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 1-10 of 10 results.

A112302 Decimal expansion of quadratic recurrence constant sqrt(1 * sqrt(2 * sqrt(3 * sqrt(4 * ...)))).

Original entry on oeis.org

1, 6, 6, 1, 6, 8, 7, 9, 4, 9, 6, 3, 3, 5, 9, 4, 1, 2, 1, 2, 9, 5, 8, 1, 8, 9, 2, 2, 7, 4, 9, 9, 5, 0, 7, 4, 9, 9, 6, 4, 4, 1, 8, 6, 3, 5, 0, 2, 5, 0, 6, 8, 2, 0, 8, 1, 8, 9, 7, 1, 1, 1, 6, 8, 0, 2, 5, 6, 0, 9, 0, 2, 9, 8, 2, 6, 3, 8, 3, 7, 2, 7, 9, 0, 8, 3, 6, 9, 1, 7, 6, 4, 1, 1, 4, 6, 1, 1, 6, 7, 1, 5, 5, 2, 8
Offset: 1

Views

Author

Michael Somos, Sep 02 2005

Keywords

Comments

From Johannes W. Meijer, Jun 27 2016: (Start)
With Phi(z, p, q) the Lerch transcendent, define LP(n) = (1/n) * sum(Phi(1/2, n-k, 1) * LP(k), k=0..n-1), with LP(0) = 1. Conjecture: Lim_{n -> infinity} LP(n) = A112302.
For similar formulas, see A090998 and A135002. For background information, see A274181.
The structure of the n! * LP(n) formulas leads to the multinomial coefficients A036039. (End)

Examples

			1.6616879496335941212958189227499507499644186350250682081897111680...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.
  • S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., AMS Chelsea 2000. See Appendix I. p. 348.

Crossrefs

Programs

  • Mathematica
    RealDigits[ Fold[ N[ Sqrt[ #2*#1], 128] &, Sqrt@ 351, Reverse@ Range@ 350], 10, 111][[1]] (* Robert G. Wilson v, Nov 05 2010 *)
    Exp[-Derivative[1, 0][PolyLog][0, 1/2]] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Apr 07 2014, after Jonathan Sondow *)
  • PARI
    {a(n) = if( n<-1, 0, n++; default( realprecision, n+2); floor( prodinf( k=1, k^2^-k)* 10^n) % 10)};
    
  • PARI
    prodinf(n=1,n^2^-n) \\ Charles R Greathouse IV, Apr 07 2013
    
  • Python
    from mpmath import polylog, diff, exp, mp
    mp.dps = 120
    somos_const = exp(-diff(lambda n: polylog(n, 1/2), 0))
    A112302 = [int(d) for d in mp.nstr(somos_const, n=mp.dps)[:-1] if d != '.']  # Jwalin Bhatt, Nov 23 2024

Formula

Equals Product_{n>=1} n^(1/2^n). - Jonathan Sondow, Apr 07 2013
Equals exp(A114124) = A188834/2 = sqrt(A259235). - Hugo Pfoertner, Nov 23 2024
From Jwalin Bhatt, Apr 02 2025: (Start)
Equals exp(-PolyLog'(0,1/2)), where PolyLog'(x,y) represents the derivative of the polylogarithm w.r.t. x.
Equals Product_{n>=1} (1+1/n)^(1/2^n).
Equals exp(Sum_{n>=2} log(n)/2^n).
Equals 2*exp(Sum_{n>=1} (log(1+1/n)-1/n)/2^n). (End)

A052129 a(0) = 1; thereafter a(n) = n*a(n-1)^2.

