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.

A123725 Numerators of fractional partial quotients appearing in a continued fraction for the power series Sum_{n>=0} x^(2^n - 1)/(n+1)^s.

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Oct 12 2006

Keywords

Examples

			Surprisingly, the following analog of the Riemann zeta function:
Z(x,s) = Sum_{n>=0} x^(2^n-1)/(n+1)^s = 1 + x/2^s + x^3/3^s +x^7/4^s+..
may be expressed by the continued fraction:
Z(x,s) = [1; f(1)^s/x, -f(2)^s/x, -f(3)^s/x,...,f(n)^s*(-1)^e(n)/x,...]
such that the (2^n-1)-th convergent = Sum_{k=0..n} x^(2^k-1)/(k+1)^s,
where f(n) = (b(n)+2)/(b(n)+1)^2 and e(n) = A073089(n+1) for n>=1,
and b(n) = A007814(n) the exponent of highest power of 2 dividing n.
Thus a(n) = (A007814(n) + 2)*(-1)^A073089(n+1) are numerators and
A123726(n) = (A007814(n) + 1)^2 are denominators of partial quotients.
Case s=1.
Sum_{n>=0} x^(2^n-1)/(n+1) = [1; 2/x, -(3/4)/x, -2/x, -(4/9)/x, 2/x,
(3/4)/x, -2/x, -(5/16)/x, 2/x, -(3/4)/x, -2/x, (4/9)/x, 2/x, (3/4)/x,
-2/x, -(6/25)/x, 2/x, -(3/4)/x, -2/x, -(4/9)/x, 2/x, (3/4)/x, -2/x,...].
Note that (2^n-1)-th convergents exactly equal n-th partial sums:
[1;2/x] = 1 + x/2;
[1;2/x,-(3/4)/x,-2/x] = 1 + x/2 + x^3/3;
[1;2/x,-(3/4)/x,-2/x,-(4/9)/x,2/x,(3/4)/x,-2/x] = 1 +x/2 +x^3/3 +x^7/4.
Case s=2.
Sum_{n>=0} x^(2^n-1)/(n+1)^2 = [1; 4/x, -(9/16)/x, -4/x, -(16/81)/x,
4/x, (9/16)/x, -4/x, -(25/256)/x, 4/x, -(9/16)/x, -4/x, (16/81)/x, 4/x,
(9/16)/x, -4/x, -(36/625)/x, 4/x, -(9/16)/x, -4/x, (16/81)/x, 4/x ...].
Note that (2^n-1)-th convergents exactly equal n-th partial sums:
[1;4/x] = 1 + x/4;
[1;4/x,-(9/16)/x,-4/x] = 1 + x/4 + x^3/9;
[1;4/x,-(9/16)/x,-4/x,-(16/81)/x,4/x,(9/16)/x,-4/x]=1+x/4+x^3/9+x^7/16.
Case s=3.
Sum_{n>=0} x^(2^n-1)/(n+1)^3 = [1; 8/x,-(27/64)/x,-8/x,-(64/729)/x,8/x,
(27/64)/x,-8/x,-(125/4096)/x, 8/x,-(27/64)/x,-8/x, (64/729)/x, 8/x,
(27/64)/x,-8/x,-(216/15625)/x, 8/x, -(27/64)/x, -8/x, (64/729)/x ...].
Likewise, the (2^n-1)-th convergents exactly equal n-th partial sums.
It is conjectured that these patterns continue for all s.

Links

Crossrefs

Cf. A123726 (denominators); A007814, A073089, A089080 (unsigned).

Programs

  • PARI
    {a(n)=numerator(subst(contfrac(sum(m=0,#binary(n),1/x^(2^m-1)/(m+1)),n+3)[n+1],x,1))}
  • PARI
    A007814(n) = valuation(n,2);
A073089(n) = { if(n<=1, return(0)); n-=1; my(v=2^valuation(n, 2)); return((0==bitand(n, v<<1)) != (v%2)); }; \\ From A073089
A123725(n) = if(!n,1,(A007814(n) + 2) * (-1)^A073089(n+1)); \\ Antti Karttunen, Nov 01 2018

Formula

a(n) = (A007814(n) + 2) * (-1)^A073089(n+1), n>=1, with a(0)=1.
a(n) = A089080(n+1) * (-1)^A073089(n+1) for n>=0.

A317538 Indices m for which A317413(m) = 1.

Original entry on oeis.org

1, 4, 6, 9, 12, 16, 19, 21, 24, 28, 30, 33, 37, 40, 43, 45, 48, 52, 54, 57, 60, 64, 67, 69, 73, 76, 78, 81, 85, 88, 91, 93, 96, 100, 102, 105, 108, 112, 115, 117, 120, 124, 126, 129, 133, 136, 139, 141, 145, 148, 150, 153, 156, 160, 163, 165, 169, 172, 174, 177, 181, 184, 187, 189, 192, 196, 198, 201, 204
Offset: 1

Views

Author

A.H.M. Smeets, Jul 30 2018

Keywords

Crossrefs

Cf. A073089, A317413.

