(n): return n_order(10,n//2**multiplicity(2,n)//5**multiplicity(5,n)) # Chai Wah Wu, Feb 07 2022
  • Sage
    def a(n):
        n = ZZ(n)
        rad = 2**n.valuation(2) * 5**n.valuation(5)
        return Zmod(n // rad)(10).multiplicative_order()
    [a(n) for n in range(1, 20)]
    # F. Chapoton, May 03 2020
    

Formula

Note that if n=r*s where r is a power of 2 and s is odd then a(n)=a(s). Also if n=r*s where r is a power of 5 and s is not divisible by 5 then a(n) = a(s). So we just need a(n) for n not divisible by 2 or 5. This is the smallest number m such that n divides 10^m - 1; m is a divisor of phi(n), where phi = A000010.
phi(n) = n-1 only if n is prime and since a(n) divides phi(n), a(n) can only equal n-1 if n is prime. - Scott Hemphill (hemphill(AT)alumni.caltech.edu), Nov 23 2006
a(n)=a(A132740(n)); a(A132741(n))=a(A003592(n))=1. - Reinhard Zumkeller, Aug 27 2007

Extensions

More terms from James Sellers, Feb 05 2000

A054703 Number of distinct powers of 2 modulo n.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 3, 4, 6, 5, 10, 4, 12, 4, 4, 5, 8, 7, 18, 6, 6, 11, 11, 5, 20, 13, 18, 5, 28, 5, 5, 6, 10, 9, 12, 8, 36, 19, 12, 7, 20, 7, 14, 12, 12, 12, 23, 6, 21, 21, 8, 14, 52, 19, 20, 6, 18, 29, 58, 6, 60, 6, 6, 7, 12, 11, 66, 10, 22, 13, 35, 9, 9, 37, 20, 20, 30, 13, 39, 8, 54
Offset: 1

Views

Author

Henry Bottomley, Apr 20 2000

Keywords

Crossrefs

Cf. A054704 (base 3), A054705 (4), A054706 (5), A054707 (6), A054708 (7), A054709 (8), A054717 (9), A054710 (10), A351524 (11), A054712 (12), A054713 (13), A054714 (14), A054715 (15), A054716 (16).

Programs

  • Mathematica
    a[n_] := Module[{e = IntegerExponent[n, 2]}, e + MultiplicativeOrder[2, n/2^e]]; Array[a, 100] (* Amiram Eldar, Aug 25 2024 *)

Formula

a(n) = A007814(n) + A007733(n). - Max Alekseyev, May 13 2016

Extensions

Definition improved by T. D. Noe, Jul 21 2008

A036259 Numbers k such that the multiplicative order of 2 modulo k is odd.

Original entry on oeis.org

1, 7, 23, 31, 47, 49, 71, 73, 79, 89, 103, 127, 151, 161, 167, 191, 199, 217, 223, 233, 239, 263, 271, 311, 329, 337, 343, 359, 367, 383, 431, 439, 463, 479, 487, 497, 503, 511, 529, 553, 599, 601, 607, 623, 631, 647, 713, 719, 721, 727, 743, 751
Offset: 1

Views

Author

Keywords

Comments

Odd numbers k such that A007733(k) = A002326((k-1)/2) is odd.
Closed under multiplication. - Emmanuel Vantieghem, May 07 2025

Examples

			2^3 = 1 mod 7, 3 is odd, so 7 is in the sequence.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1, 999, 2], OddQ[MultiplicativeOrder[2, #]]&] (* Jean-François Alcover, Dec 20 2017 *)
  • PARI
    is(n)=n%2 && znorder(Mod(2,n))%2 \\ Charles R Greathouse IV, Jun 24 2015
    
  • Python
    from sympy import n_order
    from itertools import count, islice
    def A036259_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:n_order(2,n)&1,count(max(startvalue,1)|1,2))
    A036259_list = list(islice(A036259_gen(),20)) # Chai Wah Wu, Feb 07 2023
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