Original entry on oeis.org

1, 1, 2, 12, 576, 1658880, 16511297126400, 1908360529573854283038720000, 29134719286683212541013468732221146917416153907200000000
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 12 2002

Keywords

Comments

Somos's quadratic recurrence sequence.
Iff n is prime (n>2), the n-adic valuation of a(2n) is 3*A001045(n) (three times the values at the prime indices of Jacobsthal numbers), which is 2^n+1. For example: the 11-adic valuation at a(22) = 2049 = 3*A001045(11)= 683. 3*683 = 2^11+1 = 2049. True because: When n is prime, n-adic valuation is 1 at A052129(n), then doubles as n-increases to 2n, at which point 1 is added; thus A052129(2n) = 2^n+1. Since 3*A001045(n) = 2^n+1, n-adic valuation of A052129(2n) = 3*A001045(n) when n is prime. - Bob Selcoe, Mar 06 2014
Unreduced denominators of: f(1) = 1, f(n) = f(n-1) + f(n-1)/(n-1). - Daniel Suteu, Jul 29 2016

Examples

			a(3) = 3*a(2)^2 = 3*(2*a(1)^2)^2 = 3*(2*(1*a(0)^2)^2)^2 = 3*(2*(1*1^2)^2)^2 = 3*(2*1)^2 = 3*4 = 12.
G.f. = 1 + x + 2*x^2 + 12*x^3 + 576*x^4 + 1658880*x^5 + 16511297126400*x^6 + ...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Mathematica
    Join[{1},RecurrenceTable[{a[1]==1,a[n]==n a[n-1]^2},a,{n,10}]]  (* Harvey P. Dale, Apr 26 2011 *)
    a[ n_] := If[ n < 1, Boole[n == 0], Product[ (n - k)^2^k, {k, 0, n - 1}]]; (* Michael Somos, May 24 2013 *)
    a[n_] := Product[ k^(2^(n - k)), {k,1,n}] (* Jonathan Sondow, Mar 17 2014 *)
    NestList[{#[[1]]+1,#[[1]]*#[[2]]^2}&,{1,1},10][[All,2]] (* Harvey P. Dale, Jul 30 2018 *)
  • PARI
    {a(n) = if( n<1, n==0, prod(k=0, n-1, (n - k)^2^k))}; /* Michael Somos, May 24 2013 */

Formula

a(n) ~ s^(2^n) / (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where s = 1.661687949633... (see A112302) and A116603. - Michael Somos, Apr 02 2006
a(n) = n * A030450(n - 1) if n>0. - Michael Somos, Oct 22 2006
a(n) = (a(n-1) + a(n-2)^2) * (a(n-1) / a(n-2))^2. - Michael Somos, Mar 20 2012
a(n) = product_{k=1..n} k^(2^(n-k)). - Jonathan Sondow, Mar 17 2014
A088679(n+1)/a(n) = n+1. -Daniel Suteu, Jul 29 2016

A123854 Denominators in an asymptotic expansion for the cubic recurrence sequence A123851.

Original entry on oeis.org

1, 4, 32, 128, 2048, 8192, 65536, 262144, 8388608, 33554432, 268435456, 1073741824, 17179869184, 68719476736, 549755813888, 2199023255552, 140737488355328, 562949953421312, 4503599627370496, 18014398509481984, 288230376151711744, 1152921504606846976
Offset: 0

Views

Author

Keywords

Comments

A cubic analog of the asymptotic expansion A116603 of Somos's quadratic recurrence sequence A052129. Numerators are A123853.
Equals 2^A004134(n); also the denominators in expansion of (1-x)^(-1/4). - Alexander Adamchuk, Oct 27 2006
All terms are powers of 2 and log_2 a(n) = A004134(n) = 3*n - A000120(n). - Alexander Adamchuk, Oct 27 2006 [Edited by Petros Hadjicostas, May 14 2020]
Is this the same sequence as A088802? - N. J. A. Sloane, Mar 21 2007
Almost certainly this is the same as A088802. - Michael Somos, Aug 23 2007
Denominators of Gegenbauer_C(2n,1/4,2). The denominators of Gegenbauer_C(n,1/4,2) give the doubled sequence. - Paul Barry, Apr 21 2009
If the Greubel formula in A088802 and the Luschny formula here are correct (they are the same), the sequence is a duplicate of A088802. - R. J. Mathar, Aug 02 2023

