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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A273580 Decimal expansion of the infinite nested radical sqrt(F_0 + sqrt(F_1 + sqrt(F_3 + ...))), where F_k are the Fermat numbers A000215.

Original entry on oeis.org

2, 5, 2, 9, 5, 4, 3, 3, 2, 6, 2, 2, 0, 3, 9, 8, 4, 3, 0, 3, 1, 0, 3, 7, 9, 1, 2, 8, 8, 5, 9, 7, 5, 3, 3, 3, 5, 1, 9, 3, 5, 3, 7, 1, 2, 4, 4, 5, 9, 3, 8, 3, 4, 1, 7, 8, 6, 5, 7, 1, 8, 7, 1, 1, 3, 9, 6, 7, 3, 0, 9, 4, 6, 5, 4, 0, 4, 8, 7, 4, 8, 2, 5, 3, 1, 0, 3, 3, 5, 4, 4, 6, 0, 7, 2, 1, 5, 0, 0, 2, 3, 8, 9, 3, 3
Offset: 1

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Author

Stanislav Sykora, May 25 2016

Keywords

Comments

The convergence of this expression follows from Vijayaraghavan's theorem, for which it represents an extreme example.
Two PARI programs to compute this constant are listed below. The first one is a brute-force implementation of the definition and allows the computation of only 13 digits before exceeding current PARI capabilities. The second one implements the following 'trick' inspired by a comment in A094885: Let us try to compute first x = a/sqrt(2). We have x = (1/sqrt(2))sqrt(3+ sqrt(5+ sqrt(17+ ... ))) = sqrt(3/2+ (1/2)sqrt(5+ sqrt(17+ ... ))) = sqrt(3/2+ sqrt(5/4+ (1/4)sqrt(17+ ... ))) = sqrt(3/2+ sqrt(5/4+ sqrt(17/16+ ... ))) = sqrt(c_0+sqrt(c_1+sqrt(c_3+...))), where c_n = (2^(2^n)+1)/2^(2^n) = 1+d_n, with d_n = 2^(-2^n). This nested radical is easy to manage to any precision. However, evaluating it up to N terms, its convergence with increasing N is no better than that of the original algorithm. To speed it up, one must notice that, since the c_n converge rapidly to 1, and since the nested radical sqrt(1+sqrt(1+...)) evaluates to the golden ratio phi (A001622), the latter is the natural best stand-in for the neglected part (terms from N+1 to infinity). With this modification, i.e., 'seeding' the iterations with phi instead of 0, the convergence becomes extremely fast (the number of valid digits more than doubles upon incrementing N by 1).

Examples

			2.5295433262203984303103791288597533351935371244593834178657187113967...
		

Crossrefs

Programs

  • PARI
    /* This function crashes PARI beyond N=28: */
    s(N)={my(r=0.0);for(k=1,N,r=sqrt(2^(2.0^(N-k))+1+r));return(r)}
    /* N is the number of terms to include in the evaluation. It turns out that the starting digits s(28) shares with s(27) are only 13 */
    
  • PARI
    /* This alternative can easily generate millions of digits: */
    d=vector(30);d[1]=0.5;for(n=2,#d,d[n]=d[n-1]^2);
    S(N)={my(r=(1+sqrt(5))/2);for(k=1,N,r=sqrt(1+d[N-k+1]+r));return(r*sqrt(2))}
    /* S(12) exceeds 1200 stable digits, S(20) goes over 150000. For the b-file, the first 2000 digits of S(13) were used, computed with the realprecision of 2100 digits */

Formula

Equals sqrt(2)*sqrt(1+1/2+sqrt(1+1/4+sqrt(1+1/16+sqrt(1+1/256+ ... )))).

A367228 Products of two consecutive Fermat numbers: a(n) = A000215(n) * A000215(n+1).

Original entry on oeis.org

15, 85, 4369, 16843009, 281479271743489, 79228162532711081671548469249, 6277101735386680764176071790128604879584176795969512275969
Offset: 0

Views

Author

Amiram Eldar, Nov 11 2023

Keywords

Comments

a(7) has 116 digits and is too large to include in the data section.
Szymiczek (1966) proved that a(n) is a super-Poulet number (A050217) for all n >= 2. All the composite Fermat numbers (A281576) are also super-Poulet numbers.

References

  • Michal Krížek, Florian Luca and Lawrence Somer, 17 Lectures on Fermat Numbers, Springer-Verlag, N.Y., 2001, p. 142.

Crossrefs

Programs

  • Mathematica
    f[n_] := 2^(2^n) + 1; a[n_] := f[n] * f[n + 1]; Array[a, 7, 0]
  • PARI
    f(n) = 2^(2^n) + 1;
    a(n) = f(n) * f(n+1);

Formula

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

A136188 Digital roots of the Fermat numbers in A000215(n).

