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.

A317932 Denominators of certain "Dirichlet Square Root" sequences: a(n) = A046644(n)/(2^A007949(n)).

Original entry on oeis.org

1, 2, 1, 8, 2, 2, 2, 16, 2, 4, 2, 8, 2, 4, 2, 128, 2, 4, 2, 16, 2, 4, 2, 16, 8, 4, 2, 16, 2, 4, 2, 256, 2, 4, 4, 16, 2, 4, 2, 32, 2, 4, 2, 16, 4, 4, 2, 128, 8, 16, 2, 16, 2, 4, 4, 32, 2, 4, 2, 16, 2, 4, 4, 1024, 4, 4, 2, 16, 2, 8, 2, 32, 2, 4, 8, 16, 4, 4, 2, 256, 8, 4, 2, 16, 4, 4, 2, 32, 2, 8, 4, 16, 2, 4, 4, 256, 2, 16, 4, 64, 2, 4, 2, 32, 4
Offset: 1

Views

Author

Antti Karttunen, Aug 11 2018

Keywords

Comments

These are denominators for rational valued sequences that are obtained as "Dirichlet Square Roots" of sequences b that satisfy the condition b(3) = 2, and b(p) = odd number for any other primes p. For example, A064989, A065769 and A234840. - Antti Karttunen, Aug 31 2018
The original definition was: Denominators of the rational valued sequence whose Dirichlet convolution with itself yields A002487, Stern's Diatomic sequence. However, this definition depends on the conjecture given in A261179.

Crossrefs

Cf. A317930, A318319, A318669 (some of the numerator sequences), A317931 (conjectured, for A002487).
Cf. A305439 (the 2-adic valuation), A318666.

Programs

Formula

a(n) = A046644(n)/A318666(n) = 2^A305439(n).
a(n) = denominator of f(n), where f(1) = 1, f(n) = (1/2) * (b(n) - Sum_{d|n, d>1, d 1, where b can be A064989, A065769 or A234840 for example, conjecturally also A002487.
Multiplicative with a(3^e) = 2^A011371(e), a(p^e) = 2^A005187(e) for any other primes. - Antti Karttunen, Sep 03 2018

Extensions

Definition changed, the original (now conjectured alternative definition) moved to the comments section by Antti Karttunen, Aug 31 2018
Keyword:mult added by Antti Karttunen, Sep 03 2018

A237646 G.f.: exp( Sum_{n>=1} A163659(n^3)*x^n/n ), where x*exp(Sum_{n>=1} A163659(n)*x^n/n) = S(x) is the g.f. of Stern's diatomic series (A002487).

Original entry on oeis.org

1, 1, 8, 7, 63, 56, 329, 273, 1736, 1463, 7511, 6048, 32585, 26537, 124440, 97903, 475287, 377384, 1658881, 1281497, 5783960, 4502463, 18825023, 14322560, 61171649, 46849089, 188181672, 141332583, 577889023, 436556440, 1696298665, 1259742225, 4970284200, 3710541975, 14019036535, 10308494560
Offset: 0

Views

Author

Paul D. Hanna, May 03 2014

Keywords

Comments

Compare to the g.f. of A195586.

Examples

			G.f.: A(x) = 1 + x + 8*x^2 + 7*x^3 + 63*x^4 + 56*x^5 + 329*x^6 + 273*x^7 +...
where
log(A(x)) = x + 15*x^2/2 - 2*x^3/3 + 127*x^4/4 + x^5/5 - 30*x^6/6 + x^7/7 + 1023*x^8/8 +...+ A237649(n)*x^n/n +...
Bisections: let A(x) = B(x^2) + x*C(x^2), then:
B(x) = 1 + 8*x + 63*x^2 + 329*x^3 + 1736*x^4 + 7511*x^5 + 32585*x^6 +...
C(x) = 1 + 7*x + 56*x^2 + 273*x^3 + 1463*x^4 + 6048*x^5 + 26537*x^6 + 97903*x^7 + 377384*x^8 + 1281497*x^9 + 4502463*x^10 +...+ A237647(n)*x^n +...
Note that C(x)^(1/7) = (1+x+x^2) * C(x^2)^(4/7) is an integer series:
C(x)^(1/7) = 1 + x + 5*x^2 + 4*x^3 + 30*x^4 + 26*x^5 + 106*x^6 + 80*x^7 + 459*x^8 + 379*x^9 + 1451*x^10 + 1072*x^11 + 5210*x^12 +...+ A237648(n)*x^n +...
Also, C(x) / (1+x+x^2)^3 = A(x)^4:
A(x)^4 = 1 + 4*x + 38*x^2 + 128*x^3 + 817*x^4 + 2536*x^5 + 12890*x^6 +...
Further, C(x)*C(x^2)^3 = A(x)^7:
A(x)^7 = 1 + 7*x + 77*x^2 + 420*x^3 + 2954*x^4 + 13986*x^5 + 78414*x^6 +...
The g.f. may be expressed by the product:
A(x) = (1+x+x^2) * (1+x^2+x^4)^7 * (1+x^4+x^8)^28 * (1+x^8+x^16)^112 * (1+x^16+x^32)^448 *...* (1 + x^(2*2^n) + x^(4*2^n))^(7*4^n) *...
		