Examples

			A123851(n) ~ c^(3^n)*n^(- 1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) - 5397/(2048*n^4) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Maple
    f:=proc(t,x) exp(sum(ln(1+m*x)/t^m,m=1..infinity)); end; for j from 0 to 29 do denom(coeff(series(f(3,x),x=0,30),x,j)); od;
    # Alternatively:
    A123854 := n -> denom(binomial(1/4,n)):
    seq(A123854(n), n=0..25); # Peter Luschny, Apr 07 2016
  • Mathematica
    Denominator[CoefficientList[Series[ 1/Sqrt[Sqrt[1-x]], {x, 0, 25}], x]] (* Robert G. Wilson v, Mar 23 2014 *)
  • PARI
    vector(25, n, n--; denominator(binomial(1/4,n)) ) \\ G. C. Greubel, Aug 08 2019
  • Sage
    # uses[A000120]
    def A123854(n): return 1 << (3*n-A000120(n))
    [A123854(n) for n in (0..25)]  # Peter Luschny, Dec 02 2012
    

Formula

From Alexander Adamchuk, Oct 27 2006: (Start)
a(n) = 2^A004134(n).
a(n) = 2^(3n - A000120(n)). (End)
a(n) = denominator(binomial(1/4,n)). - Peter Luschny, Apr 07 2016

A003504 a(0)=a(1)=1; thereafter a(n+1) = (1/n)*Sum_{k=0..n} a(k)^2 (a(n) is not always integral!).

Original entry on oeis.org

1, 1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, 7160642690122633501504, 4661345794146064133843098964919305264116096, 1810678717716933442325741630275004084414865420898591223522682022447438928019172629856
Offset: 0

Views

Author

Keywords

Comments

The sequence appears with a different offset in some other sources. - Michael Somos, Apr 02 2006
Also known as Göbel's (or Goebel's) Sequence. Asymptotically, a(n) ~ n*C^(2^n) where C=1.0478... (A115632). A more precise asymptotic formula is given in A116603. - M. F. Hasler, Dec 12 2007
Let s(n) = (n-1)*a(n). By considering the p-adic representation of s(n) for primes p=2,3,...,43, one finds that a(44) is the first nonintegral value in this sequence. Furthermore, for n>44, the valuation of s(n) w.r.t. 43 is -2^(n-44), implying that both s(n) and a(n) are nonintegral. - M. F. Hasler and Max Alekseyev, Mar 03 2009
a(44) is approximately 5.4093*10^178485291567. - Hans Havermann, Nov 14 2017.
The fractional part is simply 24/43 (see page 709 of Guy (1988)).
The more precise asymptotic formula is a(n+1) ~ C^(2^n) * (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...). - Michael Somos, Mar 17 2012

Examples

			a(3) = (1 * 2 + 2^2) / 2 = 3 given a(2) = 2.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd edition, Sect. E15.
  • Clifford Pickover, A Passion for Mathematics, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A005166, A005167, A097398, A108394, A115632, A116603 (asymptotic formula).

Programs

  • Maple
    a:=2: L:=1,1,a: n:=15: for k to n-2 do a:=a*(a+k)/(k+1): L:=L,a od:L; # Robert FERREOL, Nov 07 2015
  • Mathematica
    a[n_] := a[n] = Sum[a[k]^2, {k, 0, n-1}]/(n-1); a[0] = a[1] = 1; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Feb 06 2013 *)
    With[{n = 14}, Nest[Append[#, (#.#)/(Length[#] - 1)] &, {1, 1}, n - 2]] (* Jan Mangaldan, Mar 21 2013 *)
  • PARI
    A003504(n,s=2)=if(n-->0,for(k=1,n-1,s+=(s/k)^2);s/n,1) \\ M. F. Hasler, Dec 12 2007
    
  • Python
    a=2; L=[1,1,a]; n=15
    for k in range(1,n-1):
        a=a*(a+k)//(k+1)
        L.append(a)
    print(L) # Robert FERREOL, Nov 07 2015

Formula

a(n+1) = ((n-1) * a(n) + a(n)^2) / n if n > 1. - Michael Somos, Apr 02 2006
0 = a(n)*(+a(n)*(a(n+1) - a(n+2)) - a(n+1) - a(n+1)^2) +a(n+1)*(a(n+1)^2 - a(n+2)) if n>1. - Michael Somos, Jul 25 2016

Extensions

a(0)..a(43) are integral, but from a(44) onwards every term is nonintegral - H. W. Lenstra, Jr.
Corrected and extended by M. F. Hasler, Dec 12 2007
Further corrections from Max Alekseyev, Mar 04 2009

A123851 A cubic recurrence: a(0) = 1, a(n) = n*a(n-1)^3 for n >= 1.

