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.

A299434 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n / A(x)^((n+1)^2).

Original entry on oeis.org

1, 1, 1, 6, 77, 1451, 35730, 1082481, 38913817, 1619979291, 76724619427, 4077896446598, 240566693095072, 15609120639706252, 1105414601508493001, 84881459931003622118, 7026832554316541379141, 624014794413319426058889, 59184228450018585954486975, 5971678912361406721742217080, 638782082648832471805820934833, 72213308562202419209594988387550, 8603323896642095980014195130664418
Offset: 0

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Author

Paul D. Hanna, Feb 13 2018

Keywords

Comments

Compare to: 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.

Examples

			G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 77*x^4 + 1451*x^5 + 35730*x^6 + 1082481*x^7 + 38913817*x^8 + 1619979291*x^9 + 76724619427*x^10 +...
such that
1 = 1/A(x) + C(4,1)/4*x/A(x)^4 + C(9,2)/9*x^2/A(x)^9 + C(16,3)/16*x^3/A(x)^16 + C(25,4)/25*x^4/A(x)^25 + C(36,5)/36*x^5/A(x)^36 + C(49,6)/49*x^6/A(x)^49 + ...
more explicitly,
1 = 1/A(x) + x/A(x)^4 + 4*x^2/A(x)^9 + 35*x^3/A(x)^16 + 506*x^4/A(x)^25 + 10472*x^5/A(x)^36 + 285384*x^6/A(x)^49 + ... + A143669(n)*x^n/A(x)^((n+1)^2) + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A = Vec(sum(n=0,#A,binomial((n+1)^2,n)/(n+1)^2 * x^n/Ser(A)^((n+1)^2-1) )));G=Ser(A);A[n+1]}
    for(n=0,30,print1(a(n),", "))

A298693 G.f. A(x) satisfies: A(x) = Sum_{n>=0} binomial( n*(n+1), n)/(n+1) * x^n / A(x)^( n*(n+1) ).

Original entry on oeis.org

1, 1, 3, 22, 294, 5911, 158293, 5251690, 206696194, 9387611937, 482745371458, 27717788095397, 1757818683339028, 122058148921357056, 9212494564360610855, 751138761646263512978, 65807775099574132000968, 6166278653572358495161057, 615421469545011786309942067, 65183859793912213778457542207, 7303117991652113167690085149033
Offset: 0

Views

Author

Paul D. Hanna, Feb 03 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 22*x^3 + 294*x^4 + 5911*x^5 + 158293*x^6 + 5251690*x^7 + 206696194*x^8 + 9387611937*x^9 + 482745371458*x^10 + 27717788095397*x^11 + 1757818683339028*x^12 + 122058148921357056*x^13 + 9212494564360610855*x^14 + 751138761646263512978*x^15 + ...
such that
A(x) = 1 + C(2,1)/2*x/A(x)^2 + C(6,2)/3*x^2/A(x)^6 + C(12,3)/4*x^3/A(x)^12 + C(20,4)/5*x^4/A(x)^20 + C(30,5)/6*x^5/A(x)^30 + C(42,6)/7*x^6/A(x)^42 + C(56,7)/8*x^7/A(x)^56 + ...
more explicitly,
A(x) = 1 + x/A(x)^2 + 5*x^2/A(x)^6 + 55*x^3/A(x)^12 + 969*x^4/A(x)^20 + 23751*x^5/A(x)^30 + 749398*x^6/A(x)^42 + 28989675*x^7/A(x)^56 + ... + A135861(n)*x^n/A(x)^(n*(n+1)) + ...
		

Crossrefs

Programs

  • Mathematica
    terms = 21; A[] = 1; Do[A[x] = 1 + Sum[Binomial[n*(n+1), n]/(n+1)*x^n/ A[x]^(n*(n+1)), {n, terms}] + O[x]^terms, {terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Feb 09 2018 *)
  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A = Vec(sum(m=0,#A,binomial(m*(m+1),m)/(m+1) * x^m/Ser(A)^(m*(m+1)) )));  A[n+1]}
    for(n=0,20,print1(a(n),", "))

A298694 G.f. A(x) satisfies: A(x) = Sum_{n>=0} binomial( n*(n+1), n)/(n+1) * x^n / A(x)^(n^2).

Original entry on oeis.org

1, 1, 4, 32, 419, 8052, 207784, 6724274, 260396693, 11697865930, 596886780272, 34072732137625, 2151062784054901, 148819021611467291, 11198412956841549966, 910736443741061568539, 79616310026220269203631, 7446056807577515910468813, 741918566779386113373532994, 78467177619239380045368550016, 8779922184077661414128958823323
Offset: 0

Views

Author

Paul D. Hanna, Feb 03 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 32*x^3 + 419*x^4 + 8052*x^5 + 207784*x^6 + 6724274*x^7 + 260396693*x^8 + 11697865930*x^9 + 596886780272*x^10 + 34072732137625*x^11 + 2151062784054901*x^12 + 148819021611467291*x^13 + 11198412956841549966*x^14 + 910736443741061568539*x^15 + ...
such that
A(x) = 1 + C(2,1)/2*x/A(x) + C(6,2)/3*x^2/A(x)^4 + C(12,3)/4*x^3/A(x)^9 + C(20,4)/5*x^4/A(x)^16 + C(30,5)/6*x^5/A(x)^25 + C(42,6)/7*x^6/A(x)^36 + C(56,7)/8*x^7/A(x)^49 + ...
more explicitly,
A(x) = 1 + x/A(x) + 5*x^2/A(x)^4 + 55*x^3/A(x)^9 + 969*x^4/A(x)^16 + 23751*x^5/A(x)^25 + 749398*x^6/A(x)^36 + 28989675*x^7/A(x)^49 + ... + A135861(n)*x^n/A(x)^(n^2) + ...
		

Crossrefs

Programs

  • Mathematica
    terms = 21; A[] = 1; Do[A[x] = 1 + Sum[Binomial[n*(n+1), n]/(n+1)*x^n/ A[x]^(n^2), {n, terms}] + O[x]^terms, {terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Feb 09 2018 *)
  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A = Vec(sum(m=0, #A, binomial(m*(m+1), m)/(m+1) * x^m/Ser(A)^(m^2) ))); A[n+1]}
    for(n=0,20,print1(a(n),", "))
Showing 1-3 of 3 results.