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.

Showing 1-5 of 5 results.

A135014 Intersection of A002378 and A135013.

Original entry on oeis.org

2, 6, 342, 8372, 15006, 32942, 45156, 170982, 369056, 3008490, 3954132, 644931420, 2009953056, 12824996256, 688604539866900, 11580326725041402, 26067179544075560, 94815133778597582, 502907499251402082, 3337168950979864272, 7864766939956291580
Offset: 1

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Author

N. J. A. Sloane, Feb 10 2008

Keywords

Crossrefs

Programs

  • Maple
    A000265 := proc(n) local nshft ; if n mod 2 = 1 then RETURN(n); else nshft := n ; while nshft mod 2 = 0 do nshft := nshft/2 ; od ; RETURN(nshft) ; fi ; end: A135013 := proc(n) option remember ; if n = 1 then RETURN(1) ; else RETURN( A135013(n-1)+A000265(n)) ; fi ; end: isA002378 := proc(n) RETURN( issqr(1+4*n) ) ; end: for n from 1 do a135013 := A135013(n) ; if isA002378(a135013) then printf("%d, ",a135013) ; fi ; od: # R. J. Mathar, Feb 11 2008

Extensions

More terms from R. J. Mathar, Feb 11 2008
a(15)-a(19) from Donovan Johnson, Aug 08 2010
a(20)-a(21) from Donovan Johnson, Jul 24 2011

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 1, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 1, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77
Offset: 1

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Author

Keywords

Comments

When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).
Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002
Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004
Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005
From Marco Matosic, Jun 29 2005: (Start)
"The sequence can be arranged in a table:
1
1 3 1
1 5 3 7 1
1 9 5 11 3 13 7 15 1
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31 1
Every new row is the previous row interspaced with the continuation of the odd numbers.
Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)
This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005
2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005
It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005
In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009
This sequence appears in the analysis of A160469 and A156769, which resemble the numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009
Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013
From Alexander R. Povolotsky, Dec 17 2014: (Start)
With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:
1
1 3
1 5 3 7
1 9 5 11 3 13 7 15
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31
and so on ... .
The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)
Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017
From Paul Curtz, Feb 19 2019: (Start)
This sequence is the truncated triangle:
1, 1;
3, 1, 5;
3, 7, 1, 9;
5, 11, 3, 13, 7;
15, 1, 17, 9, 19, 5;
21, 11, 23, 3, 25, 13, 27;
7, 29, 15, 31, 1, 33, 17, 35;
...
The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.
c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019
a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

Examples

			G.f. = x + x^2 + 3*x^3 + x^4 + 5*x^5 + 3*x^6 + 7*x^7 + x^8 + 9*x^9 + 5*x^10 + 11*x^11 + ...
		

References

  • 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. A049606 (partial products), A135013 (partial sums), A099545 (mod 4), A326937 (Dirichlet inverse).
Cf. A026741 (map), A001511 (converging steps), A038550 (prime index).
Cf. A195056 (Dgf at s=3).

Programs

  • Haskell
    a000265 = until odd (`div` 2)
    -- Reinhard Zumkeller, Jan 08 2013, Apr 08 2011, Oct 14 2010
    
  • Java
    int A000265(n){
        while(n%2==0) n>>=1;
        return n;
    }
    /* Aidan Simmons, Feb 24 2019 */
    
  • Julia
    using IntegerSequences
    [OddPart(n) for n in 1:77] |> println  # Peter Luschny, Sep 25 2021
    
  • Magma
    A000265:= func< n | n/2^Valuation(n,2) >;
    [A000265(n): n in [1..120]]; // G. C. Greubel, Jul 31 2024
    
