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 1-8 of 8 results.

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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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

A048896 a(n) = 2^(A000120(n+1) - 1), n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4
Offset: 0

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Comments

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]
Row sums of triangle A128937. - Philippe Deléham, May 02 2007
a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009
a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011
For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012
a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015
From Gus Wiseman, Oct 30 2022: (Start)
Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).
Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.
(End)

Examples

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,2;
  1,2,2,4;
  1,2,2,4,2,4,4,8;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32;
  ...,
the first half-rows converge to Gould's sequence A001316.
(End)
		

Crossrefs

This is Guy Steele's sequence GS(3, 5) (see A135416).
Equals first right hand column of triangle A160468.
Equals A160469(n+1)/A002425(n+1).
Standard compositions are listed by A066099.
The opposite version (counting refinements) is A080100.
The version for Heinz numbers of partitions is A317141.

Programs

  • Haskell
    a048896 n = a048896_list !! n
    a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x,2*x])
    -- Reinhard Zumkeller, Mar 07 2011
    
  • Haskell
    import Data.List (transpose)
    a048896 = a000079 . a000120
    a048896_list = 1 : concat (transpose
       [zipWith (-) (map (* 2) a048896_list) a048896_list,
        map (* 2) a048896_list])
    -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    [Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014
  • Maple
    a := n -> 2^(add(i,i=convert(n+1,base,2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009
  • Mathematica
    NestList[Flatten[#1 /. a_Integer -> {a, 2 a}] &, {1}, 4] // Flatten (* Robert G. Wilson v, Aug 01 2012 *)
    Table[Numerator[2^n / (n + 1)!], {n, 0, 200}] (* Vincenzo Librandi, Apr 12 2014 *)
    Denominator[Table[BernoulliB[2*n] / (Zeta[2*n]/Pi^(2*n)), {n, 1, 100}]] (* Terry D. Grant, May 29 2017 *)
    Table[Denominator[((2 n)!/2^(2 n + 1)) (-1)^n], {n, 1, 100}]/4 (* Terry D. Grant, May 29 2017 *)
    2^IntegerExponent[CatalanNumber[Range[0,100]],2] (* Harvey P. Dale, Apr 30 2018 *)
  • PARI
    a(n)=if(n<1,1,if(n%2,a(n/2-1/2),2*a(n-1)))
    
  • PARI
    a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022
    

Formula

a(n) = 2^A048881(n).
a(n) = 2^k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n).
It appears that a(n) = Sum_{k=0..n} binomial(2*(n+1), k) mod 2. - Christopher Lenard (c.lenard(AT)bendigo.latrobe.edu.au), Aug 20 2001
a(0) = 1; a(2*n) = 2*a(2*n-1); a(2*n+1) = a(n).
a(n) = (1/2) * A001316(n+1). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
It appears that a(n) = Sum_{k=0..2n} floor(binomial(2n+2, k+1)/2)(-1)^k = 2^n - Sum_{k=0..n+1} floor(binomial(n+1, k)/2). - Paul Barry, Dec 24 2004
a(n) = Sum_{k=0..n} (T(n,k) mod 2) where T = A039598, A053121, A052179, A124575, A126075, A126093. - Philippe Deléham, May 02 2007
a(n) = numerator(b(n)), where sin(x)^2/x = Sum_{n>0} b(n)*(-1)^n x^(2*n-1). - Vladimir Kruchinin, Feb 06 2013
a((2*n+1)*2^p-1) = A001316(n), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 12 2013
a(n) = numerator(2^n / (n+1)!). - Vincenzo Librandi, Apr 12 2014
a(2n) = (2n+1)!/(n!n!)/A001803(n). - Richard Turk, Aug 23 2017
a(2n-1) = (2n-1)!/(n!(n-1)!)/A001790(n). - Richard Turk, Aug 23 2017

Extensions

New definition from N. J. A. Sloane, Mar 01 2008

A002425 Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*zeta(2n)).