Original entry on oeis.org

3, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8
Offset: 0

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Author

Ant King, Dec 24 2007

Keywords

Comments

As 2^(2^n)+1=5 (mod 9) for odd values of n and 2^(2^n)+1=8 (mod 9) for even values of n>0, it follows that the digital roots of the Fermat numbers form a cyclic sequence, with the 5's corresponding to odd values of n and the 8's to even values of n.
Decimal expansion of 71/198. - Enrique Pérez Herrero, Nov 13 2021

Examples

			2^(2^3) + 1 = 257. This has digital root 5 and hence a(3) = 5.
		

Crossrefs

Essentially the same as A010719.

Programs

  • Mathematica
    FermatNumber[n_]:=2^(2^n)+1;DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[ # ]&,n];DigitalRoot/@(FermatNumber[ # ] &/@Range[0,25])
  • PARI
    a(n)=if(n,if(n%2,5,8),3) \\ Charles R Greathouse IV, May 01 2016

Formula

a(n) = A010888(A000215(n)).

A162647 Numerators associated with denominators A000215(n) approximating the complementary Thue-Morse constant.

Original entry on oeis.org

2, 3, 10, 151, 38506, 2523490711, 10838310072981296746, 199931532107794273605284333428918544791, 68033174967769840440887906939858451149105560803546820641877549596291376780906
Offset: 0

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Author

Vladimir Shevelev, Jul 08 2009, Jul 14 2009

Keywords

Comments

If in the sequence of numbers N for which A010060(N+2^n)=1-A010060(N) the odious (evil) terms are
replaced by 1's (0's), we obtain a 2^(n+1)-periodic binary sequence. These are the post-period
binary (base-2) digits of the complementary Thue-Morse constant 1-A014571 = 0.58754596635989240221663...,
which has a continued fraction and convergents 3/5, 7/12, 10/17, 47/80, 151/257, 801/1365,...
The a(n) are numerators of the convergents selected with denominators taken from A000215.

Crossrefs

Formula

a(n)=A000215(n)-A162634(n). For n>=1, a(n+1)=1+(2^(2^n)-1)*a(n) = 1+A051179(n)*a(n).

Extensions

Edited by R. J. Mathar, Sep 23 2009

A242866 Union of 2*A019434 and the semiprimes in A000215.

Original entry on oeis.org

6, 10, 34, 514, 131074, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937
Offset: 1

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Author

Juri-Stepan Gerasimov, May 24 2014

Keywords

Comments

The old definition was "Semiprimes of the form 2^(2^k) + 1 or 2^(2^k) + 2 for some k >= 0", but that does not include 10. So I replaced the old definition with the old comment, which seems a better fit. Question: is the sequence defined by the old definition in the OEIS? If not it should be added. - N. J. A. Sloane, Sep 03 2023

Crossrefs

Extensions

Definition edited by N. J. A. Sloane, Sep 02 2023

A262348 a(n) = A000215(n) mod A000045(n).

Original entry on oeis.org

0, 0, 1, 2, 2, 1, 10, 17, 19, 17, 5, 17, 75, 17, 257, 983, 987, 1481, 4178, 4652, 7433, 3488, 10665, 19169, 51372, 538, 257, 65537, 422030, 17, 111134, 430622, 592107, 1036631, 4104877, 11311649, 4963932, 23430341, 16834037, 65537, 67016878, 53547017, 316617399
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 18 2015

Keywords

Crossrefs

Programs

  • Magma
    [(2^(2^n) + 1) mod(Fibonacci(n)): n in [1..29]]; // Vincenzo Librandi, Sep 19 2015
    
  • Mathematica
    f[n_] := Mod[ PowerMod[2, 2^n, Fibonacci[n]] + 1, Fibonacci[n]]; Array[f, 43] (* Robert G. Wilson v, Sep 18 2015 *)
  • PARI
    a(n) = my (fn = fibonacci(n)); lift(Mod(Mod(2, fn)^(2^n) + 1, fn)); \\ Michel Marcus, Sep 19 2015

Extensions

More terms from Robert G. Wilson v, Sep 18 2015

A275159 Primes p such that p-1 is the value of totient function of a product of distinct Fermat numbers (A000215).

Original entry on oeis.org

2, 3, 5, 17, 257, 65537, 548898078721, 1151122703583805441, 77370970260794891965562881, 632834090662785970268956262401, 1327149278901642923121482163604684801, 2787593149816327845958662202634335514787841, 91343852333181430856373443055921906148567941121
Offset: 1

Views

Author

Jaroslav Krizek, Nov 13 2016

Keywords

Comments

Primes p such that p-1 = phi(A001317(x)) has solution.

Examples

			Prime 548898078721 is in the sequence because 548898078720 = phi(1095216660735) = phi(3*5*17*4294967297); all numbers 3, 5, 17 and 4294967297 are terms of A000215 (Fermat numbers).
		

Crossrefs

Supersequence of A019434 (Fermat primes) and A092506 (primes of the form 2^n+1).

Programs

  • Magma
    Set(Sort([EulerPhi(k)+1: k in [A001317(n)] | IsPrime(EulerPhi(k)+1)]));

Extensions

a(13) from Jinyuan Wang, Nov 01 2020

A321213 a(n) is the number of divisors of n-th Fermat number (A000215).