  • Maple
    A000265:=proc(n) local t1,d; t1:=1; for d from 1 by 2 to n do if n mod d = 0 then t1:=d; fi; od; t1; end: seq(A000265(n), n=1..77);
    A000265 := n -> n/2^padic[ordp](n,2): seq(A000265(n), n=1..77); # Peter Luschny, Nov 26 2010
  • Mathematica
    a[n_Integer /; n > 0] := n/2^IntegerExponent[n, 2]; Array[a, 77] (* Josh Locker *)
    a[ n_] := If[ n == 0, 0, n / 2^IntegerExponent[ n, 2]]; (* Michael Somos, Dec 17 2014 *)
  • PARI
    {a(n) = n >> valuation(n, 2)}; /* Michael Somos, Aug 09 2006, edited by M. F. Hasler, Dec 18 2014 */
    
  • Python
    from _future_ import division
    def A000265(n):
        while not n % 2:
            n //= 2
        return n # Chai Wah Wu, Mar 25 2018
    
  • Python
    def a(n):
        while not n&1: n >>= 1
        return n
    print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Jun 26 2025
    
  • SageMath
    def A000265(n): return n//2^valuation(n,2)
    [A000265(n) for n in (1..121)] # G. C. Greubel, Jul 31 2024
  • Scheme
    (define (A000265 n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2))))) ;; Antti Karttunen, Apr 15 2017
    

Formula

a(n) = if n is odd then n, otherwise a(n/2). - Reinhard Zumkeller, Sep 01 2002
a(n) = n/A006519(n) = 2*A025480(n-1) + 1.
Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. - David W. Wilson, Aug 01 2001
a(n) = Sum_{d divides n and d is odd} phi(d). - Vladeta Jovovic, Dec 04 2002
G.f.: -x/(1 - x) + Sum_{k>=0} (2*x^(2^k)/(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))). - Ralf Stephan, Sep 05 2003
(a(k), a(2k), a(3k), ...) = a(k)*(a(1), a(2), a(3), ...) In general, a(n*m) = a(n)*a(m). - Josh Locker (jlocker(AT)mail.rochester.edu), Oct 04 2005
a(n) = Sum_{k=0..n} A127793(n,k)*floor((k+2)/2) (conjecture). - Paul Barry, Jan 29 2007
Dirichlet g.f.: zeta(s-1)*(2^s - 2)/(2^s - 1). - Ralf Stephan, Jun 18 2007
a(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller, Aug 27 2007
a(n) = 2*A003602(n) - 1. - Franklin T. Adams-Watters, Jul 02 2009
a(n) = n/gcd(2^n,n). (This also shows that the true offset is 0 and a(0) = 0.) - Peter Luschny, Nov 14 2009
a(-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2011
From Reinhard Zumkeller, May 01 2012: (Start)
A182469(n, k) = A027750(a(n), k), k = 1..A001227(n).
a(n) = A182469(n, A001227(n)). (End)
a((2*n-1)*2^p) = 2*n - 1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
G.f.: G(0)/(1 - 2*x^2 + x^4) - 1/(1 - x), where G(k) = 1 + 1/(1 - x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))/(x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2))) + (1 - 2*x^(2^(k+2)) + x^(2^(k+3)))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 06 2013
a(n) = A003961(A064989(n)). - Antti Karttunen, Apr 15 2017
Completely multiplicative with a(2) = 1 and a(p) = p for prime p > 2, i.e., the sequence b(n) = a(n) * A008683(n) for n > 0 is the Dirichlet inverse of a(n). - Werner Schulte, Jul 08 2018
From Peter Bala, Feb 27 2019: (Start)
O.g.f.: F(x) - F(x^2) - F(x^4) - F(x^8) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + ..., where L(x) = log(1/(1 - x)).
Sum_{n >= 1} x^n/a(n) = 1/2*log(G(x)), where G(x) = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 + ... is the o.g.f. of A000123. (End)
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
a(n) = A049606(n) / A049606(n-1). - Flávio V. Fernandes, Dec 08 2020
a(n) = numerator of n/2^(floor(n/2)). - Federico Provvedi, Dec 14 2021
a(n) = Sum_{d divides n} (-1)^(d+1)*phi(2*n/d). - Peter Bala, Jan 14 2024
a(n) = A030101(A030101(n)). - Darío Clavijo, Sep 19 2024

Extensions

Additional comments from Henry Bottomley, Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified by David A. Corneth, Apr 15 2017

A000918 a(n) = 2^n - 2.

