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.

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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.

A276907 L.g.f.: Sum_{n>=1} [ Sum_{k>=1} k^n * x^(2*k-1) ]^n / n.

Original entry on oeis.org

1, 1, 7, 17, 56, 199, 890, 4649, 27817, 195946, 1684398, 17397323, 208799982, 2932164012, 49785808832, 1022745137705, 24671296028079, 695270673553051, 23526126768837873, 965093874912658722, 46827415587504280547, 2655503102769481320544, 179856174616910379655073, 14761130793635395568878091, 1439881917495260610082685956, 164363140573098989525137162900, 22322323085863965044351721067969
Offset: 1

Views

Author

Paul D. Hanna, Sep 28 2016

Keywords

Comments

L.g.f. equals the logarithm of the g.f. of A276906.

Examples

			L.g.f.: A(x) = x + x^2/2 + 7*x^3/3 + 17*x^4/4 + 56*x^5/5 + 199*x^6/6 + 890*x^7/7 + 4649*x^8/8 + 27817*x^9/9 + 195946*x^10/10 + 1684398*x^11/11 + 17397323*x^12/12 +...
such that A(x) equals the series:
A(x) = Sum_{n>=1} (x + 2^n*x^3 + 3^n*x^5 +...+ k^n*x^(2*k-1) +...)^n/n.
This logarithmic series can be written using the Eulerian numbers like so:
A(x) = x/(1-x^2)^2 + (x + x^3)^2/(1-x^2)^6/2 + (x + 4*x^3 + x^5)^3/(1-x^2)^12/3 + (x + 11*x^3 + 11*x^5 + x^7)^4/(1-x^2)^20/4 + (x + 26*x^3 + 66*x^5 + 26*x^7 + x^9)^5/(1-x^2)^30/5 + (x + 57*x^3 + 302*x^5 + 302*x^7 + 57*x^9 + x^11)^6/(1-x^2)^42/6 +...+ [ Sum_{k=1..n} A008292(n,k) * x^(2*k-1) ]^n / (1-x^2)^(n*(n+1))/n +...
where
exp(A(x)) = 1 + x + x^2 + 3*x^3 + 7*x^4 + 18*x^5 + 53*x^6 + 188*x^7 + 799*x^8 + 4001*x^9 + 24050*x^10 + 179248*x^11 + 1639637*x^12 +...+ A276906(n)*x^n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = n * polcoeff( sum(m=1, n, sum(k=1, n, k^m * x^(2*k-1) +x*O(x^n))^m/m ), n)}
    for(n=1, 30, 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^(2*k-1)/(1-x^2 +Oxn)^(m+1) )^m / m ); n*polcoeff(A, n)}
    for(n=1, 30, print1(a(n), ", "))

Formula

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