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 1-3 of 3 results.

A134652 Indices for which A097344 differs from A097345.

Original entry on oeis.org

59, 1519, 7814, 17225, 39079, 950619, 977019, 1280699
Offset: 1

Views

Author

M. F. Hasler, Jan 25 2008

Keywords

Comments

The terms 59 and 1519 were found by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008.
a(6) > 10^5.
These are the numbers m such that f(m) = Sum_{k=0..m} binomial(m,k)/(k+1)^2 (binomial transform of 1/(k+1)^2) has the same numerator as g(m) = Sum_{k=0..m} (2^(k+1) - 1)/(k+1) (which are also the partial sums of the binomial transformation of 1/(k+1)).
Obviously, f(m) = Sum_{k=0..m} binomial(m+1, k+1)/((k+1)*(m+1)) and since g(m) = (m+1) f(m) (cf. notes by R. J. Mathar on A097345), g(m) = Sum_{k=1..m+1} binomial(m+1,k)/k.
We have the equivalences: numerator(g(n)) = numerator(f(n)) <=> (n+1) | denominator(f(n)) <=> gcd(numerator(g(n)), n+1) = 1.
Therefore this sequence can be alternatively defined in either of the following two ways: numbers n such that the denominator of f(n) is not divisible by (n+1); numbers n such that the numerator of g(n) is not coprime to (n+1).
In terms of M = m+1, the characterization reads: a(n)+1 = numbers M such that denominator(Sum_{k=1..M} binomial(M-1, k-1)/k^2) is not a multiple of M = numbers M such that numerator(Sum_{k=1..M} (2^k - 1)/k) is not coprime to M.

Crossrefs

Programs

  • Mathematica
    Reap[ For[n = 1, n < 10^5, n++, If[ !Divisible[ Denominator[ HypergeometricPFQ[{1, 1, -n}, {2, 2}, -1]], n+1], Print[n]; Sow[n] ] ] ][[2, 1]] (* Jean-François Alcover, Oct 15 2013 *)
  • PARI
    t=1; for( n=2,10^5, gcd( numerator(t+=(1<1 & print(n-1))

Extensions

a(6)-a(8) from Amiram Eldar, Apr 08 2019

A097344 Numerators in binomial transform of 1/(n+1)^2.

Original entry on oeis.org

1, 5, 29, 103, 887, 1517, 18239, 63253, 332839, 118127, 2331085, 4222975, 100309579, 184649263, 1710440723, 6372905521, 202804884977, 381240382217, 13667257415003, 25872280345103, 49119954154463, 93501887462903, 4103348710010689, 7846225754967739, 75162749477272151
Offset: 0

Views

Author

Paul Barry, Aug 06 2004

Keywords

Comments

Is this identical to A097345? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.
If the putative formula a(n)=A081528(n) sum{k=0..n, binomial(n, k)/(k+1)^2} were true, then this sequence coincides with A097345 according to Mathar's notes. However, the term n=9 in the binomial transform of 1/(n+1)^2 has the denominator 5040=A081528(9)/4=A081528(10)/5. So the formula cannot be true. - M. F. Hasler, Jan 25 2008
a(n) is also the numerator of u(n+1) with u(n) = (1/n)*Sum_{k=1..n} (2^k-1)/k and we have the formula: polylog(2,x/(1-x)) = Sum_{n>=1} u(n)*x^n on the interval [-1/2, 1/2]. - Groux Roland, Feb 01 2009

Examples

			The first values of the binomial transform of 1/(n+1)^2 are 1, 5/4, 29/18, 103/48, 887/300, 1517/360, 18239/2940, 63253/6720, 332839/22680, 118127/5040, 2331085/60984, ...
		

Crossrefs

Programs

  • Maple
    f:=n->numer(add( binomial(n,k)/(k+1)^2, k=0..n));
  • Mathematica
    Table[HypergeometricPFQ[{1, 1, -n}, {2, 2}, -1] // Numerator, {n, 0, 24}] (* Jean-François Alcover, Oct 14 2013 *)
  • Maxima
    a(n):=if n<0 then 1 else 1/((n+1)^2)*((n)*(3*n+1)*a(n-1)-2*(n-1)*(n)*a(n-2)+1);
    makelist(num(a(n),n,0,10); /* Vladimir Kruchinin, Jun 01 2016 */
    
  • PARI
    A097344(n)=numerator(sum(k=0,n,binomial(n,k)/(k+1)^2)) \\ M. F. Hasler, Jan 25 2008
    
  • Python
    from fractions import Fraction
    A097344_list, tlist = [1], [Fraction(1,1)]
    for i in range(1,100):
        for j in range(len(tlist)):
            tlist[j] *= Fraction(i,i-j)
        tlist += [Fraction(1,(i+1)**2)]
        A097344_list.append(sum(tlist).numerator) # Chai Wah Wu, Jun 04 2015
    
  • Sage
    def A097344_list(size):
        R, L = [1], [1]
        inc = sqr = 1
        for i in range(1, size):
            for j in range(i):
                L[j] *= i / (i - j)
            inc += 2; sqr += inc
            L.extend(1 / sqr)
            R.append(sum(L).numerator())
        return R
    A097344_list(50) # (after Chai Wah Wu) Peter Luschny, Jun 05 2016

Formula

a(n) = numerator(b(n)), b(n) = 1/((n+1)^2)*((n)*(3*n+1)*b(n-1)-2*(n-1)*(n)*b(n-2)+1). - Vladimir Kruchinin, May 31 2016

Extensions

Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008 and M. F. Hasler, Jan 25 2008
Moved comment on numerators of a logarithmic g.f. over to A097345 - R. J. Mathar, Mar 04 2010

A097346 Numerators of a Fibonacci related transform of 1/(n+1).

Original entry on oeis.org

1, 2, 9, 24, 170, 240, 2415, 7000, 30744, 45360, 741510, 1108800, 21673080, 32744712, 49684635, 151351200, 3932288360, 6028102080, 176061013128, 271314186000, 419047770960, 648560072160, 23130491854470, 35929017476352
Offset: 0

Views

Author

Paul Barry, Aug 06 2004

Keywords

Crossrefs

Cf. A097345.

Formula

a(n)=A003418(n)*sum{k=0..floor(n/2), binomial(n-k, k)/(k+1)}
Showing 1-3 of 3 results.