Programs

  • Python
    n, f, i, p, q, base = 1, 1, 0, 0, 1, 2
while i < 10000000:
    i, p, q = i+1, p*base, q*base
    if i == f:
        p, n = p+1, n+1
        f = f*n
n, a, j = 0, 0, 0
while p%q > 0:
    a, f, p, q = a+1, p//q, q, p%q
    if f == 1:
        n = n+1
        print(n, a-1)

Formula

a(n) = 3*(n-1) + 1 - A073089(n).

A123726 Denominators of fractional partial quotients appearing in a continued fraction for the power series Sum_{n>=0} x^(2^n - 1)/(n+1)^s.

Original entry on oeis.org

1, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 36, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 49, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 25, 1, 4, 1, 9, 1, 4, 1, 16, 1, 4, 1, 9, 1, 4, 1, 36, 1, 4
Offset: 0

Views

Author

Paul D. Hanna, Oct 12 2006

Keywords

Comments

A123725(n) = (A007814(n) + 2)*(-1)^A073089(n+1) are the numerators of the partial quotients.

Examples

			Surprisingly, the following analog of the Riemann zeta function:
Z(x,s) = Sum_{n>=0} x^(2^n-1)/(n+1)^s = 1 + x/2^s + x^3/3^s +x^7/4^s+..
may be expressed by the continued fraction:
Z(x,s) = [1; f(1)^s/x, -f(2)^s/x, -f(3)^s/x,...,f(n)^s*(-1)^e(n)/x,...]
such that the (2^n-1)-th convergent = Sum_{k=0..n} x^(2^k-1)/(k+1)^s,
where f(n) = (b(n)+2)/(b(n)+1)^2 and e(n) = A073089(n+1) for n>=1,
and b(n) = A007814(n) the exponent of highest power of 2 dividing n.

Links

Crossrefs

Cf. A123725 (numerators), A007814, A073089, A001511.

Programs

  • Magma
    [1] cat [(Valuation(n, 2) + 1)^2: n in [1..50]]; // G. C. Greubel, Nov 01 2018
  • Mathematica
    Join[{1}, Table[(1 + IntegerExponent[n, 2])^2, {n, 1, 50}]] (* G. C. Greubel, Nov 01 2018 *)
  • PARI
    {a(n)=denominator(subst(contfrac(sum(m=0,#binary(n),1/x^(2^m-1)/(m+1)),n+3)[n+1],x,1))}
  • PARI
    /* a(n) = (A007814(n)+1)^2: */ {a(n)=if(n==0,1,(valuation(n,2)+1)^2)}

Formula

a(n) = (A007814(n) + 1)^2 = A001511(n)^2 for n>=1, with a(0)=1, where A007814(n) is the exponent of the highest power of 2 dividing n.
Multiplicative with a(2^e) = (e + 1)^2, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Jul 31 2018
From Amiram Eldar, Dec 29 2022: (Start)
Dirichlet g.f.: zeta(s)*(4^s+2^s)/(2^s-1)^2.
Sum_{k=1..n} a(k) ~ 6*n. (End)

Extensions

Ref to A001511 added by Franklin T. Adams-Watters, Dec 22 2013

A110037 Signed version of A090678 and congruent to A088567 mod 2.

Original entry on oeis.org

1, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, -1, 1, 0, -1, 0, 0, 1, 0, -1, 0
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2005

Keywords

Comments

a(n) = (-1)^[n/2]*A090678(n) = A088567(n) mod 2, where A088567(n) equals the number of "non-squashing" partitions of n. a(n) = -A110036(n)/2 for n>=2, where the A110036 gives the partial quotients of the continued fraction expansion of 1 + Sum_{n>=0} 1/x^(2^n).

Crossrefs

Cf. A110036, A090678, A088567, A073089.

Programs

  • PARI
    {a(n)=polcoeff(A=1+x-x^2*(1+x)/(1+x^2)+ sum(k=1,#binary(n),x^(3*2^(k-1))/prod(j=0,k,1+x^(2^j)+x*O(x^n))),n)}

Formula

G.f.: A(x) = 1+x - x^2*(1+x)/(1+x^2) + Sum_{k>=1} x^(3*2^(k-1))/Product_{j=0..k} (1+x^(2^j)).
Conjecture: a(n) = A073089(n) - A073089(n+1) for n >= 2. - Alan Michael Gómez Calderón, Aug 19 2025

A081769 a(n)-th term of the continued fraction for sum(k>=0,1/2^(2^k)) is 2.

Original entry on oeis.org

5, 13, 18, 23, 25, 30, 38, 43, 45, 53, 58, 60, 65, 70, 78, 83, 85, 93, 98, 103, 105, 110, 118, 120, 125, 133, 138, 140, 145, 150, 158, 163, 165, 173, 178, 183, 185, 190, 198, 203, 205, 213, 218, 220, 225, 230, 238, 240, 245, 253, 258, 263, 265, 270, 278, 280
Offset: 1

Views

Author

Benoit Cloitre, Apr 09 2003

Keywords

Examples

			The continued fraction for sum(k>=0,1/2^(2^k)) is : 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1... hence a(1)=5

Crossrefs

Cf. A006466, A073089.

Formula

a(n)=5*n + 0 or + 3 and it appears that a(n)=5*n+3*A073089(n+2)
Showing 1-5 of 5 results.

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