Original entry on oeis.org

2, 2, 2, 2, 2, 4, 4, 4, 4, 8, 16, 32
Offset: 0

Views

Author

Jinyuan Wang, Oct 31 2018

Keywords

Examples

			A000215(n) is prime for n=0 to 4, so a(n)= 2 for n=0 to 4.
		

Crossrefs

Programs

  • GAP
    List(List([0..11],n->2^(2^n)+1),i->Number(DivisorsInt(i))); # Muniru A Asiru, Nov 03 2018
  • Magma
    [DivisorSigma(0, 2^2^n + 1): n in [1..100]]
    
  • Mathematica
    Table[DivisorSigma[0, 2^2^n + 1], {n, 120}]
  • PARI
    a(n) = numdiv(2^2^n+1)
    

Formula

a(n) = A000005(A000215(n)). - Omar E. Pol, Oct 31 2018
a(n) = 2^A046052(n) for squarefree A000215(n). - Amiram Eldar, Oct 31 2018

Extensions

a(10)-a(11) from Amiram Eldar, Oct 31 2018

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

Original entry on oeis.org

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

Views

Author

John Tromp, Dec 11 1996

Keywords

Comments

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3 is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

Examples

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
  • K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A086463 (Dgf at s=2).

Programs

  • Haskell
    a007814 n = if m == 0 then 1 + a007814 n' else 0
                where (n', m) = divMod n 2
    -- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011
    
  • Haskell
    a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
    --  Walt Rorie-Baety, Mar 22 2013
    
  • Magma
    [Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013
    
  • Maple
    ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
    A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010
  • Mathematica
    Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
    IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
    p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
    DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
    Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
    Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)
  • PARI
    A007814(n)=valuation(n,2);
    
  • Python
    import math
    def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017
    
  • Python
    def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022
    
  • R
    sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013
    
  • Scheme
    (define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Extensions

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010
Edited by Ralf Stephan, Feb 08 2014

A019434 Fermat primes: primes of the form 2^(2^k) + 1, for some k >= 0.

Original entry on oeis.org

3, 5, 17, 257, 65537
Offset: 1

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It is conjectured that there are only 5 terms. Currently it has been shown that 2^(2^k) + 1 is composite for 5 <= k <= 32 (see Eric Weisstein's Fermat Primes link). - Dmitry Kamenetsky, Sep 28 2008
No Fermat prime is a Brazilian number. So Fermat primes belong to A220627. For proof see Proposition 3 page 36 in "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
This sequence and A001220 are disjoint (see "Other theorems about Fermat numbers" in Wikipedia link). - Felix Fröhlich, Sep 07 2014
Numbers n > 1 such that n * 2^(n-2) divides (n-1)! + 2^(n-1). - Thomas Ordowski, Jan 15 2015
From Jaroslav Krizek, Mar 17 2016: (Start)
Primes p such that phi(p) = 2*phi(p-1); primes from A171271.
Primes p such that sigma(p-1) = 2p - 3.
Primes p such that sigma(p-1) = 2*sigma(p) - 5.
For n > 1, a(n) = primes p such that p = 4 * phi((p-1) / 2) + 1.
Subsequence of A256444 and A256439.
Conjectures:
1) primes p such that phi(p) = 2*phi(p-2).
2) primes p such that phi(p) = 2*phi(p-1) = 2*phi(p-2).
3) primes p such that p = sigma(phi(p-2)) + 2.
4) primes p such that phi(p-1) + 1 divides p + 1.
5) numbers n such that sigma(n-1) = 2*sigma(n) - 5. (End)
Odd primes p such that ratio of the form (the number of nonnegative m < p such that m^q == m (mod p))/(the number of nonnegative m < p such that -m^q == m (mod p)) is a divisor of p for all nonnegative q. - Juri-Stepan Gerasimov, Oct 13 2020
Numbers n such that tau(n)*(number of distinct ratio (the number of nonnegative m < n such that m^q == m (mod n))/(the number of nonnegative m < n such that -m^q == m (mod n))) for nonnegative q is equal to 4. - Juri-Stepan Gerasimov, Oct 22 2020
The numbers of primitive roots for the five known terms are 1, 2, 8, 128, 32768. - Gary W. Adamson, Jan 13 2022
Prime numbers such that every residue is either a primitive root or a quadratic residue. - Keith Backman, Jul 11 2022
If there are only 5 Fermat primes, then there are only 31 odd order groups which have a 2-group automorphism group. See the Miles Englezou link for a proof. - Miles Englezou, Mar 10 2025

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 137-141, 197.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see Table 1, p. 458.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • Richard K. Guy, Unsolved Problems in Number Theory, A3.
  • Hardy and Wright, An Introduction to the Theory of Numbers, bottom of page 18 in the sixth edition, gives an heuristic argument that this sequence is finite.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 7, 70.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Subsequence of A147545 and of A334101. Cf. also A333788, A334092.
Cf. A045544.

Programs

Formula

a(n+1) = A180024(A049084(a(n))). - Reinhard Zumkeller, Aug 08 2010
a(n) = 1 + A001146(n-1), if 1 <= n <= 5. - Omar E. Pol, Jun 08 2018
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