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 1-4 of 4 results.

A132317 a(n) = [x^(2^n)] Product_{i=0..n} (1 + x^(2^i) )^(2^(n-i)); equals column 1 of triangle A132318.

Original entry on oeis.org

1, 2, 15, 1024, 7048181, 469389728563470, 2954306864416502250656677496683, 165756604793755389851497802171770083459242616940095659925793836
Offset: 0

Views

Author

Paul D. Hanna, Aug 19 2007

Keywords

Comments

Next term, a(8), has 126 digits.

Examples

			a(0) = [x] (1+x) = 1;
a(1) = [x^2] (1+x)^2*(1+x^2) = 2;
a(2) = [x^4] (1+x)^4*(1+x^2)^2*(1+x^4) = 15;
a(3) = [x^8] (1+x)^8*(1+x^2)^4*(1+x^4)^2*(1+x^8) = 1024;
a(4) = [x^16] (1+x)^16*(1+x^2)^8*(1+x^4)^4*(1+x^8)^2*(1+x^16) = 7048181.

Crossrefs

Cf. A132318, A132316.

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[(1 + x^(2^j))^(2^(n-j)),{j,0,n}],{x,0,2^n}], {n,0,10}] (* Vaclav Kotesovec, Oct 09 2020 *)
  • PARI
    {a(n)=polcoeff(prod(i=0,n,(1 + x^(2^i) +x*O(x^(2^n)))^(2^(n-i))), 2^n)}

A132316 a(n) = [x^n] Product_{i=0..n} (1 + x^(2^i) )^(2^(n-i)).

Original entry on oeis.org

1, 2, 8, 88, 2812, 284832, 96344064, 112162777984, 458279216351168, 6667184111642112512, 349410072608155198029824, 66605152356815910201401874432, 46557942811582437260863430233248768
Offset: 0

Views

Author

Paul D. Hanna, Aug 18 2007

Keywords

Examples

			a(2) = [x^2] (1+x)^4*(1+x^2)^2*(1+x^4) = 8;
a(3) = [x^3] (1+x)^8*(1+x^2)^4*(1+x^4)^2*(1+x^8) = 88;
a(4) = [x^4] (1+x)^16*(1+x^2)^8*(1+x^4)^4*(1+x^8)^2*(1+x^16) = 2812.

Crossrefs

Cf. A132317, A132318.

Programs

  • Mathematica
    Table[SeriesCoefficient[Product[(1 + x^(2^j))^(2^(n-j)),{j,0,n}],{x,0,n}], {n,0,15}] (* Vaclav Kotesovec, Oct 09 2020 *)
  • PARI
    {a(n)=polcoeff(prod(i=0,#binary(n),(1 + x^(2^i) +x*O(x^n))^(2^(n-i))), n)}

Formula

a(n) ~ 2^(n^2) / n!. - Vaclav Kotesovec, Oct 09 2020

A333209 a(n) is the denominator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

Original entry on oeis.org

2, 36, 7392, 1688148, 197412831, 21085413226416, 101768454084335346, 60343478516053297339236, 73240105330540144095414793632, 1956470757376233684880813258936380492, 32802418997525523144166495047229414174839, 202042966989952174292936124782341088713724476716231
Offset: 1

Views

Author

A.H.M. Smeets, Mar 11 2020

Keywords

Comments

The numerators are given in A333208.
See A333088 and A333089 for similar fractions for infinite sums of Fibonacci numbers.
Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)) is a fraction for n > 0.
Sum_{i >= 0} 1/Lucas(i)^2 = 1/4 + A105394, i.e., the n = 0 case, is believed to be transcendental.

Examples

			These infinite sums begin: 1/2, 7/36, 551/7392, ...

Links

Crossrefs

Cf. A000032, A105394, A132318, A333088, A333089, A333208.

Programs

  • Mathematica
    a[n_] := Denominator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* Amiram Eldar, Mar 11 2020 *)
  • Python
    from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum,sden*f0)
    print(n,sden*f0//d)
f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020
  • Python
    from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum,sden*f0)
    print(n,sden*f0//d)
    f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020

Formula

a(n) = denominator of (1/Fibonacci(2n)) * Sum_{i = 1..n} 1/(Lucas(2i-2)*Lucas(2i-1)).
Lim_{n -> inf} (A333208(n)/a(n)) / (A333208(n-1)/a(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.
The following generalization holds: (Start)
Let H_(a,b) (n) be defined by H_(a,b) (0) = a, H_(a,b) (1) = b and H_(a,b) (n) = H_(a,b) (n-1) + H_(a,b) (n-2) for n > 1, then
Sum_{i >= 0} 1/(H_(a,b) (i)*H_(a,b) (i+2n)) = (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(H_(a,b) (2i-2)*H_(a,b) (2i-1)) for n > 0, and are thus all fractions. Specially, H_(0,1) are the Fibonacci numbers A000045, H_(2,1) as here, are the Lucas numbers A000032, and H_(3,1) are the Pibonacci numbers A104449. (End)

A333208 a(n) is the numerator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

Original entry on oeis.org

1, 7, 551, 48091, 2148268, 87644575267, 161577754532123, 36595152483523582367, 16965509829762630129638831, 173107561150078104051618631740949, 1108595900580419409151086339986148307, 2608169750203411467722731179728125652086612772
Offset: 1

Views

Author

A.H.M. Smeets, Mar 11 2020

Keywords

Comments

The denominators are given in A333209.
See A333088 and A333089 for similar fractions for infinite sums of Fibonacci numbers.
Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)) is a fraction for n > 0.
Sum_{i >= 0} 1/Lucas(i)^2 = 1/4 + A105394, i.e., the n = 0 case, is believed to be transcendental.

Examples

			These infinite sums begin: 1/2, 7/36, 551/7392, ...

Crossrefs

Cf. A000032, A105394, A132318, A333088, A333089, A333209.

Programs

  • Mathematica
    a[n_] := Numerator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* Amiram Eldar, Mar 11 2020 *)

Formula

a(n) = numerator of (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(Lucas(2i-2)*Lucas(2i-1)).
Lim_{n -> inf} (a(n)/A333209(n)) / (a(n-1)/A333209(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.
The following generalization holds: (Start)
Let H_(a,b) (n) be defined by H_(a,b) (0) = a, H_(a,b) (1) = b and H_(a,b) (n) = H_(a,b) (n-1) + H_(a,b) (n-2) for n > 1, then
Sum_{i >= 0} 1/(H_(a,b) (i)*H_(a,b) (i+2n)) = (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(H_(a,b) (2i-2)*H_(a,b) (2i-1)) for n > 0, and are thus all fractions. Specially, H_(0,1) are the Fibonacci numbers A000045, H_(2,1) as here, are the Lucas numbers A000032, and H_(3,1) are the Pibonacci numbers A104449. (End)
Showing 1-4 of 4 results.

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