Original entry on oeis.org

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 968383680827, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961
Offset: 1

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Comments

Differs from the absolute values of A275994 the first time at index 60.
Consider the C(k)-summation process for divergent series: the series Sum((-1)^n*(n+1)^k) == 1 - 2^k + 3^k - 4^k + ..., summable C(1) to the value 1/2 for k = 0, is for each k >= 1 exactly summable C(k+1) to the sum s(k+1) = (2^(k+1)-1)*B(k+1)/ (k+1) and so a(n) = abs(numerator(s(2n))). - Benoit Cloitre, Apr 27 2002
Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - Ralf Stephan, Dec 21 2004
(-1)^n*a(n+1) is the numerator of Euler(2n+1,1). - N. J. A. Sloane, Nov 10 2009 (a misprint corrected by Vladimir Shevelev, Sep 18 2017)
a(n) is the absolute value of the constant term of the Euler polynomial E_{2n-1} times the even part of 2n. - Peter Luschny, Nov 26 2010
From Vladimir Shevelev, Aug 31 2017: (Start)
Let E_m(x) = x^m + Sum_{odd k=1..m} e_k(m)*x^(m-k) be the Euler polynomial, let 2*n-1 <= m. Show that the expression c(m,n) = |e_(2*n-1)(m)|/binomial(m,2*n-1) does not depend on m and c(m,n) = a(n)/A006519(2*n). Indeed, by the formula in the Shevelev link |e_(2*n-1)(m)| = binomial(m,2*n-1)*(4^n-1)*B_(2*n)/n. On the other hand, by Cloitre's formula, we have a(n) = (4^n-1)*|B_(2*n)|*2^A001511(n) /n. Taking into account that 2^A001511 = A006519(2*n) we obtain the claimed equality. Since sign(e_k(n)) = (-1)^((k+1)/2), we have the following application of the sequence: e_k(n) = (-1)^((k+1)/2))*a((k+1)/2)*binomial(n,k)/A006519(k+1). (End)

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 73.
  • S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.
  • Konrad Knopp, Theory and application of infinite series, Divergent series, Dover, p. 479
  • L. Oettinger, Archiv. Math. Phys., 26 (1856), see esp. p. 5.
  • 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

Numerator given by A037239.
Different from A089171, A275994.

Programs

  • Magma
    [Denominator(4*n/((4^n-1)*Bernoulli(2*n))): n in [1..20]]; // G. C. Greubel, Jul 03 2019
  • Maple
    A002425 := n -> (-1)^n*euler(2*n-1,0)*2^padic[ordp](2*n,2); # Peter Luschny, Nov 26 2010
    A002425_list := proc(n) 1/(1+1/exp(z)); series(%,z,2*n+4);
    seq(numer((-1)^i*(2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end;
    A002425_list(17); # Peter Luschny, Jul 12 2012
  • Mathematica
    a[n_]:= (-1)^(n-1) * Numerator[EulerE[2n-1, 1]]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Sep 20 2011, after N. J. A. Sloane's comment *)
    a[n_]:= If[n<1, 0, With[{m = 2n-1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* Michael Somos, Sep 14 2013 *)
    Table[2*(4^n-1)*Zeta[1-2n] // Abs // Numerator, {n, 1, 20}] (* Jean-François Alcover, Oct 16 2013 *)
  • PARI
    for(n=1,20,print1(abs(numerator(2*bernfrac(2*n)*(4^n-1)/(2*n))),","))
    
  • PARI
    a(n)=if(n<1,0,(-1)^n/n*(1-4^n)*bernfrac(2*n)*2^valuation(2*n,2))
    
  • PARI
    a(n)=(-1)^n*4*bitand(n,-n)*polylog(1-2*n,-1); \\ Peter Luschny, Nov 22 2012
    
