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 1-2 of 2 results.

A276750 L.g.f.: Sum_{n>=1} [ Sum_{k>=1} k^n * x^k ]^n / n.

Original entry on oeis.org

1, 5, 22, 117, 821, 7796, 101417, 1810093, 44561794, 1515368605, 71428667861, 4677209119632, 426268582440013, 54220470799325101, 9632796180856419722, 2397253932245127919389, 835827069207839232602401, 409329501365419311969616628, 281600921299273941316256813501, 272632759803890415543364253988037, 371636574592049013061911521355729422, 713832787857018847209335427225631327093
Offset: 1

Views

Author

Paul D. Hanna, Sep 17 2016

Keywords

Comments

L.g.f. equals the logarithm of the g.f. of A156170.

Examples

			L.g.f.: A(x) = x + 5*x^2/2 + 22*x^3/3 + 117*x^4/4 + 821*x^5/5 + 7796*x^6/6 + 1810093*x^7/7 + 44561794*x^8/8 + 1515368605*x^9/9 + 71428667861*x^10/10 +...
such that A(x) equals the series:
A(x) = Sum_{n>=1} (x + 2^n*x^2 + 3^n*x^3 +...+ k^n*x^k +...)^n/n.
This logarithmic series can be written using the Eulerian numbers like so:
A(x) = x/(1-x)^2 + (x + x^2)^2/(1-x)^6/2 + (x + 4*x^2 + x^3)^3/(1-x)^12/3 + (x + 11*x^2 + 11*x^3 + x^4)^4/(1-x)^20/4 + (x + 26*x^2 + 66*x^3 + 26*x^4 + x^5)^5/(1-x)^30/5 + (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)^6/(1-x)^42/6 +...+ [ Sum_{k=1..n} A008292(n,k) * x^k ]^n / (1-x)^(n*(n+1))/n +...
where
exp(A(x)) = 1 + x + 3*x^2 + 10*x^3 + 41*x^4 + 219*x^5 + 1602*x^6 + 16635*x^7 + 247171*x^8 + 5242108*x^9 + 157390565*x^10 +...+ A156170(n)*x^n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = n * polcoeff( sum(m=1, n, sum(k=1, n, k^m*x^k +x*O(x^n))^m/m ), n)}
    for(n=1, 20, print1(a(n), ", "))
    
  • PARI
    {A008292(n, k) = sum(j=0, k, (-1)^j * (k-j)^n * binomial(n+1, j))}
    {a(n) = my(A=1, Oxn=x*O(x^n)); A = sum(m=1, n+1, sum(k=1, m, A008292(m, k)*x^k/(1-x +Oxn)^(m+1) )^m / m ); n*polcoeff(A, n)}
    for(n=1, 20, print1(a(n), ", "))

Formula

L.g.f.: Sum_{n>=1} [ Sum_{k=1..n} A008292(n,k) * x^k / (1-x)^(n+1) ]^n / n, where A008292 are the Eulerian numbers.

A276754 L.g.f.: Sum_{n>=1} [ Sum_{k>=1} k^(2*n) * x^k ]^n / n.

Original entry on oeis.org

1, 9, 76, 1157, 33291, 1792296, 196919213, 39766253741, 16931726147956, 13298466280839329, 22076711237844558263, 69166686377284889199104, 448760359479425463648647769, 5685081590883001302122022078913, 144528951819771627855280850227089996, 7431791795502279858136165452572662669213, 743200333842768450767851829731370148558347843, 154769006272445896954868694741314742556915451805336
Offset: 1

Views

Author

Paul D. Hanna, Sep 17 2016

Keywords

Comments

L.g.f. equals the logarithm of the g.f. of A276752.

Examples

			L.g.f.: A(x) = x + 9*x^2/2 + 76*x^3/3 + 1157*x^4/4 + 33291*x^5/5 + 1792296*x^6/6 + 196919213*x^7/7 + 39766253741*x^8/8 + 16931726147956*x^9/9 + 13298466280839329*x^10/10 +...
such that A(x) equals the series:
A(x) = Sum_{n>=1} (x + 2^(2*n)*x^2 + 3^(2*n)*x^3 +...+ k^(2*n)*x^k +...)^n/n.
This logarithmic series can be written using the Eulerian numbers like so:
A(x) = (x + x^2)/(1-x)^3 + (x + 11*x^2 + 11*x^3 + x^4)^2/(1-x)^10/2 + (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)^3/(1-x)^21/3 + (x + 247*x^2 + 4293*x^3 + 15619*x^4 + 15619*x^5 + 4293*x^6 + 247*x^7 + x^8)^4/(1-x)^36/4 + (x + 1013*x^2 + 47840*x^3 + 455192*x^4 + 1310354*x^5 + 1310354*x^6 + 455192*x^7 + 47840*x^8 + 1013*x^9 + x^10)^5/(1-x)^55/5 + (x + 4083*x^2 + 478271*x^3 + 10187685*x^4 + 66318474*x^5 + 162512286*x^6 + 162512286*x^7 + 66318474*x^8 + 10187685*x^9 + 478271*x^10 + 4083*x^11 + x^12)^6/(1-x)^78/6 +...+ [ Sum_{k=1..2*n} A008292(2*n,k) * x^k ]^n / (1-x)^(2*n^2+n) /n +...
where
exp(A(x)) = 1 + x + 5*x^2 + 30*x^3 + 327*x^4 + 7085*x^5 + 307280*x^6 + 28472653*x^7 + 5000661017*x^8 + 1886425568702*x^9 + 1331753751874235*x^10 +...+ A276752(n)*x^n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = n * polcoeff( sum(m=1, n, sum(k=1, n, k^(2*m)*x^k +x*O(x^n))^m/m ), n)}
    for(n=1, 20, print1(a(n), ", "))
    
  • PARI
    {A008292(n, k) = sum(j=0, k, (-1)^j * (k-j)^n * binomial(n+1, j))}
    {a(n) = my(A=1, Oxn=x*O(x^n)); A = sum(m=1, n+1, sum(k=1, 2*m, A008292(2*m, k)*x^k/(1-x +Oxn)^(2*m+1) )^m / m ); n * polcoeff(A, n)}
    for(n=1, 20, print1(a(n), ", "))

Formula

L.g.f.: Sum_{n>=1} [ Sum_{k=1..2*n-1} A008292(2*n,k) * x^k / (1-x)^(2*n+1) ]^n / n, where A008292 are the Eulerian numbers.
Showing 1-2 of 2 results.