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.
The first few rows of the triangle are: [1] [1, 1] [1, 5, 1] [1, 15, 18, 1] [1, 37, 129, 58, 1] [1, 83, 646, 877, 179, 1]
A156920 := proc(n,m): if n=m then 1; elif m=0 then 1 ; elif m<0 or m>n then 0; else (m+1)*procname(n-1, m)+(2*n-2*m+1)*procname(n-1, m-1) ; end if; end proc: seq(seq(A156920(n, m), m=0..n), n=0..8); RHCnr:=5; RHCmax:=10; RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr); LHCnr:=5; LHCmax:=10: LHCend:=LHCnr+LHCmax: for k from LHCnr to LHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!)*x^p/(1-4*x)^(p+1),p=0..k)/ (((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); for n from 0 to nmax do d[n]:= coeff(fx,x,n)/2^n end do: LHC[n]:=d[LHCnr-1] end do: a:=n-> LHC[n]: seq(a(n), n=LHCnr..LHCend-1);
T[, 0] = 1; T[n, n_] = 1; T[n_, m_] := T[n, m] = (m + 1)*T[n - 1, m] + (2*n - 2*m + 1)*T[n - 1, m - 1]; Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017 *)
f := n -> sum(j^n*(j!)^2*2^j/(2*j)!, j = 1..infinity): seq(f(n), n = 0..5); # gives # [1+(1/2)*Pi, 3+Pi, 11+(7/2)*Pi, 55+(35/2)*Pi, 355+113*Pi, 2807+(1787/2)*Pi].
Table[Expand[FunctionExpand[FullSimplify[Sum[j^n*2^j/Binomial[2*j, j], {j, 1, Infinity}]]]][[1]], {n, 0, 20}] (* Vaclav Kotesovec, May 14 2020 *)
N=20; x='x+O('x^N); f=sqrt(exp(x)/(2-exp(x))); Vec(serlaplace(deriv(f*intformal(f)))) \\ Seiichi Manyama, Oct 22 2019
# An alternative version of the sequence starts (for n >= 0): # 0, 1, 3, 11, ..., or in terms of the approximation: [(1/2)*Pi, 1+(1/2)*Pi, # 3+Pi, 11+(7/2)*Pi, ...]. Similar to the formula of Detlef Meya above, the # sequence then can be computed (without a special initial case) as: from functools import cache from math import comb as binomial @cache def a(n): return n + sum((binomial(n, j) - 1) * a(n - j) for j in range(1, n)) print([a(n) for n in range(23)]) # Peter Luschny, Jun 09 2023
N:= 60; # to get a(n) for n <= N S:= series(exp(x)/sqrt(2-exp(2*x)), x, N+1): seq(coeff(S,x,j), j=0..N); # Robert Israel, May 19 2014
CoefficientList[Series[E^x/Sqrt[2-E^(2*x)]-1, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 03 2013 *)
{a(n)=local(A=1+x+x*O(x^n)); for(i=0,n,A=1+intformal(A+A^3)); n!*polcoeff(A,n)} \\ Paul D. Hanna, Oct 04 2008
CoefficientList[Series[(E^x/(4-3*E^x))^(1/4), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 15 2013 *)
a(n)=n!*polcoeff((exp(x +x*O(x^n))/(4-3*exp(x +x*O(x^n))))^(1/4),n)
/* As solution to integral equation: */ a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A^4*exp(-x+x*O(x^n))));n!*polcoeff(A,n)
CoefficientList[Series[(E^x/(5-4*E^x))^(1/5), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 22 2013 *)
a(n)=n!*polcoeff((exp(x +x*O(x^n))/(5-4*exp(x +x*O(x^n))))^(1/5),n)
/* As solution to integral equation: */ a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A^6*exp(-x+x*O(x^n))));n!*polcoeff(A,n)
E.g.f.: A(x) = 1 + x + 3/2*x^2 + 17/6*x^3 + 139/24*x^4 + 1481/120*x^5 +...
With[{nn=20},CoefficientList[Series[(Exp[x]/(3-2Exp[x]))^(1/3),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jan 26 2013 *)
{a(n) = n!*polcoeff((exp(x +x*O(x^n))/(3-2*exp(x +x*O(x^n))))^(1/3),n)}
for(n=0,25,print1(a(n),", "))
/* As solution to integral equation: */
{a(n) = local(A=1+x+x*O(x^n)); for(i=0,n, A = 1 + intformal(A^4*exp(-x+x*O(x^n)))); n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))
sqrt(1/(2*exp(x)-2*x*exp(x)-1)) = 1 + x^2/2! + 2*x^3/3! + 12*x^4/4! + 64*x^5/5! + 485*x^6/6! + 4038*x^7/7! + 39991*x^8/8! + 441992*x^9/9! + ...
a(13) = Sum_{k=1..6} A014307(k)*A008306(13,k) = 18381446004.
A014307(1)*A008306(13,1) == -1 (mod 13), because A014307(1) = 1 and A008306(13,1) = (13-1)!
For k>=2, A008306(13,k) == 0 (mod 13), result a(13) == -1 (mod 13).
A014307 := proc(n) option remember; `if`(n=0, 1 , 1+add((-1+binomial(n, k))*A014307(k), k=1..n-1)) end: A008306 := proc(n, k): if k=1 then (n-1)! ; elif n<=2*k-1 then 0; else (n-1)*procname(n-1, k)+(n-1)*procname(n-2, k-1) ; end if; end proc: a := n-> add((A014307(k)*A008306(n,k)), k=1..floor(n/2)):a(0):=1 ; seq(a(n), n=0..24); # second program: a := series(sqrt((1/(2*exp(x)-2*x*exp(x)-1))), x=0, 25): seq(n!*coeff(a, x, n), n=0..24);
CoefficientList[Series[Sqrt[1/(2*E^x-2*x*E^x-1)], {x, 0, 24}], x] * Range[0, 24]!
my(x='x+O('x^25)); Vec(serlaplace(sqrt(1 / (2*exp(x) - 2*x*exp(x) -1)))) \\ Michel Marcus, Jun 24 2021
Square array begins: 1, 1, 1, 1, 1, 1, 1, 1, 1, ... 1, 2, 3, 3, 3, 3, 3, 3, 3, ... 1, 3, 7, 12, 15, 15, 15, 15, 15, ... 1, 5, 16, 35, 63, 90, 105, 105, 105, ... 1, 8, 38, 105, 226, 417, 645, 840, 945, ... 1, 13, 89, 329, 841, 1787, 3348, 5445, 7665, ... 1, 21, 209, 1014, 3251, 7938, 16717, 31647, 53250, ... 1, 34, 491, 3116, 12483, 36500, 86311, 180560, 344403, ... 1, 55, 1153, 9610, 47481, 167631, 459803, 1062435, 2211181, ... ...
Triangle begins: 1; 1, 0; 1, 1, 0; 1, 5, 1, 0; 1, 18, 15, 1, 0; 1, 58, 129, 37, 1, 0; ...
T[0, 0] := 1; T[n_, -1] := 0; T[n_, n_] := 0; T[n_, k_] := T[n, k] = (n - k)*T[n - 1, k - 1] + (2*k + 1)*T[n - 1, k]; Join[{1}, Table[If[k < 0, 0, If[k >= n, 0, T[n, k]]], {n, 1, 5}, {k, 0, n}] // Flatten] (* G. C. Greubel, Jun 30 2017 *)
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