Original entry on oeis.org

-1, 0, 2, 6, 14, 30, 62, 126, 254, 510, 1022, 2046, 4094, 8190, 16382, 32766, 65534, 131070, 262142, 524286, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 67108862, 134217726, 268435454, 536870910, 1073741822, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 68719476734, 137438953470
Offset: 0

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Author

Keywords

Comments

For n > 1, a(n) is the expected number of tosses of a fair coin to get n-1 consecutive heads. - Pratik Poddar, Feb 04 2011
For n > 2, Sum_{k=1..a(n)} (-1)^binomial(n, k) = A064405(a(n)) + 1 = 0. - Benoit Cloitre, Oct 18 2002
For n > 0, the number of nonempty proper subsets of an n-element set. - Ross La Haye, Feb 07 2004
Numbers j such that abs( Sum_{k=0..j} (-1)^binomial(j, k)*binomial(j + k, j - k) ) = 1. - Benoit Cloitre, Jul 03 2004
For n > 2 this formula also counts edge-rooted forests in a cycle of length n. - Woong Kook (andrewk(AT)math.uri.edu), Sep 08 2004
For n >= 1, conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4, 5, 6 and 7 as a digit. - Alexandre Wajnberg, Apr 25 2005
Beginning with a(2) = 2, these are the partial sums of the subsequence of A000079 = 2^n beginning with A000079(1) = 2. Hence for n >= 2 a(n) is the smallest possible sum of exactly one prime, one semiprime, one triprime, ... and one product of exactly n-1 primes. A060389 (partial sums of the primorials, A002110, beginning with A002110(1)=2) is the analog when all the almost primes must also be squarefree. - Rick L. Shepherd, May 20 2005
From the second term on (n >= 1), the binary representation of these numbers is a 0 preceded by (n - 1) 1's. This pattern (0)111...1110 is the "opposite" of the binary 2^n+1: 1000...0001 (cf. A000051). - Alexandre Wajnberg, May 31 2005
The numbers 2^n - 2 (n >= 2) give the positions of 0's in A110146. Also numbers k such that k^(k + 1) = 0 mod (k + 2). - Zak Seidov, Feb 20 2006
Partial sums of A155559. - Zerinvary Lajos, Oct 03 2007
Number of surjections from an n-element set onto a two-element set, with n >= 2. - Mohamed Bouhamida, Dec 15 2007
It appears that these are the numbers n such that 3*A135013(n) = n*(n + 1), thus answering Problem 2 on the Mathematical Olympiad Foundation of Japan, Final Round Problems, Feb 11 1993 (see link Japanese Mathematical Olympiad).
Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x is a proper subset of y or y is a proper subset of x and x and y are disjoint. Then a(n+1) = |R|. - Ross La Haye, Mar 19 2009
The permutohedron Pi_n has 2^n - 2 facets [Pashkovich]. - Jonathan Vos Post, Dec 17 2009
First differences of A005803. - Reinhard Zumkeller, Oct 12 2011
For n >= 1, a(n + 1) is the smallest even number with bit sum n. Cf. A069532. - Jason Kimberley, Nov 01 2011
a(n) is the number of branches of a complete binary tree of n levels. - Denis Lorrain, Dec 16 2011
For n>=1, a(n) is the number of length-n words on alphabet {1,2,3} such that the gap(w)=1. For a word w the gap g(w) is the number of parts missing between the minimal and maximal elements of w. Generally for words on alphabet {1,2,...,m} with g(w)=g>0 the e.g.f. is Sum_{k=g+2..m} (m - k + 1)*binomial((k - 2),g)*(exp(x) - 1)^(k - g). a(3)=6 because we have: 113, 131, 133, 311, 313, 331. Cf. A240506. See the Heubach/Mansour reference. - Geoffrey Critzer, Apr 13 2014
For n > 0, a(n) is the minimal number of internal nodes of a red-black tree of height 2*n-2. See the Oct 02 2015 comment under A027383. - Herbert Eberle, Oct 02 2015
Conjecture: For n>0, a(n) is the least m such that A007814(A000108(m)) = n-1. - L. Edson Jeffery, Nov 27 2015
Actually this follows from the procedure for determining the multiplicity of prime p in C(n) given in A000108 by Franklin T. Adams-Watters: For p=2, the multiplicity is the number of 1 digits minus 1 in the binary representation of n+1. Obviously, the smallest k achieving "number of 1 digits" = k is 2^k-1. Therefore C(2^k-2) is divisible by 2^(k-1) for k > 0 and there is no smaller m for which 2^(k-1) divides C(m) proving the conjecture. - Peter Schorn, Feb 16 2020
For n >= 0, a(n) is the largest number you can write in bijective base-2 (a.k.a. the dyadic system, A007931) with n digits. - Harald Korneliussen, May 18 2019
The terms of this sequence are also the sum of the terms in each row of Pascal's triangle other than the ones. - Harvey P. Dale, Apr 19 2020
For n > 1, binomial(a(n),k) is odd if and only if k is even. - Charlie Marion, Dec 22 2020
For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 2. - David desJardins, Oct 27 2022
For n >= 1, a(n+1) is the number of roots of unity in the unique degree-n unramified extension of the 2-adic field Q_2. Note that for each p, the unique degree-n unramified extension of Q_p is the splitting field of x^(p^n) - x, hence containing p^n - 1 roots of unity for p > 2 and 2*(2^n - 1) for p = 2. - Jianing Song, Nov 08 2022

