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.

A207969 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^4 * x^n/n ).

Original entry on oeis.org

1, 5, 15, 60, 295, 1625, 9430, 56465, 345010, 2139595, 13419500, 84926105, 541398665, 3472389210, 22385362895, 144945232375, 942089445030, 6143582084115, 40181143112035, 263482860974570, 1731780213622125, 11406235045261205, 75268685723935940
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2012

Keywords

Comments

Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k>=0.
Note that exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k+1) * x^n/n ) is not an integer series for integers k.
Note that exp( Sum_{n>=1} Fibonacci(n)^(2*k) * x^n/n ) is not an integer series for integers k.

Examples

			G.f.: A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + 295*x^4 + 1625*x^5 + 9430*x^6 +...
such that
log(A(x))/5 = x + x^2/2 + 2^4*x^3/3 + 3^4*x^4/4 + 5^4*x^5/5 + 8^4*x^6/6 + 13^4*x^7/7 +...+ Fibonacci(n)^4*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^4*x^k/k)+x*O(x^n)),n)}
    for(n=0,25,print1(a(n),", "))

Formula

The o.g.f. A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + ... is an algebraic function: A(x)^5 = (1 + 3*x + x^2)^4/( (1 - 7*x + x^2)*(1 - 2*x + x^2)^3 ). Cf. A203804. - Peter Bala, Apr 03 2014
a(n) ~ 2^(4/5) * 5^(1/10) * phi^(4*n) / (Gamma(1/5) * 3^(1/5) * n^(4/5)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 18 2020

A207970 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^6 * x^n/n ).

Original entry on oeis.org

1, 5, 15, 140, 1505, 21875, 319620, 4936985, 77358485, 1236083870, 19982821875, 326511608255, 5379199407890, 89249496596015, 1489580814490755, 24988546214618750, 421055477328447620, 7122346563647277860, 120891417096833214485, 2058225554792946621495
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2012

Keywords

Comments

Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k >= 0.
Note that exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k+1) * x^n/n ) is not an integer series for integers k.
Note that exp( Sum_{n>=1} Fibonacci(n)^(2*k) * x^n/n ) is not an integer series for integers k.

Examples

			G.f.: A(x) = 1 + 5*x + 15*x^2 + 140*x^3 + 1505*x^4 + 21875*x^5 + 319620*x^6 + ...
such that
log(A(x))/5 = x + x^2/2 + 2^6*x^3/3 + 3^6*x^4/4 + 5^6*x^5/5 + 8^6*x^6/6 + 13^6*x^7/7 + ... + Fibonacci(n)^6*x^n/n + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^6*x^k/k)+x*O(x^n)),n)}
    for(n=0,31,print1(a(n),", "))

Formula

The o.g.f. A(x) = 1 + 5*x + 15*x^2 + 140*x^3 + ... is an algebraic function: A(x)^25 = ( (1 + 2*x + x^2)^10*(1 + 7*x + x^2)^6 )/( (1 - 3*x + x^2)^15*(1 - 18*x + x^2) ). Cf. A203806. - Peter Bala, Apr 03 2014
a(n) ~ 2^(17/25) * 5^(13/50) * phi^(6*n) / (Gamma(1/25) * 3^(3/5) * n^(24/25)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 18 2020

A207972 Expansion of g.f.: exp( Sum_{n>=1} 5*Fibonacci(n^2) * x^n/n ).

Original entry on oeis.org

1, 5, 20, 115, 1665, 82650, 12847310, 5620114060, 6659421195205, 21082748688390045, 177217804775828062850, 3941798437750184226876305, 231505293200405380457355524620, 35848160499603817968830380832049915, 14619744406297572472084577939841875791890
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2012

Keywords

Comments

Moss and Ward prove that this is an integral sequence. - Peter Bala, Nov 28 2022
Let A(x) be the g.f. for this sequence. Note that the expansion of A(x)^(1/5) = exp( Sum_{n>=1} Fibonacci(n^2) * x^n/n ) does not have integer coefficients.

Examples

			G.f.: A(x) = 1 + 5*x + 20*x^2 + 115*x^3 + 1665*x^4 + 82650*x^5 + ...
such that
log(A(x))/5 = x + 3*x^2/2 + 34*x^3/3 + 987*x^4/4 + 75025*x^5/5 + 14930352*x^6/6 + 7778742049*x^7/7 + ... + Fibonacci(n^2)*x^n/n + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k^2)*x^k/k)+x*O(x^n)),n)}
    for(n=0,16,print1(a(n),", "))

A207835 G.f.: exp( Sum_{n>=1} 5*L(n)*x^n/n ), where L(n) = Fibonacci((n-1)^2) + Fibonacci((n+1)^2).

Original entry on oeis.org

1, 15, 200, 3525, 134355, 16781664, 6730280105, 7679335074975, 23795707614699850, 197148338964056588955, 4337960355881995023988299, 252594793852565664429620014530, 38838042059493582778244565420563025, 15744729667082405326504405819215652913325
Offset: 0

Views

Author

Paul D. Hanna, Feb 20 2012

Keywords

Comments

Given g.f. A(x), note that A(x)^(1/5) is not an integer series.
Compare the definition to the g.f. of the Fibonacci numbers:
1/(1-x-x^2) = exp( Sum_{n>=1} Lucas(n)*x^n/n ), where Lucas(n) = Fibonacci(n-1) + Fibonacci(n+1).

Examples

			G.f.: A(x) = 1 + 15*x + 200*x^2 + 3525*x^3 + 134355*x^4 + 16781664*x^5 +...
such that, by definition,
log(A(x))/5 = 3*x + 35*x^2/2 + 990*x^3/3 + 75059*x^4/4 + 14931339*x^5/5 + 7778817074*x^6/6 +...+ (Fibonacci((n-1)^2) + Fibonacci((n+1)^2))*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {L(n)=fibonacci((n-1)^2)+fibonacci((n+1)^2)}
    {a(n)=polcoeff(exp(sum(m=1,n,5*L(m)*x^m/m)+x*O(x^n)),n)}
    for(n=0,21,print1(a(n),", "))

A207971 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*n) * x^n/n ).

Original entry on oeis.org

1, 5, 15, 140, 8795, 9808325, 57315191130, 2812698182891585, 894119494320160426760, 2048089587570930007354766745, 32079229816919862900907520464756250, 3500720882833094608324749707338857577696205, 2633228648869966875007549667526201212159637714889015
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2012

Keywords

Comments

Give g.f. A(x), note that A(x)^(1/5) is not an integer series.

Examples

			G.f.: A(x) = 1 + 5*x + 15*x^2 + 140*x^3 + 8795*x^4 + 9808325*x^5 +...
such that
log(A(x))/5 = x + x^2/2 + 2^6*x^3/3 + 3^8*x^4/4 + 5^10*x^5/5 + 8^12*x^6/6 + 13^14*x^7/7 +...+ Fibonacci(n)^(2*n)*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^(2*k)*x^k/k)+x*O(x^n)),n)}
    for(n=0,31,print1(a(n),", "))
Showing 1-5 of 5 results.