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.

A156170 G.f.: exp( Sum_{n>=1} [Sum_{k>=1} k^n*x^k]^n/n ), a power series in x with integer coefficients.

Original entry on oeis.org

1, 1, 3, 10, 41, 219, 1602, 16635, 247171, 5242108, 157390565, 6663089873, 396778864166, 33200932308437, 3906922702271961, 646161881511137940, 150482521507292513413, 49318093291540113084965, 22790150225552744270503692, 14843990673285561887923674163, 13646527810852572644275538963207, 17710656073227095563348293151121448
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2009

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 41*x^4 + 219*x^5 + 1602*x^6 +...
log(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 +...+ A276750(n)*x^n/n +...
The logarithm of g.f. A(x) equals the series:
log(A(x)) = Sum_{n>=1} (x + 2^n*x^2 + 3^n*x^3 +...+ k^n*x^k +...)^n/n,
or,
log(A(x)) = (x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 +...) +
(x + 2^2*x^2 + 3^2*x^3 + 4^2*x^4 + 5^2*x^5 +...)^2/2 +
(x + 2^3*x^2 + 3^3*x^3 + 4^3*x^4 + 5^3*x^5 +...)^3/3 +
(x + 2^4*x^2 + 3^4*x^3 + 4^4*x^4 + 5^4*x^5 +...)^4/4 + ...
This logarithmic series can be written using the Eulerian numbers like so:
log(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^2+n)/n +...
		

Crossrefs

Programs

  • PARI
    {a(n) = polcoeff( exp( sum(m=1,n, sum(k=1,n, k^m*x^k +x*O(x^n))^m/m ) ),n)}
    for(n=0,30,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 = exp( sum(m=1,n+1, sum(k=1,m, A008292(m,k)*x^k/(1-x +Oxn)^(m+1) )^m / m ) ); polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: exp( Sum_{n>=1} [ Sum_{k=1..n} A008292(n,k) * x^k / (1-x)^(n+1) ]^n / n ), where A008292 are the Eulerian numbers. - Paul D. Hanna, Sep 13 2016
Conjecture: log(a(n)) ~ n^2 * log(2)/4. - Vaclav Kotesovec, Sep 02 2017

A276743 G.f.: Sum_{n>=0} [ Sum_{k>=1} k^n * x^k ]^n.

Original entry on oeis.org

1, 1, 3, 12, 63, 447, 4286, 56185, 1008317, 24917676, 849963761, 40142633815, 2633061525012, 240207555735097, 30578843349537575, 5434894746337720676, 1352812180415380719387, 471689727423751377883607, 230943183470327388401886858, 158839247095790148049487792081, 153694547774391577758847456894905
Offset: 0

Views

Author

Paul D. Hanna, Sep 29 2016

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 63*x^4 + 447*x^5 + 4286*x^6 + 56185*x^7 + 1008317*x^8 + 24917676*x^9 + 849963761*x^10 +...
such that
A(x) = Sum_{n>=0} (x + 2^n*x^2 + 3^n*x^3 +...+ k^n*x^k +...)^n.
Explicitly,
A(x) = 1 + (x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 +...) +
(x + 4*x^2 + 9*x^3 + 16*x^4 + 25*x^5 + 36*x^6 +...)^2 +
(x + 8*x^2 + 27*x^3 + 64*x^4 + 125*x^5 + 216*x^6 +...)^3 +
(x + 16*x^2 + 81*x^3 + 256*x^4 + 625*x^5 + 1296*x^6 +...)^4 +
(x + 32*x^2 + 243*x^3 + 1024*x^4 + 3125*x^5 + 7776*x^6 +...)^5 +
...
The g.f. can be written using the Eulerian numbers like so:
A(x) = 1 + x/(1-x)^2 + (x + x^2)^2/(1-x)^6 + (x + 4*x^2 + x^3)^3/(1-x)^12 + (x + 11*x^2 + 11*x^3 + x^4)^4/(1-x)^20 + (x + 26*x^2 + 66*x^3 + 26*x^4 + x^5)^5/(1-x)^30 + (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)^6/(1-x)^42 +...+ [ Sum_{k=1..n} A008292(n,k) * x^k ]^n / (1-x)^(n*(n+1)) +...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=1);
    A = sum(m=0,n+1, sum(k=1,n+1, k^m * x^k +x*O(x^n))^m); polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* Using Eulerian numbers A008292 */
    {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=0, n+1, sum(k=1, m, A008292(m, k)*x^k/(1-x +Oxn)^(m+1) )^m ); polcoeff(A, n)}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: Sum_{n>=0} [ Sum_{k=1..n} A008292(n,k) * x^k ]^n / (1-x)^(n*(n+1)), where A008292 are the Eulerian numbers.