Examples

			a(4) = 14 because the 14 = 6 + 4 + 4 rationals (in lowest terms) for n = 3 are (see the Jun 14 2017 formula above): 1/2, 1, 3/2, 2, 5/2, 3; 1/4, 3/4, 5/4, 7/4; 1/8, 3/8, 5/8, 7/8. - _Wolfdieter Lang_, Jun 14 2017
		

References

  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134. - Mohammad K. Azarian, Oct 27 2011
  • S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2009 page 86, Exercise 3.16.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • 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).
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.

Crossrefs

Row sums of triangle A026998.
Cf. A058809 (3^n-3, n>0).

Programs

  • Haskell
    a000918 = (subtract 2) . (2 ^)
    a000918_list = iterate ((subtract 2) . (* 2) . (+ 2)) (- 1)
    -- Reinhard Zumkeller, Apr 23 2013
    
  • Magma
    [2^n - 2: n in [0..40]]; // Vincenzo Librandi, Jun 23 2011
    
  • Maple
    seq(2^n-2,n=0..20) ;
  • Mathematica
    Table[2^n - 2, {n, 0, 29}] (* Alonso del Arte, Dec 01 2012 *)
  • PARI
    a(n)=2^n-2 \\ Charles R Greathouse IV, Jun 16 2011
    