  • Sage
    def A002425_list(n):
        T = [0]*n; T[0] = 1; S = [0]*n; k2 = 0
        for k in (1..n-1): T[k] = k*T[k-1]
        for k in (1..n):
            if is_odd(k): S[k-1] = 4*k2; k2 += 1
            else: S[k-1] = S[k2-1]+2*k2-1
            for j in (k..n-1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return [T[j]>>S[j] for j in (0..n-1)]
    A002425_list(20)  # Peter Luschny, Nov 17 2012
    
  • Sage
    [denominator(4*n/((4^n-1)*bernoulli(2*n))) for n in (1..20)] # G. C. Greubel, Jul 03 2019
    

Formula

a(n) = (-1)^n/n*(1 - 4^n)*B(2*n)*2^A001511(n) where B(k) denotes the k-th Bernoulli number. - Benoit Cloitre, Dec 30 2003
This is different from the sequence of numerators of the expansion of cosec(x) - cot(x) - see A089171.
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))).
Equals A160469(n)/A048896(n-1).
Equals A089171(n)*A089170(n-1). (End)
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - Peter Luschny, Jul 12 2012
a(n) = numerator(abs(2*(4^n-1)*zeta(1-2*n))). - Jean-François Alcover, Oct 16 2013
For every positive integers n,k we have a(n) = (-1)^(n+k)*N(2*n-1,k) + 2*(-1)^(n-1)*A006519(2*n)*(1^(2*n-1)-2^(2*n-1)+..+(-1)^k*(k-1)^(2*n-1)), where N(n,k) is the numerator of Euler(n,k). So, the right hand side is an invariant of k. - Vladimir Shevelev, Sep 19 2017
a(n) = numerator(r(n)) where r(n) = (-1)^binomial(2*n, 2)*Sum_{k=1..2*n}(-1)^k*Stirling2(2*n, k)*2^(-k)*(k-1)!. - Peter Luschny, May 24 2020
a(n) = 2*(-1)^n*A335956(2*n)*zeta(1-2*n). - Peter Luschny, Aug 30 2020

Extensions

The n=15 term was formerly incorrectly given as 86125672563301143.
Formula and cross-references edited by Johannes W. Meijer, May 21 2009

A002430 Numerators in Taylor series for tan(x). Also from Taylor series for tanh(x).

Original entry on oeis.org

1, 1, 2, 17, 62, 1382, 21844, 929569, 6404582, 443861162, 18888466084, 113927491862, 58870668456604, 8374643517010684, 689005380505609448, 129848163681107301953, 1736640792209901647222, 418781231495293038913922
Offset: 1

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Author

Keywords

Comments

a(n) appears to be a multiple of A046990(n) (checked up to n=250). - Ralf Stephan, Mar 30 2004
The Taylor series for tan(x) appears to be identical to the sequence of quotients A160469(n)/A156769(n). - Johannes W. Meijer, May 24 2009

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... =
  x + (1/3)*x^3 + (2/15)*x^5 + (17/315)*x^7 + (62/2835)*x^9 + ... =
  Sum_{n >= 1} (2^(2n) - 1) * (2x)^(2n-1) * |bernoulli_2n| / (n*(2n-1)!).
tanh(x) = x - (1/3)*x^3 + (2/15)*x^5 - (17/315)*x^7 + (62/2835)*x^9 - (1382/155925)*x^11 + ...
		

References

  • G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 74.
  • H. A. Rothe, in C. F. Hindenburg, editor, Sammlung Combinatorisch-Analytischer Abhandlungen, Vol. 2, Chap. XI. Fleischer, Leipzig, 1800, p. 329.
  • 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. A036279 (denominators), A000182, A099612, A160469, A156769.