Crossrefs

Programs

  • PARI
    {A163659(n)=if(n<1, 0, if(n%3, 1, -2)*sigma(2^valuation(n, 2)))}
    {a(n)=polcoeff(exp(sum(k=1, n, A163659(k^3)*x^k/k)+x*O(x^n)), n)}
    for(n=0, 40, print1(a(n), ", "))

Formula

G.f.: exp( Sum_{n>=1} A237649(n)*x^n/n ), where A237649(n) = A163659(n^3).
G.f. A(x) satisfies:
(1) A(x) = (1+x+x^2) * (1+x^2+x^4)^3 * A(x^2)^4.
(2) A(x) = (1+x+x^2) * Product_{n>=0} ( 1 + x^(2*2^n) + x^(4*2^n) )^(7*4^n).
(3) A(x) / A(-x) = (1+x+x^2) / (1-x+x^2).
Bisections: let A(x) = B(x^2) + x*C(x^2), then
(4) B(x) = (1+x) * C(x).
(5) C(x) = (1+x+x^2)^7 * C(x^2)^4.
(6) A(x) = (1+x+x^2) * C(x^2).
(7) A(x)^7 = C(x) * C(x^2)^3.
(8) A(x)^4 = C(x) / (1+x+x^2)^3.
(9) A(x)^3 = ( C(x)/A(x) - C(x^2)^4/A(x^2)^4 ) / (6*x + 14*x^3 + 6*x^5).

A195587 a(n) = A163659(n^2), where A163659 is the logarithmic derivative of Stern's diatomic series (A002487).

Original entry on oeis.org

1, 7, -2, 31, 1, -14, 1, 127, -2, 7, 1, -62, 1, 7, -2, 511, 1, -14, 1, 31, -2, 7, 1, -254, 1, 7, -2, 31, 1, -14, 1, 2047, -2, 7, 1, -62, 1, 7, -2, 127, 1, -14, 1, 31, -2, 7, 1, -1022, 1, 7, -2, 31, 1, -14, 1, 127, -2, 7, 1, -62, 1, 7, -2, 8191, 1, -14, 1, 31, -2, 7, 1, -254, 1, 7, -2, 31, 1, -14, 1, 511, -2, 7, 1, -62, 1, 7, -2, 127, 1, -14, 1, 31, -2, 7, 1, -4094
Offset: 1

Views

Author

Paul D. Hanna, Sep 20 2011

Keywords

Comments

Multiplicative because A163659 is. - Andrew Howroyd, Jul 26 2018

Examples

			L.g.f.: L(x) = x + 7*x^2/2 - 2*x^3/3 + 31*x^4/4 + x^5/5 - 14*x^6/6 + x^7/7 + 127*x^8/8 +...
where
exp(L(x)) = 1 + x + 4*x^2 + 3*x^3 + 15*x^4 + 12*x^5 + 37*x^6 + 25*x^7 +...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Times @@ (Function[{p, e}, Which[p == 2, 2^(e+1) - 1, p == 3, -2, True, 1]] @@@ FactorInteger[n^2]);
    a /@ Range[1, 100] (* Jean-François Alcover, Sep 20 2019 *)
  • PARI
    {A163659(n)=if(n<1,0,if(n%3,1,-2)*sigma(2^valuation(n,2)))}
    {a(n)=A163659(n^2)}
    for(n=1, 64, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(X=x+x*O(x^n), A); A=log(1+X+X^2) + sum(k=0, #binary(n), 3*2^k*log(1 + X^(2*2^k) + X^(4*2^k))); n*polcoeff(A, n)}
    for(n=1, 64, print1(a(n), ", "))

Formula

L.g.f.: log(1+x+x^2) + Sum_{n>=0} 3*2^n * log(1 + x^(2*2^n) + x^(4*2^n)) = Sum_{n>=1} a(n)*x^n/n. - Paul D. Hanna, May 04 2014
G.f.: x*(1+2*x)/(1+x+x^2) + Sum_{n>=0} 6*4^n * x^(2*2^n) * (1 + 2*x^(2*2^n)) / (1 + x^(2*2^n) + x^(4*2^n)). - Paul D. Hanna, May 04 2014
Dirichlet g.f.: zeta(s) * (1 - 3^(1-s)) * (2^s + 2) / (2^s - 4). - Amiram Eldar, Oct 24 2023
Showing 1-3 of 3 results.