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 1-3 of 3 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 +...

Links

Crossrefs

Cf. A276750, A276751, A276752, A156171, A155200, A008292, A292500.

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

A292500 G.f.: exp( Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n / n ).

Original entry on oeis.org

1, 1, 4, 18, 122, 1382, 26992, 967860, 59207134, 6539607238, 1225903048760, 407719392472476, 233686070341415140, 233030334505100451484, 407716349332865096406960, 1219594666823043463552070760, 6484753389847998264537623184230, 58288150472645787928029816422705798, 936721167715228772497787011017302901192, 25340260842241991639562678352357479545874188
Offset: 0

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).
Note that exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ] / (1-x)^(n+1) * x^n/n ) does not yield an integer series.
Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - Vaclav Kotesovec, Oct 17 2020

Examples

			G.f.: 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 + 1219594666823043463552070760*x^15 +...
RELATED SERIES.
log(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 +...+ A292502(n)*x^n/n +...
The logarithm of g.f. A(x) equals the series:
log(A(x)) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 +...+ (2*k-1)^n*x^k +...)^n/n,
or,
log(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:
log(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 +...

Links

Crossrefs

Cf. A292501, A292502 (log), A156170, A276751, A276752, A156171, A155200, A060187.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[Sum[2^(k^2) * x^k * LerchPhi[x, -k, 1/2]^k/k, {k, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 17 2020 *)
  • PARI
    {a(n) = polcoeff( exp( sum(m=1, n+1, sum(k=1, n+1, (2*k-1)^m * x^k +x*O(x^n))^m/m ) ), n)}
for(n=0, 30, print1(a(n), ", "))
  • 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 = exp( 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 ) );
polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: exp( 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.

A292501 G.f.: exp( Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n * (1-x)^n / n ).

Original entry on oeis.org

1, 1, 3, 13, 91, 1119, 23235, 879361, 55447631, 6274018595, 1192773105789, 400761393446831, 231147252957096671, 231434829013884972151, 406000810484101907916927, 1216355994930424625967455929, 6474418584620388915674215696687, 58229572245447428847208518694227279, 936163501254507409972001699357677028097, 25330794407893091120626418701416294765820223, 1224635875718403110628189182372406488768960029317
Offset: 0

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).
Note that exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ] / (1-x)^n * x^n/ n ) does not yield an integer series.
Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - Vaclav Kotesovec, Oct 17 2020

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 91*x^4 + 1119*x^5 + 23235*x^6 + 879361*x^7 + 55447631*x^8 + 6274018595*x^9 + 1192773105789*x^10 + 400761393446831*x^11 + 231147252957096671*x^12 + 231434829013884972151*x^13 + 406000810484101907916927*x^14 + 1216355994930424625967455929*x^15 +...
RELATED SERIES.
log(A(x)) = x + 5*x^2/2 + 31*x^3/3 + 305*x^4/4 + 5041*x^5/5 + 131477*x^6/6 + 5973311*x^7/7 + 436089793*x^8/8 + 55949083681*x^9/9 + 11863792842885*x^10/10 + 4395111080551775*x^11/11 + 2768928615166879025*x^12/12 + 3005637312940054635857*x^13/13 + 5680764740993004611483477*x^14/14 + 18239242940612856315412499071*x^15/15 +...
The logarithm of g.f. A(x) equals the series:
log(A(x)) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 +...+ (2*k-1)^n*x^k +...)^n * (1-x)^n/n,
or,
log(A(x)) = (x + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 +...) * (1-x) +
(x + 3^2*x^2 + 5^2*x^3 + 7^2*x^4 + 9^2*x^5 +...)^2 * (1-x)^2/2 +
(x + 3^3*x^2 + 5^3*x^3 + 7^3*x^4 + 9^3*x^5 +...)^3 * (1-x)^3/3 +
(x + 3^4*x^2 + 5^4*x^3 + 7^4*x^4 + 9^4*x^5 +...)^4 * (1-x)^4/4 + ...
This logarithmic series can be written using the Eulerian numbers of type B like so:
log(A(x)) = (x + x^2) / (1-x) +
(x + 6*x^2 + x^3)^2 / (1-x)^4/2 +
(x + 23*x^2 + 23*x^3 + x^4)^3 / (1-x)^9/3 +
(x + 76*x^2 + 230*x^3 + 76*x^4 + x^5)^4 / (1-x)^16/4 +
(x + 237*x^2 + 1682*x^3 + 1682*x^4 + 237*x^5 + x^6)^5 / (1-x)^25/5 +
(x + 722*x^2 + 10543*x^3 + 23548*x^4 + 10543*x^5 + 722*x^6 + x^7)^6 / (1-x)^36/6 +
(x + 2179*x^2 + 60657*x^3 + 259723*x^4 + 259723*x^5 + 60657*x^6 + 2179*x^7 + x^8)^7 / (1-x)^49/7 +...+
[ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n  / (1-x)^(n^2) * x^n/n +...

Links

Crossrefs

Cf. A292500, A156170, A276751, A276752, A156171, A155200, A060187.

Programs

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

Formula

G.f.: exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2) * x^n/n ), where A060187 are the Eulerian numbers of type B.
Showing 1-3 of 3 results.

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