Programs

  • Magma
    [Numerator( (-1)^(n-1)*4^n*(4^n-1)*Bernoulli(2*n)/Factorial(2*n) ): n in [1..20]]; // G. C. Greubel, Jul 03 2019
    
  • Maple
    R := n -> (-1)^floor(n/2)*(4^n-2^n)*Zeta(1-n)/(n-1)!:
    seq(numer(R(2*n)), n=1..20); # Peter Luschny, Aug 25 2015
  • Mathematica
    a[n_]:= (-1)^Floor[n/2]*(4^n - 2^n)*Zeta[1-n]/(n-1)!; Table[Numerator@ a[2n], {n, 20}] (* Michael De Vlieger, Aug 25 2015 *)
  • PARI
    a(n) = numerator( (-1)^(n-1)*4^n*(4^n-1)*bernfrac(2*n)/(2*n)! ); \\ G. C. Greubel, Jul 03 2019
    
  • Sage
    [numerator( (-1)^(n-1)*4^n*(4^n-1)*bernoulli(2*n)/factorial(2*n)  ) for n in (1..20)] # G. C. Greubel, Jul 03 2019

Formula

a(n) is the numerator of (-1)^(n-1)*2^(2*n)*(2^(2*n) -1)* Bernoulli(2*n)/(2*n)!. - Johannes W. Meijer, May 24 2009
Let R(x) = (cos(x*Pi/2) + sin(x*Pi/2))*(4^x - 2^x)*Zeta(1-x)/(x-1)!. Then a(n) = numerator(R(2*n)) and A036279(n) = denominator(R(2*n)). - Peter Luschny, Aug 25 2015

Extensions

More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Jan 29 2003

A036279 Denominators in the Taylor series for tan(x).

Original entry on oeis.org

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 2900518163668125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875
Offset: 1

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Author

Keywords

Comments

The n-th coefficient of Taylor series for tan(x) appears to be identical to the quotient A160469(n)/A156769(n). - Johannes W. Meijer, May 24 2009

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + (1/3)*x^3 + (2/15)*x^5 + (17/315)*x^7 + (62/2835)*x^9 + ... =
  Sum_{n >= 1} (2^(2n) - 1) * (2x)^(2n-1) * |Bernoulli_2n| / (n*(2n-1)!).
The coefficients in the expansion of tan x are 0, 1, 0, 1/3, 0, 2/15, 0, 17/315, 0, 62/2835, 0, 1382/155925, 0, 21844/6081075, 0, 929569/638512875, 0, ... = A002430/A036279
tanh(x) = x - (1/3)*x^3 + (2/15)*x^5 - (17/315)*x^7 + (62/2835)*x^9 - (1382/155925)*x^11 + ...
The coefficients in the expansion of tanh x are 0, 1, 0, -1/3, 0, 2/15, 0, -17/315, 0, 62/2835, 0, -1382/155925, 0, 21844/6081075, 0, -929569/638512875, 0, 6404582/10854718875, 0, -443861162/1856156927625, ... = A002430/A036279
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).
  • G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 74.
  • H. A. Rothe, in C. F. Hindenburg, editor, Sammlung Combinatorisch-Analytischer Abhandlungen, Vol. 2, Chap. XI. Fleischer, Leipzig, 1800, p. 329.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 34, equation 34:6:1 at page 323.

Crossrefs

Programs

  • Maple
    R := n -> (-1)^floor(n/2)*(4^n-2^n)*Zeta(1-n)/(n-1)!:
    seq(denom(R(2*n)), n=1..18); # Peter Luschny, Aug 25 2015
  • Mathematica
    f[n_] := (-1)^Floor[n/2] (4^n - 2^n) Zeta[1 - n]/(n - 1)!; Table[Denominator@ f[2 n], {n, 17}] (* Michael De Vlieger, Aug 25 2015 *)

Formula

a(n) = denominator((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)*Bernoulli(2*n)/(2*n)!). - Johannes W. Meijer, May 24 2009
Let R(x) = (cos(x*Pi/2) + sin(x*Pi/2))*(4^x-2^x)*Zeta(1-x)/(x-1)!. Then a(n) = denominator(R(2*n)) and A002430(n) = numerator(R(2*n)). - Peter Luschny, Aug 25 2015

Extensions

Incorrect comment by Stephen Crowley deleted by Johannes W. Meijer, Jan 19 2009

A156769 a(n) = denominator(2^(2*n-2)/factorial(2*n-1)).