A292502 L.g.f.: Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n / n.

Original entry on oeis.org

1, 7, 43, 399, 6091, 151255, 6550307, 465127199, 58293976795, 12191724780647, 4471204259257363, 2799295142330495151, 3026340345288168023883, 5704756586858875194533367, 18287793731664040419412785283, 103736521111190203113027053903423, 990788254951454647260121962687606203, 16859931481746848392491523248553253264263, 481447154976629475966161111465088882379644147
Offset: 1

Views

Author

Paul D. Hanna, Sep 17 2017

Keywords

Comments

A060187(n,k) = Sum_{j=1..k} (-1)^(k-j) * binomial(n,k-j) * (2*j-1)^(n-1).
L.g.f. equals the logarithm of the g.f. of A292500.
Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - Vaclav Kotesovec, Oct 17 2020

Examples

			L.g.f: A(x) = x + 7*x^2/2 + 43*x^3/3 + 399*x^4/4 + 6091*x^5/5 + 151255*x^6/6 + 6550307*x^7/7 + 465127199*x^8/8 + 58293976795*x^9/9 + 12191724780647*x^10/10 + 4471204259257363*x^11/11 + 2799295142330495151*x^12/12 + 3026340345288168023883*x^13/13 + 5704756586858875194533367*x^14/14 + ...
The l.g.f. A(x) equals the series:
A(x) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 + ... + (2*k-1)^n*x^k + ...)^n/n,
or,
A(x) = (x + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 + ...) +
(x + 3^2*x^2 + 5^2*x^3 + 7^2*x^4 + 9^2*x^5 + ...)^2/2 +
(x + 3^3*x^2 + 5^3*x^3 + 7^3*x^4 + 9^3*x^5 + ...)^3/3 +
(x + 3^4*x^2 + 5^4*x^3 + 7^4*x^4 + 9^4*x^5 + ...)^4/4 + ...
This logarithmic series can be written using the Eulerian numbers of type B like so:
A(x) = (x + x^2) / (1-x)^2 +
(x + 6*x^2 + x^3)^2 / (1-x)^6/2 +
(x + 23*x^2 + 23*x^3 + x^4)^3 / (1-x)^12/3 +
(x + 76*x^2 + 230*x^3 + 76*x^4 + x^5)^4 / (1-x)^20/4 +
(x + 237*x^2 + 1682*x^3 + 1682*x^4 + 237*x^5 + x^6)^5 / (1-x)^30/5 +
(x + 722*x^2 + 10543*x^3 + 23548*x^4 + 10543*x^5 + 722*x^6 + x^7)^6 / (1-x)^42/6 +
(x + 2179*x^2 + 60657*x^3 + 259723*x^4 + 259723*x^5 + 60657*x^6 + 2179*x^7 + x^8)^7 / (1-x)^56/7 + ... +
[ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n  / (1-x)^(n^2+n) * x^n/n + ...
Exponentiation yields the g.f. of A292500:
exp(A(x)) = 1 + x + 4*x^2 + 18*x^3 + 122*x^4 + 1382*x^5 + 26992*x^6 + 967860*x^7 + 59207134*x^8 + 6539607238*x^9 + 1225903048760*x^10 + 407719392472476*x^11 + 233686070341415140*x^12 + 233030334505100451484*x^13 + 407716349332865096406960*x^14 + ... + A292500(n)*x^n + ...
which is an integer series.
		

Crossrefs

Programs

  • PARI
    {A060187(n, k) = sum(j=1, k, (-1)^(k-j) * binomial(n, k-j) * (2*j-1)^(n-1))}
    {a(n) = my(A=1, Oxn=x*O(x^n));
    A = sum(m=1, n+1, sum(k=0, m, A060187(m+1, k+1)*x^k)^m /(1-x +Oxn)^(m^2+m) * x^m/m );
    n*polcoeff(A, n)}
    for(n=1, 30, print1(a(n), ", "))
    
  • PARI
    {a(n) = n*polcoeff( sum(m=1, n+1, sum(k=1, n, (2*k-1)^m * x^k +x*O(x^n))^m/m ), n)}
    for(n=1, 30, print1(a(n), ", "))

Formula

L.g.f.: Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2+n) * x^n/n, where A060187 are the Eulerian numbers of type B.

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 1-4 of 4 results.