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.

A173217 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2) / 2^(n+1).

Original entry on oeis.org

1, 3, 36, 744, 21606, 807912, 36948912, 1997801520, 124666314300, 8817945612300, 697162848757056, 60925366551278592, 5831682410241684192, 606763511537812563648, 68184018356901256320192, 8229830886505821175612416, 1061871008421711265790015880, 145851902823090076435152800208, 21247730059665104564252809209792
Offset: 0

Views

Author

Paul D. Hanna, Mar 05 2010

Keywords

Comments

Variant of A104209, which enumerates labeled directed multigraphs.
Number of labeled digraphs with n edges and no vertices of degree zero, in which loops are permitted but not duplicate edges. - David Bevan, Apr 22 2013

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[k^2, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *)
    Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -2*j, 0]/2, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *)
  • PARI
    {a(n)=local(A=sum(m=0,n^2+100,(1+x +O(x^(n+2)))^(m^2)/2^(m+1)));round(polcoeff(A,n))}
    
  • PARI
    /* Continued fraction expression: */
    {a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(2 - q^(4*n-4*k+1)/(1 - q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 18 2018

Formula

G.f.: 1/(2 - q/(1 - q*(q^2-1)/(2 - q^5/(1 - q^3*(q^4-1)/(2 - q^9/(1 - q^5*(q^6-1)/(2 - q^13/(1 - q^7*(q^8-1)/(2 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity. - Paul D. Hanna, Mar 18 2018
G.f.: Sum_{n>=0} 1/2^(n+1) * (1+x)^n * Product_{k=1..n} (2 - (1+x)^(4*k-3)) / (2 - (1+x)^(4*k-1)), due to a q-series identity. - Paul D. Hanna, Mar 18 2018
a(n) ~ 2^(2*n - 1/2 - log(2)/8) * n^n / (exp(n) * log(2)^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

A303920 G.f.: A(x,y) = (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2).

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 6, 6, 0, 0, 0, 4, 56, 56, 4, 0, 0, 0, 1, 117, 722, 722, 117, 1, 0, 0, 0, 0, 126, 2982, 12012, 12012, 2982, 126, 0, 0, 0, 0, 0, 84, 6916, 79548, 246092, 246092, 79548, 6916, 84, 0, 0, 0, 0, 0, 36, 10900, 312880, 2322000, 6002824, 6002824, 2322000, 312880, 10900, 36, 0, 0, 0, 0, 0, 9, 12717, 864009, 13617765, 74916306, 170048394, 170048394, 74916306, 13617765, 864009, 12717, 9, 0, 0, 0, 0, 0, 1, 11421, 1825786, 57282026, 604000555, 2669115383, 5489377628, 5489377628, 2669115383, 604000555, 57282026, 1825786, 11421, 1, 0, 0
Offset: 0

Views

Author

Paul D. Hanna, May 02 2018

Keywords

Comments

G.f. A(x,y) = Sum_{n>=0} Sum_{k=0..2*n} T(n,k) * x^n*y^k, where T(n,k) is the term of this triangle at position k in row n.