Original entry on oeis.org

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 49308808782358125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875
Offset: 1

Views

Author

Johannes W. Meijer, Feb 15 2009

Keywords

Comments

Resembles A036279, the denominators in the Taylor series for tan(x). The first difference occurs at a(12).
The numerators of the two formulas for this sequence lead to A001316, Gould's sequence.
Stephen Crowley indicated on Aug 25 2008 that a(n) = denominator(Zeta(2*n)/Zeta(1-2*n)) and here numerator((Zeta(2*n)/Zeta(1-2*n))/(2*(-1)^(n)*(Pi)^(2*n))) leads to Gould's sequence.
This sequence appears in the Eta and Zeta triangles A160464 and A160474. Its resemblance to the sequence of the denominators of the Taylor series for tan(x) led to the conjecture A156769(n) = A036279(n)*A089170(n-1). - Johannes W. Meijer, May 24 2009

Crossrefs

Cf. A036279 Denominators in Taylor series for tan(x).
Cf. A001316 Gould's sequence appears in the numerators.
Cf. A000265, A036279, A089170, A117972, A160464, A160469 (which resembles the numerators of the Taylor series for tan(x)), A160474. - Johannes W. Meijer, May 24 2009

Programs

  • Magma
    [Denominator(4^(n-1)/Factorial(2*n-1)): n in [1..25]]; // G. C. Greubel, Jun 19 2021
    
  • Maple
    a := n ->(2*n-1)!*2^(add(i,i=convert(n-1,base,2))-2*n+2); # Peter Luschny, May 02 2009
  • Mathematica
    a[n_] := Denominator[4^(n-1)/(2n-1)!];
    Array[a, 15] (* Jean-François Alcover, Jun 20 2018 *)
  • Sage
    [denominator(4^(n-1)/factorial(2*n-1)) for n in (1..25)] # G. C. Greubel, Jun 19 2021

Formula

a(n) = denominator( Product_{k=1..n-1} 2/(k*(2*k+1)) ).
G.f.: (1/2)*z^(1/2)*sinh(2*z^(1/2)).
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = abs(A117972(n))/A000265(n).
a(n) = A036279(n)*A089170(n-1). (End)
a(n) = A049606(2*n-1). - Zhujun Zhang, May 29 2019

A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

Original entry on oeis.org

1, 1, 2, 1, 17, 26, 2, 62, 192, 60, 1, 1382, 7192, 5097, 502, 2, 21844, 171511, 217186, 55196, 2036, 2, 929569, 10262046, 20376780, 9893440, 1089330, 16356, 4, 6404582, 94582204, 271154544, 215114420, 48673180, 2567568, 16376, 1
Offset: 1

Views

Author

Johannes W. Meijer, May 24 2009

Keywords

Comments

In A160464 we defined the ES1 matrix by ES1[2*m-1,n=1] and in A094665 it was shown that the n-th term of the coefficients of matrix row ES1[1-2*m,n] for m >= 1 can be generated with the RES1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRES1(z,1-2*m) = sum(RES1(1-2*m,n)*z^(n-1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,1-2*m) = (-1)*RE(z,1-2*m)/(2*p(m-1)*(z-1)^(m)). The p(m-1), m >= 1, sequence is Gould's sequence A001316.
The coefficients of the RE(z,1-2*m) polynomials lead to the triangle given above.
The E(z,n) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
Some of our results are conjectures based on numerical evidence.

