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 1-9 of 9 results.

A112943 Logarithmic derivative of A112942 such that a(n)=(1/6)*A112942(n+1) for n>0, where A112942 equals the INVERT transform (with offset) of sextuple factorials A008543.

Original entry on oeis.org

1, 11, 181, 4031, 114001, 3917771, 158531941, 7380184511, 388385146081, 22791211333451, 1475182111403221, 104384110708795391, 8015356365346614961, 663741406196190241931, 58957686544170035607301
Offset: 1

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			log(1+x + 6*x*[x + 11*x^2 + 181*x^3 + 4031*x^4 + 114001*x^5 +...])
= x + 11/2*x^2 + 181/3*x^3 + 4031/4*x^4 + 114001/5*x^5 + ...

Crossrefs

Cf. A008543, A112942; A112934, A112935, A112936, A112937, A112938, A112939, A112940, A112941.

Programs

  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+6*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

Formula

G.f.: log(1+x + 6*x*[Sum_{n>=1} a(n)]) = Sum_{n>=1} a(n)/n*x^n.

A112934 a(0) = 1; a(n+1) = Sum_{k=0..n} a(k)*A001147(n-k), where A001147 = double factorial numbers.

Original entry on oeis.org

1, 1, 2, 6, 26, 158, 1282, 13158, 163354, 2374078, 39456386, 737125446, 15279024026, 347786765150, 8621313613954, 231139787526822, 6663177374810266, 205503866668090750, 6751565903597571842
Offset: 0

Views

Author

Philippe Deléham and Paul D. Hanna, Oct 09 2005

Keywords

Examples

			A(x) = 1 + x + 2*x^2 + 6*x^3 + 26*x^4 + 158*x^5 + 1282*x^6 + ...
1/A(x) = 1 - x - x^2 - 3*x^3 - 15*x^4 - 105*x^5 - ... - A001147(n)*x^(n+1) - ...
a(4) = a(3+1) = Sum_{k=0..3} a(k)*A001147(3-k) = a(0)*5!! + a(1)*3!! + a(2)*1 + a(3)*1 = 1*15 + 1*3 + 2*1 + 6*1 = 26. - _Michael B. Porter_, Jul 22 2016

Links

Crossrefs

Cf. A001147, A112935 (log derivative); A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

Programs

  • Maple
    a_list := proc(len) local A, n; A[0] := 1; A[1] := 1;
for n from 2 to len-1 do A[n] := (2*n-1)*A[n-1] - add(A[j]*A[n-j], j=1..n-1) od;
convert(A, list) end: a_list(19); # Peter Luschny, May 22 2017
# Alternative:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
else (n - k) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Oct 02 2023
  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[a[k]*(2n - 2k - 3)!!, {k, 0, n - 1}]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+2*x^2*deriv(F)/F); return(polcoeff(F,n,x))}
  • PARI
    {a(n) = local(A); if( n<1, n==0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = (2*k - 1) * A[k-1] - sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* Michael Somos, Jul 23 2011 */