Examples

			G.f.: A(x,y) = (1-y) * ( 1 + y*(1 + x*(1-y)^2) + y^2*(1 + x*(1-y)^2)^4 + y^3*(1 + x*(1-y)^2)^9 + y^4*(1 + x*(1-y)^2)^16 + y^5*(1 + x*(1-y)^2)^25 + ... ).
Explicitly,
A(x,y) = 1 + x*(y + y^2) + x^2*(6*y^2 + 6*y^3) + x^3*(4*y^2 + 56*y^3 + 56*y^4 + 4*y^5) + x^4*(y^2 + 117*y^3 + 722*y^4 + 722*y^5 + 117*y^6 + y^7) + x^5*(126*y^3 + 2982*y^4 + 12012*y^5 + 12012*y^6 + 2982*y^7 + 126*y^8) + x^6*(84*y^3 + 6916*y^4 + 79548*y^5 + 246092*y^6 + 246092*y^7 + 79548*y^8 + 6916*y^9 + 84*y^10) + x^7*(36*y^3 + 10900*y^4 + 312880*y^5 + 2322000*y^6 + 6002824*y^7 + 6002824*y^8 + 2322000*y^9 + 312880*y^10 + 10900*y^11 + 36*y^12) + x^8*(9*y^3 + 12717*y^4 + 864009*y^5 + 13617765*y^6 + 74916306*y^7 + 170048394*y^8 + 170048394*y^9 + 74916306*y^10 + 13617765*y^11 + 864009*y^12 + 12717*y^13 + 9*y^14) + x^9*(y^3 + 11421*y^4 + 1825786*y^5 + 57282026*y^6 + 604000555*y^7 + 2669115383*y^8 + 5489377628*y^9 + 5489377628*y^10 + 2669115383*y^11 + 604000555*y^12 + 57282026*y^13 + 1825786*y^14 + 11421*y^15 + y^16) + ...
This triangle begins:
[1];
[0, 1, 1];
[0, 0, 6, 6, 0];
[0, 0, 4, 56, 56, 4, 0];
[0, 0, 1, 117, 722, 722, 117, 1, 0];
[0, 0, 0, 126, 2982, 12012, 12012, 2982, 126, 0, 0];
[0, 0, 0, 84, 6916, 79548, 246092, 246092, 79548, 6916, 84, 0, 0];
[0, 0, 0, 36, 10900, 312880, 2322000, 6002824, 6002824, 2322000, 312880, 10900, 36, 0, 0];
[0, 0, 0, 9, 12717, 864009, 13617765, 74916306, 170048394, 170048394, 74916306, 13617765, 864009, 12717, 9, 0, 0];
[0, 0, 0, 1, 11421, 1825786, 57282026, 604000555, 2669115383, 5489377628, 5489377628, 2669115383, 604000555, 57282026, 1825786, 11421, 1, 0, 0]; ...
		

Crossrefs

Cf. A303921 (diagonal), A303922 (column sums), A001813 (row sums), A265936 (y=2), A173217.

Programs

  • PARI
    /* G.f. by Definition: */
    {T(n,k) = my(A = (1-y) * sum(m=0,2*n, y^m * (1 + x*(1-y)^2  +x*O(x^(2*n)) )^(m^2))); polcoeff(polcoeff(A, n,x),k,y)}
    for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))
    
  • PARI
    /* Continued fraction expression: */
    {T(n,k) = my(CF=1, q = 1 + x*(1-y)^2 +x*O(x^(2*n))); for(k=0, n, CF = 1/(1 - q^(4*n-4*k+1)*y/(1 - q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*y*CF)) ); polcoeff(polcoeff((1-y)*CF, n,x),k,y)}
    for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))
    
  • PARI
    /* G.f. by q-series identity: */
    {T(n,k) = my(A =1, q = 1 + x*(1-y)^2 +x*O(x^(2*n))); A = (1-y) * sum(m=0,2*n, y^m*q^m * prod(k=1,m, (1 - y*q^(4*k-3)) / (1 - y*q^(4*k-1) +x*O(x^(2*n))) )); polcoeff(polcoeff(A, n,x),k,y)}
    for(n=0, 10, for(k=0,2*n, print1(T(n,k), ", ")); print(""))