Examples

			The first few rows are:
[1]
[1]
[2, 1]
[17, 26, 2]
[62, 192, 60, 1]
The first few polynomials RE(z,m) are:
RE(z,-1) = 1
RE(z,-3) = 1
RE(z,-5) = 2+z
RE(z,-7) = 17+26*z+2*z^2
The first few GFRES1(z,m) are:
GFRES1(z,-1) = -(1/1)*(1)/(2*(z-1)^1)
GFRES1(z,-3) = -(1/2)*(1)/(2*(z-1)^2)
GFRES1(z,-5) = -(1/2)*(2+z)/(2*(z-1)^3)
GFRES1(z,-7) = -(1/4)*(17+26*z+2*z^2)/(2*(z-1)^4)
		

Crossrefs

For the Eulerian numbers E(n, k) see A008292.
The p(n) sequence equals Gould's sequence A001316.
The first right hand column of the triangle equals A048896.
The first left hand column equals A160469.
The row sums equal the absolute values of A117972.

Programs

  • Maple
    nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i-1, x), x): for na from 0 to dgr do c(na) := coeff(T(i-1, x), x, na) od: T(i-1, x+1) := 0: for nb from 0 to dgr do T(i-1, x+1) := T(i-1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i-1, nc+1) := coeff(T(i-1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i-1, x+1) - 2*x^2*T(i-1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((-1)^(m-q)*(q^k)*binomial(k+1, m-q), q=1..m) od: fx(p) := (-1)^(p+1) * (sum(E(r, k)*z^(k-r), r=1..k))/(z-1)^(p+1): GF(-(2*p+1)) := sort(simplify(((-1)^p* 1/2^(p+1)) * sum(ECGP(k-1, k-s)*fx(k-s), s=0..k-1)), ascending): NUMGF(-(2*p+1)) := -numer(GF(-(2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF(-(2*p+1)), z, n-1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n-1)/2+m+1) := A(n+1, m) end do end do: a := n-> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);
  • Mathematica
    T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
    a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m - 1}, x Sum[ a[k] a[n - k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n - 1}]] (* Michael Somos, Apr 22 2020 *)

Extensions

Edited by Johannes W. Meijer, Sep 23 2012

A089170 Numerator of 2*BernoulliB[2*(n+1)]*(4^(n+1)-1)/(2*(n+1))] divided by numerator of the series coefficients of 1/(1 + Cosh[x]).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 527, 1, 1, 1, 1, 31, 1, 1, 731, 1, 41, 1, 1, 1, 37, 1333, 17, 1, 1, 1, 31, 1, 1, 1, 17, 73, 1, 1, 1, 43, 1271, 59, 629, 1, 73, 2759, 43, 1, 1, 1, 17, 1, 67, 7519, 1, 31, 89, 1, 289, 1, 29020032511, 1, 10573, 1, 1, 1, 2227, 486029
Offset: 0

Views

Author

Wouter Meeussen, Dec 07 2003

Keywords

Comments

Ratios of two similar sequences.
This sequence is related to the sequences of the numerators and denominators of the Taylor series for tan(x), i.e., A002430 and A036279, and the similar sequences A160469 and A156769. - Johannes W. Meijer, May 24 2009

Crossrefs

Cf. A002425.
From Johannes W. Meijer, May 24 2009: (Start)
Equals A160469(n+1)/A002430(n+1).
Equals A156769(n+1)/A036279(n+1).
(End)

Programs

  • Maple
    seq(numer(2*bernoulli(2*n)*(4^n-1)/(2*n))/numer((4^n-1)*bernoulli(2*n)/(2*n)!),n=1..100); # C. Ronaldo
  • Mathematica
    Table[Numerator[2*BernoulliB[2*n]*(4^n -1)/(2*n)]/Numerator[SeriesCoefficient[Series[1/(1+Cosh[x]), {x, 0, 2n}], 2n-2]], {n, 1, 128}]

Formula

For n>=0, a(n)=c(n+1) where c(n)=numerator((4^n-1)*B(2*n)/n)/numerator((4^n-1)*B(2*n)/(2*n)!), B(k) denotes the k-th Bernoulli number. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 19 2004
Showing 1-8 of 8 results.