Formula

INVERT transform of double factorials (A001147), shifted right one place, where g.f. A(x) satisfies: A(x) = 1 + x*[d/dx x*A(x)^2]/A(x)^2.
G.f. A(x) satisfies: A(x) = 1+x + 2*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +2*x^2/(1-3*x -2*2*1*x^2/(1-7*x -2*3*3*x^2/(1-11*x -2*4*5*x^2/(1-15*x - ... -2*n*(2*n-3)*x^2/(1-(4*n-1)*x - ...)))) (continued fraction).
G.f.: A(x) = 1/(1-x/(1 -1*x/(1-2*x/(1 -3*x/(1-4*x(1 - ...))))))) (continued fraction).
From Paul Barry, Dec 04 2009: (Start)
The g.f. of a(n+1) is 1/(1-2x/(1-x/(1-4x/(1-3x/(1-6x/(1-5x/(1-.... (continued fraction).
The Hankel transform of a(n+1) is A137592. (End)
a(n) = Sum_{k=0..n} A111106(n,k). - Philippe Deléham, Jun 20 2006
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) is the upper left term in M^n, M = the production matrix:
1, 1;
1, 1, 2;
1, 1, 2, 3;
1, 1, 2, 3, 4;
1, 1, 2, 3, 4, 5;
... (End)
From Gary W. Adamson, Jul 21 2016: (Start)
Another production matrix Q is:
1, 1, 0, 0, 0, ...
1, 0, 3, 0, 0, ...
1, 0, 0, 5, 0, ...
1, 0, 0, 0, 7, ...
...
The sequence is generated by extracting the upper left term of powers of Q. By extracting the top row of Q^n, we obtain a triangle with the sequence in the left column and row sums = (1, 2, 6, 26, 158, ...): (1), (1, 1), (2, 1, 3), (6, 2, 3, 15), (26, 6, 6, 15, 105), ... (End)
a(n) = (2*n - 1) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: 1 / (1 - b(0)*x / (1 - b(1)*x / ...)) where b = A028310. - Michael Somos, Mar 31 2012
From Sergei N. Gladkovskii, Aug 11 2012, Aug 12 2012, Dec 26 2012, Mar 20 2013, Jun 02 2013, Aug 14 2013, Oct 22 2013: (Start) Continued fractions:
G.f. 1/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1).
G.f. 1 + x/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1).
G.f.: A(x) = 1 + x/(G(0) - x) where G(k) = 1 + (2*k+1)*x - x*(2*k+2)/G(k+1).
G.f.: Q(0) where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+2)/Q(k+1)).
G.f.: 2/G(0) where G(k) = 1 + 1/(1 - x/(x + 1/(2*k-1)/G(k+1))).
G.f.: 3*x - G(0) where G(k) = 3*x - 2*x*k - 1 - x*(2*k-1)/G(k+1).
G.f.: 1 + x*Q(0) where Q(k) = 1 - x*(2*k+2)/(x*(2*k+2) - 1/(1 - x*(2*k+1)/(x*(2*k+1) - 1/Q(k+1)))). (End)
a(n) ~ n^(n-1) * 2^(n-1/2) / exp(n). - Vaclav Kotesovec, Feb 22 2014

A112936 INVERT transform (with offset) of triple factorials (A008544), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^3]/A(x)^3.

Original entry on oeis.org

1, 1, 3, 15, 111, 1131, 14943, 243915, 4742391, 106912131, 2739347103, 78569371275, 2492748594471, 86650852740531, 3274367635513263, 133625238021647835, 5856377114106629751, 274320168321004350531
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			A(x) = 1 + x + 3*x^2 + 15*x^3 + 111*x^4 + 1131*x^5 + 14943*x^6 +...
1/A(x) = 1 - x - 2*x^2 - 10*x^3 - 80*x^4 - 880*x^5 -...-A008544(n)*x^(n+1)-...

Crossrefs

Cf. A008544, A112937 (log derivative); A112934, A112935, A112938, A112939, A112940, A112941, A112942, A112943.

Programs

  • Mathematica
    a = ConstantArray[0,20]; a[[1]]=1; Do[a[[n]] = (3*n-2)*a[[n-1]] - Sum[a[[k]]*a[[n-k]],{k,1,n-1}],{n,2,20}]; Flatten[{1,a}] (* Vaclav Kotesovec after Michael Somos, Feb 22 2014 *)
CoefficientList[Series[1/(1+(1/3*ExpIntegralE[2/3,-1/(3*x)])/E^(1/(3*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+3*x^2*deriv(F)/F); return(polcoeff(F,n,x))}
  • PARI
    {a(n) = my(A); if( n<1, n==0, A = vector(n, k, 1); for(k=2, n, A[k] = (3*k - 2)*A[k-1] - sum(j=1, k-1, A[j] * A[k-j])); A[n])}; /* Michael Somos, Jul 23 2011 */
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( 1 / sum(k=0, n, x^k * prod(i=1, k, 3*i - 4), x * O(x^n)), n))}; /* Michael Somos, Oct 17 2016 */
  • PARI
    {a(n) = my(A); if( n<0, 0, A = O(x); for(k=0, n, A = (x + sqrt(x^2 + 4*x^5*A')) / 2); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */
  • PARI
    {a(n) = my(A); if( n<1, n==0, A = x; for(k=1, n, A = truncate(A) + O(x^(3*k + 4)); A += A + x^4*A' - A^2/x); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */

Formula

G.f. satisfies: A(x) = 1+x + 3*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +3*x^2/(1-5*x -3*2*2*x^2/(1-11*x -3*3*5*x^2/(1-17*x -3*4*8*x^2/(1-23*x -... -3*n*(3*n-4)*x^2/(1-(6*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-x/(1 -2*x/(1-3*x/(1 -5*x/(1-6*x/(1 -8*x/(1-9*x/(1 -...)))))))) (continued fraction).
a(n) = (3*n - 2) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: Q(0) where Q(k) = 1 - x*(3*k-1)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: 2/G(0)+4*x, where G(k)= 1 + 1/(1 - x*(3*k+3)/(x*(3*k+5) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
G.f.: 2/G(0), where G(k)= 1 + 1/(1 - x*(3*k-1)/(x*(3*k-1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: -4/Sum_{n>=0} [Product_{k=0..n} (3*k-4)]*x^n. - Sergei N. Gladkovskii, Jun 06 2013
a(n) ~ n! * 3^(n-1) / (GAMMA(2/3) * n^(4/3)) . - Vaclav Kotesovec, Feb 22 2014
Given g.f. A(x), then y = x * A(x^3) satisfies y^2 = x*y + x^5*y'. - Michael Somos, Oct 17 2016

A112938 INVERT transform (with offset) of quadruple factorials (A008545), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^4]/A(x)^4.

Original entry on oeis.org

1, 1, 4, 28, 292, 4156, 75844, 1694812, 44835172, 1369657468, 47422855300, 1834403141788, 78377228106148, 3664969183404220, 186134931067171012, 10201887125268108508, 600142156513333537252, 37713563573426417361148
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			A(x) = 1 + x + 4*x^2 + 28*x^3 + 292*x^4 + 4156*x^5 + ...
1/A(x) = 1 - x - 3*x^2 - 21*x^3 - 231*x^4 -... -A008545(n)*x^(n+1)-...

Links

Crossrefs

Cf. A008545, A112939 (log derivative); A112934, A112935, A112936, A112937, A112940, A112941, A112942, A112943.

Programs

  • Mathematica
    CoefficientList[Series[1/(1 + 1/4*ExpIntegralE[3/4,-1/(4*x)]/E^(1/(4*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+4*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

Formula

G.f. satisfies: A(x) = 1+x + 4*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +4*x^2/(1-7*x -4*2*3*x^2/(1-15*x -4*3*7*x^2/(1-23*x -4*4*11*x^2/(1-31*x -... -4*n*(4*n-5)*x^2/(1-(8*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-1*x/(1 -3*x/(1-4*x/(1 -7*x/(1-8*x/(1 -11*x/(1-12*x/(1 -...)))))))) (continued fraction).
G.f.: Q(0) where Q(k) = 1 - x*(4*k-1)/(1 - x*(4*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: 1 + 2*x/G(0), where G(k)= 1 + 1/(1 - 2*x*(4*k+4)/(2*x*(4*k+4) - 1 + 2*x*(4*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
a(n) ~ (n-1)! * 4^(n-1) / (GAMMA(3/4) * n^(1/4)). - Vaclav Kotesovec, Feb 22 2014

A112940 INVERT transform (with offset) of quintuple factorials (A008546), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^5]/A(x)^5.

Original entry on oeis.org

1, 1, 5, 45, 605, 11045, 257005, 7288245, 243870205, 9401560645, 410141056205, 19966451812245, 1072718714991005, 63033317759267045, 4020725747388170605, 276661592017425909045, 20424931173615717011005
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			A(x) = 1 + x + 5*x^2 + 45*x^3 + 605*x^4 + 11045*x^5 +...
1/A(x) = 1 - x - 4*x^2 - 36*x^3 - 504*x^4 -... -A008546(n)*x^(n+1) -...

Links

Crossrefs

Cf. A008546, A112941 (log derivative); A112934, A112935, A112936, A112937, A112938, A112939, A112942, A112943.

Programs

  • Mathematica
    CoefficientList[Series[1/(1 + 1/5*ExpIntegralE[4/5, -1/(5*x)]/E^(1/(5*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+5*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

Formula

G.f. satisfies: A(x) = 1+x + 5*x^2*[d/dx A(x)]/A(x) (log derivative). G.f.: A(x) = 1+x +5*x^2/(1-9*x -5*2*4*x^2/(1-19*x -5*3*9*x^2/(1-29*x -5*4*13*x^2/(1-39*x -... -5*n*(5*n-6)*x^2/(1-(10*n-1)*x -...)))) (continued fraction). G.f.: A(x) = 1/(1-1*x/(1 -4*x/(1-5*x/(1 -9*x/(1-10*x/(1 -14*x/(1-15*x/(1 -...)))))))) (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x*(5*k+4)/(1 - x*(5*k+5)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
a(n) ~ (n-1)! * 5^(n-1) / (GAMMA(4/5) * n^(1/5)). - Vaclav Kotesovec, Feb 22 2014

A112935 Logarithmic derivative of A112934 such that a(n)=(1/2)*A112934(n+1) for n>0, where A112934 equals the INVERT transform of double factorials A001147.