Formula

GENERATING FUNCTIONS.
(1) A(x,y) = (1-y) * Sum_{n>=0} y^n * (1 + x*(1-y)^2)^(n^2).
(2) A(x,y) = (1-y) * Sum_{n>=0} y^n * q^n * Product_{k=1..n} (1 - q^(4*k-3)*y) / (1 - q^(4*k-1)*y), where q = 1 + x*(1-y)^2, due to a q-series identity.
(3) A(x,y) = (1-y)/(1 - q*y/(1 - q*(q^2-1)*y/(1 - q^5*y/(1 - q^3*(q^4-1)*y/(1 - q^9*y/(1- q^5*(q^6-1)*y/(1 - q^13*y/(1 - q^7*(q^8-1)*y/(1 - ...))))))))), where q = 1 + x*(1-y)^2, a continued fraction due to an identity of a partial elliptic theta function.
FORMULAS INVOLVING TERMS.
Sum_{k=0..2*n} T(n,k) = (2*n)!/n!, for n>=0 (row sums = A001813).
Sum_{k=0..2*n} T(n,k) * (-1)^k = 0, for n>=1 (symmetric rows).
Sum_{k=0..2*n} T(n,k) * 2^k = A265936(n), for n>=1.
Sum_{k=0..2*n} T(n,k) / 2^k = A173217(n) / 4^n, for n>=0.
Sum_{j=0..k^2} T(j,k) = A303922(k), for k>=0 (column sums).
T(n,n) = A303921(n), for n>=0 (diagonal).

A265937 G.f.: Sum_{n>=0} (1 + x)^(n*(n+1)/2) / 2^n.

Original entry on oeis.org

2, 4, 24, 248, 3600, 67296, 1538672, 41593920, 1297683360, 45891815040, 1814072216864, 79263667304640, 3793393788125760, 197339219789611200, 11087608251010390080, 669127189486395204544, 43167108189991530605184, 2964541208087967215725440, 215934869210274766223069440, 16627513858173093851116296960, 1349582577808759197056647917696, 115158206188199564942934814336896, 10305721256666828267464573643658240
Offset: 0

Views

Author

Paul D. Hanna, Dec 23 2015

Keywords

Examples

			G.f.: A(x) = 2 + 4*x + 24*x^2 + 248*x^3 + 3600*x^4 + 67296*x^5 + 1538672*x^6 + 41593920*x^7 + 1297683360*x^8 + 45891815040*x^9 + 1814072216864*x^10 +...
where
A(x) = 1 + (1+x)/2 + (1+x)^3/2^2 + (1+x)^6/2^3 + (1+x)^10/2^4 + (1+x)^15/2^5 + (1+x)^21/2^6 + (1+x)^28/2^7 + (1+x)^36/2^8 +...+ (1+x)^(n*(n+1)/2)/2^n +...
		

Crossrefs

Cf. A265936.

Programs

  • Mathematica
    Table[Sum[StirlingS1[n, j] * Sum[Binomial[j, s] * HurwitzLerchPhi[1/2, -j - s, 0], {s, 0, j}] / 2^j, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2019 *)
  • PARI
    /* Informal listing of terms: */
    {Vec( round( sum(n=0, 600, (1+x +O(x^31))^(n*(n+1)/2)/2^n * 1.) ) )}
    {Vec( round( sum(n=0, 200, (1.+x)^n/2^n * prod(k=1, n, (2 - (1+x)^(2*k-1)) / (2 - (1+x)^(2*k)) +O(x^21) ) ) ) )}

Formula

G.f.: Sum_{n>=0} (1+x)^n/2^n * Product_{k=1..n} (2 - (1+x)^(2*k-1))/(2 - (1+x)^(2*k)) due to a q-series identity.
G.f.: 1/(1 - (1+x)/2 /(1 - (1+x)*((1+x)-1)/2 /(1 - (1+x)^3/2 /(1 - (1+x)^2*((1+x)^2-1)/2 /(1 - (1+x)^5/2 /(1 - (1+x)^3*((1+x)^3-1)/2 /(1 - (1+x)^7/2 /(1 - (1+x)^4*((1+x)^4-1)/2 /(1 - ...))))))))), a continued fraction due to a partial elliptic theta function identity.
a(n) = Sum_{k>=(sqrt(8*n+1)-1)/2} binomial(k*(k+1)/2,n) / 2^k.
a(n) = 2*A173219(n). - Vaclav Kotesovec, Oct 08 2019
a(n) ~ 2^(n+1) * n^n / (2^(log(2)/4) * log(2)^(2*n+1) * exp(n)). - Vaclav Kotesovec, Oct 08 2019
Showing 1-3 of 3 results.