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.

A135861 a(n) = binomial(n*(n+1),n)/(n+1).

Original entry on oeis.org

1, 1, 5, 55, 969, 23751, 749398, 28989675, 1329890705, 70625252863, 4263421511271, 288417894029200, 21616536107173175, 1778197364075525550, 159297460456229992380, 15438280311293473537331, 1609484153977526457766689, 179612918129148904884024975
Offset: 0

Views

Author

Paul D. Hanna, Dec 02 2007

Keywords

Links

Crossrefs

Cf. A014068, A107863, A135860, A135862, A295763.

Programs

  • Maple
    A135861:=n->binomial(n*(n+1),n)/(n+1); seq(A135861(n), n=0..15); # Wesley Ivan Hurt, May 08 2014
  • Mathematica
    Table[Binomial[n*(n + 1), n]/(n + 1), {n, 0, 15}]
  • PARI
    a(n)=binomial(n*(n+1),n)/(n+1)

Formula

a(n) = A135860(n)/(n+1).
a(n) = [x^(n^2)] 1/(1 - x)^n. - Ilya Gutkovskiy, Oct 10 2017
a(p) == 1 ( mod p^4 ) for prime p >= 5 and a(2*p) == 4*p + 1 ( mod p^4 ) for prime p >= 5 (apply Mestrovic, equation 37). - Peter Bala, Feb 23 2020
a(n) ~ exp(n + 1/2) * n^(n - 3/2) / sqrt(2*Pi). - Vaclav Kotesovec, Oct 17 2020

A295765 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n/A(x)^n.

Original entry on oeis.org

1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, 15487290535, 859400072837, 53749578759526, 3743585586509849, 287496351622105328, 24143937833744911767, 2201703647718624364913, 216700738558116024114289, 22900073562659910815354339, 2586409916780162599516986945, 310947096149155992699450689912, 39650252031533561961437812566315
Offset: 0

Views

Author

Paul D. Hanna, Jan 06 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 25*x^3 + 369*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + 859400072837*x^10 + ...
such that
A(x) = 1 + x/A(x) + 4*(x/A(x))^2 + 35*(x/A(x))^3 + 506*(x/A(x))^4 + 10472*(x/A(x))^5 + 285384*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)^2*(x/A(x))^n + ...
RELATED SERIES.
Define B(x) = A(x*B(x)) and A(x) = B(x/A(x)) then B(x) begins
B(x) = 1 + x + 4*x^2 + 35*x^3 + 506*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + ... + binomial((n+1)^2,n)/(n+1)^2*x^n + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, ...];
  [1, 2, 7, 56, 797, 16650, 460291, 15862152, 655825337, ...];
  [1, 3, 12, 94, 1293, 26409, 719922, 24587202, 1010428347, ...];
  [1, 4, 18, 140, 1867, 37272, 1001476, 33887832, 1384043365, ...];
  [1, 5, 25, 195, 2530, 49366, 1306860, 43802060, 1777652015, ...];
  [1, 6, 33, 260, 3294, 62832, 1638166, 54370836, 2192294775, ...];
  [1, 7, 42, 336, 4172, 77826, 1997688, 65638294, 2629075183, ...];
  [1, 8, 52, 424, 5178, 94520, 2387940, 77652024, 3089164371, ...];
  [1, 9, 63, 525, 6327, 113103, 2811675, 90463365, 3573805950, ...]; ...
in which the main diagonal begins:
  [1, 2, 12, 140, 2530, 62832, 1997688, ..., binomial((n+1)^2,n)/(n+1), ...].

Links

Crossrefs

cf. A295764, A295763, A143669.

Programs

  • Mathematica
    terms = 21; A[] = 1; Do[A[x] = Sum[Binomial[(n+1)^2, n]/(n+1)^2*x^n/ A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)
  • PARI
    {a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m)/(m+1) - V[m+1])/(m+1);); A[n+1]}
for(n=0,20,print1(a(n),", "))

Formula

G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n)/(n+1) for n>=0.
a(n) ~ c * exp(n) * n^(n - 5/2), where c = exp(3/2 - exp(-2)) / sqrt(2*Pi) = 1.56162380971247949723297... - Vaclav Kotesovec, Oct 17 2020, updated Apr 20 2024

A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

Original entry on oeis.org

1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, 130495143060, 8002598818976, 548083639109204, 41481408076217480, 3439242379130298624, 310088460754461374608, 30213233395705566500248, 3164041937175571224372118, 354458174755809132668847604, 42301583754989091491716706272, 5358139267082652649000120167456
Offset: 0

Views

Author

Paul D. Hanna, Jan 06 2018

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 84*x^3 + 1522*x^4 + 38940*x^5 + 1278880*x^6 + 51136512*x^7 + 2407921070*x^8 + 130495143060*x^9 + 8002598818976*x^10 +...
such that
A(x) = 1 + 2*(x/A(x)) + 12*(x/A(x))^2 + 140*(x/A(x))^3 + 2530*(x/A(x))^4 + 62832*(x/A(x))^5 + 1997688*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)*(x/A(x))^n +...
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, ...];
  [1, 4, 20, 200, 3444, 85312, 2744928, 108267280, 5049708672, ...];
  [1, 6, 36, 356, 5862, 140508, 4424984, 172064160, 7946443350, ...];
  [1, 8, 56, 560, 8888, 206176, 6350112, 243284064, 11121338640, ...];
  [1, 10, 80, 820, 12650, 284252, 8556240, 322780800, 14599990830, ...];
  [1, 12, 108, 1144, 17292, 376992, 11084864, 411518448, 18410660208, ...];
  [1, 14, 140, 1540, 22974, 487004, 13983816, 510586400, 22584587382, ...];
  [1, 16, 176, 2016, 29872, 617280, 17308096, 621216192, 27156348512, ...];
  [1, 18, 216, 2580, 38178, 771228, 21120768, 744800256, 32164253550, ...]; ...
in which the main diagonal begins:
  [1, 4, 36, 560, 12650, 376992, 13983816, ..., binomial((n+1)^2,n), ...],
thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^((n+1)^2) for n>=0.

Links

Crossrefs

Cf. A295765, A295763, A070780.

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = Sum[Binomial[(n+1)^2, n]/(n+1)*x^n/A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)
  • PARI
    {a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m) - V[m+1])/(m+1);); A[n+1]}
for(n=0,20,print1(a(n),", "))

Formula

G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n) for n>=0.
a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(3/2 - 2*exp(-2)) / sqrt(2*Pi) = 1.3639581816297942453935... - Vaclav Kotesovec, Aug 11 2021, updated Apr 20 2024
Showing 1-3 of 3 results.

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