Original entry on oeis.org

1, 3, 13, 79, 641, 6579, 81677, 1187039, 19728193, 368562723, 7639512013, 173893382575, 4310656806977, 115569893763411, 3331588687405133, 102751933334045375, 3375782951798785921, 117693183724386637635
Offset: 1

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			log(1+x + 2*x*[x + 3*x^2 + 13*x^3 + 79*x^4 + 641*x^5 +...])
= x + 3/2*x^2 + 13/3*x^3 + 79/4*x^4 + 641/5*x^5 +...

Crossrefs

Cf. A001147, A112934; A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

Programs

  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+2*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

Formula

G.f.: log(1 + x + 2*x*[Sum_{n>=1} a(n)*x^n]) = Sum_{k>=1} a(n)/n*x^n.
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1/(x*G(0)) - 1/(2*x), where G(k)= 1 + 1/(1 - 2*x*(2*k+2)/(2*x*(2*k+2) - 1 + 2*x*(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013

A112937 Logarithmic derivative of A112936 such that a(n)=(1/3)*A112936(n+1) for n>0, where A112936 equals the INVERT transform (with offset) of triple factorials A008544.

Original entry on oeis.org

1, 5, 37, 377, 4981, 81305, 1580797, 35637377, 913115701, 26189790425, 830916198157, 28883617580177, 1091455878504421, 44541746007215945, 1952125704702209917, 91440056107001450177, 4558596081095404198741
Offset: 1

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			log(1+x + 3*x*[x + 5*x^2 + 37*x^3 + 377*x^4 + 4981*x^5 +...])
= x + 5/2*x^2 + 37/3*x^3 + 377/4*x^4 + 4981/5*x^5 + ...

Crossrefs

Cf. A008544, A112936; A112934, A112935, A112938, A112939, A112940, A112941, A112942, A112943.

Programs

  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+3*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

Formula

G.f.: log(1+x + 3*x*[Sum_{k>=1} a(n)]) = Sum_{k>=1} a(n)/n*x^n.

A112939 Logarithmic derivative of A112938 such that a(n)=(1/4)*A112938(n+1) for n>0, where A112938 equals the INVERT transform (with offset) of quadruple factorials A008545.

Original entry on oeis.org

1, 7, 73, 1039, 18961, 423703, 11208793, 342414367, 11855713825, 458600785447, 19594307026537, 916242295851055, 46533732766792753, 2550471781317027127, 150035539128333384313, 9428390893356604340287, 630318228814408172573761
Offset: 1

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			log(1+x + 4*x*[x + 7*x^2 + 73*x^3 + 1039*x^4 + 18961*x^5 +...])
= x + 7/2*x^2 + 73/3*x^3 + 1039/4*x^4 + 18961/5*x^5 + ...

Crossrefs

Cf. A008545, A112938; A112934, A112935, A112936, A112937, A112940, A112941, A112942, A112943.

Programs

  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+4*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

Formula

G.f.: log(1+x + 4*x*[Sum_{k>=1} a(n)]) = Sum_{k>=1} a(n)/n*x^n.
G.f.: 1/x - G(0)/(2*x), where G(k)= 1 + 1/(1 - x*(4*k-1)/(x*(4*k-3) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013

A112941 Logarithmic derivative of A112940 such that a(n)=(1/5)*A112940(n+1) for n>0, where A112940 equals the INVERT transform (with offset) of quintuple factorials A008546.

Original entry on oeis.org

1, 9, 121, 2209, 51401, 1457649, 48774041, 1880312129, 82028211241, 3993290362449, 214543742998201, 12606663551853409, 804145149477634121, 55332318403485181809, 4084986234723143402201, 322064057582671115832449
Offset: 1

Views

Author

Paul D. Hanna, Oct 09 2005

Keywords

Examples

			log(1+x + 5*x*[x + 9*x^2 + 121*x^3 + 2209*x^4 + 51401*x^5 +...])
= x + 9/2*x^2 + 121/3*x^3 + 2209/4*x^4 + 51401/5*x^5 + ...

Crossrefs

Cf. A008546, A112940; A112934, A112935, A112936, A112937, A112938, A112939, A112942, A112943.

Programs

  • PARI
    {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+5*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

Formula

G.f.: log(1+x + 5*x*[Sum_{n>=1} a(n)]) = Sum_{n>=1} a(n)/n*x^n.
Showing 1-9 of 9 results.

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