  • Python
    def A000918(n): return (1<Chai Wah Wu, Jun 10 2025

Formula

a(n) = 2*A000225(n-1).
G.f.: 1/(1 - 2*x) - 2/(1 - x), e.g.f.: (e^x - 1)^2 - 1. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
For n >= 1, a(n) = A008970(n + 1, 2). - Philippe Deléham, Feb 21 2004
G.f.: (3*x - 1)/((2*x - 1)*(x - 1)). - Simon Plouffe in his 1992 dissertation for the sequence without the leading -1
a(n) = 2*a(n - 1) + 2. - Alexandre Wajnberg, Apr 25 2005
a(n) = A000079(n) - 2. - Omar E. Pol, Dec 16 2008
a(n) = A058896(n)/A052548(n). - Reinhard Zumkeller, Feb 14 2009
a(n) = A164874(n - 1, n - 1) for n > 1. - Reinhard Zumkeller, Aug 29 2009
a(n) = A173787(n,1); a(n) = A028399(2*n)/A052548(n), n > 0. - Reinhard Zumkeller, Feb 28 2010
a(n + 1) = A027383(2*n - 1). - Jason Kimberley, Nov 02 2011
G.f.: U(0) - 1, where U(k) = 1 + x/(2^k + 2^k/(x - 1 - x^2*2^(k + 1)/(x*2^(k + 1) - (k + 1)/U(k + 1) ))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n+1) is the sum of row n in triangle A051601. - Reinhard Zumkeller, Aug 05 2013
a(n+1) = A127330(n,0). - Reinhard Zumkeller, Nov 16 2013
a(n) = Sum_{k=1..n-1} binomial(n, k) for n > 0. - Dan McCandless, Nov 14 2015
From Miquel Cerda, Aug 16 2016: (Start)
a(n) = A000225(n) - 1.
a(n) = A125128(n-1) - A000325(n).
a(n) = A095151(n) - A125128(n) - 1. (End)
a(n+1) = 2*(n + Sum_{j=1..n-1} (n-j)*2^(j-1)), n >= 1. This is the number of the rationals k/2, k = 1..2*n for n >= 1 and (2*k+1)/2^j for j = 2..n, n >= 2, and 2*k+1 < n-(j-1). See the example for n = 3 below. Motivated by the proposal A287012 by Mark Rickert. - Wolfdieter Lang, Jun 14 2017

Extensions

Maple programs fixed by Vaclav Kotesovec, Dec 13 2014

A157001 Fractions x/y, with 1<=x, y<=n, that reduce to (odd)/(even).

Original entry on oeis.org

0, 1, 2, 5, 7, 11, 14, 21, 25, 32, 37, 47, 53, 63, 70, 85, 93, 106, 115, 132, 142, 158, 169, 191, 203, 222, 235, 259, 273, 295, 310, 341, 357, 382, 399, 430, 448, 476, 495, 532, 552, 583, 604, 642, 664, 698, 721, 767, 791, 828, 853, 898, 924, 964, 991, 1043, 1071
Offset: 1

Views

Author

Adam McDougall (mcdougall.adam(AT)gmail.com), Feb 20 2009

Keywords

Comments

Equivalently, a(n) is the number of fractions x/y, with 1<=x, y<=n, that reduce to (even)/(odd).

Crossrefs

a(n) = (n^2-A135013)/2.

A174989 Partial sums of A003602.

Original entry on oeis.org

1, 2, 4, 5, 8, 10, 14, 15, 20, 23, 29, 31, 38, 42, 50, 51, 60, 65, 75, 78, 89, 95, 107, 109, 122, 129, 143, 147, 162, 170, 186, 187, 204, 213, 231, 236, 255, 265, 285, 288, 309, 320, 342, 348, 371, 383, 407, 409, 434, 447, 473, 480, 507, 521, 549, 553, 582, 597
Offset: 1

Views

Author

Jonathan Vos Post, Apr 03 2010

Keywords

Comments

I conjecture that infinitely many terms are prime. For n<=10^5, exactly 5115 terms are prime. For n<=10^7, there are 352704 prime terms. The largest prime for n<10^10 is at n=9999999983, a(n)=16666666618226308891. Below 10^100, n=(10^100)-345. Below 10^500, n=(10^500)-2414. - Griffin N. Macris, May 04 2016
Since (n^2+3n)/6 < a(n) < (n^2+5n+4)/6, the sum of reciprocals of this sequence converges to a value between 13/6 and 11/3, approximately 2.888. - Griffin N. Macris, May 07 2016

Crossrefs

Programs

  • Mathematica
    a[0]:=0;
    a[n_]:=Ceiling[n/2](1+Ceiling[n/2])/2 + a[Floor[n/2]];
    Array[a,50] (* Griffin N. Macris, May 04 2016 *)

Formula

a(n) = Sum{i=1..n} A003602(i) = Sum_{i=1..n} (A000265(i) + 1)/2.
From Griffin N. Macris, May 04 2016 (Start)
a(0) = 0; a(n) = A000217(ceiling(n/2)) + a(floor(n/2)).
Asymptotically, a(n) ~ (n^2+3n)/6. (End)
a(n) = (A135013(n) + n)/2. - Amiram Eldar, Dec 27 2022
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