Original entry on oeis.org

1, 1, 2, 24, 55296, 845378412871680, 3624972460853492659595005581182702601633792000
Offset: 0

Views

Author

Keywords

Comments

A cubic analog of Somos's quadratic recurrence sequence A052129.
Terms a(7) onward are too big to include in data section. - G. C. Greubel, Aug 10 2019

Examples

			a(3) = 3*a(2)^3 = 3*(2*a(1)^3)^3 = 3*(2*(1*a(0)^3)^3)^3 = 3*(2*(1*1^3)^3)^3 = 3*(2*1)^3 = 3*8 = 24.
G.f. = 1 + x + 2*x^2 + 24*x^3 + 55296*x^4 + 845378412871680*x^5 + ...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • GAP
    List([0..8], n-> Product([0..n-1], k-> (n-k)^(3^k)) ); # G. C. Greubel, Aug 10 2019
  • Magma
    [n eq 0 select 1 else (&*[(n-k)^(3^k): k in [0..n-1]]):n in [0..8]]; // G. C. Greubel, Aug 10 2019
    
  • Mathematica
    a[n_]:= If[n==0, 1, n*a[n-1]^3]; Table[a[n], {n,0,7}]
    nxt[{n_,a_}]:={n+1, (n+1)a^3}; NestList[nxt,{0,1},7][[All,2]] (* Harvey P. Dale, May 25 2019 *)
  • PARI
    {a(n) = if( n<1, n==0, prod(k=0, n-1, (n - k)^3^k))}; /* Michael Somos, Aug 07 2016 */
    
  • Sage
    [1]+[prod((n-k)^(3^k) for k in (0..n-1)) for n in (1..8)] # G. C. Greubel, Aug 10 2019
    

Formula

a(n) ~ c^(3^n)*n^(-1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.

Extensions

Corrected by Harvey P. Dale, May 25 2019

A123852 Decimal expansion of (1*(2*(3*...)^(1/3))^(1/3))^(1/3).

Original entry on oeis.org

1, 1, 5, 6, 3, 6, 2, 6, 8, 4, 3, 3, 2, 2, 6, 9, 7, 1, 6, 8, 5, 3, 3, 7, 0, 3, 2, 2, 8, 8, 7, 3, 6, 9, 3, 5, 6, 5, 1, 3, 0, 1, 4, 5, 4, 3, 8, 9, 1, 8, 8, 8, 6, 3, 7, 9, 9, 9, 2, 5, 9, 5, 9, 8, 9, 8, 3, 1, 7, 7, 8, 1, 6, 0, 7, 2, 8, 2, 6, 1, 9, 4, 6, 0, 7, 9, 0, 8, 1, 3, 3, 8, 2, 0, 3, 7, 8, 3, 1, 7
Offset: 1

Views

Author

Keywords

Comments

Cubic recurrence constant (see A123851): a cubic analog of Somos's quadratic recurrence constant A112302.

Examples

			1.156362684332269716853370322887369356513014543891888637999259598983177816...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Mathematica
    Take[RealDigits[Product[N[n^3^-n,200], {n,400}]][[1]], 100]
    RealDigits[Exp[-D[PolyLog[n, 1/3], n]/.n->0], 10, 100][[1]] (* Jean-François Alcover, Jan 28 2014 *)
  • PARI
    prodinf(n=1, n^(1/3^n)) \\ Michel Marcus, Aug 03 2019

Formula

Equals Product_{n>=1} n^(1/3^n).

Extensions

References updated by R. J. Mathar, Aug 12 2010

A123853 Numerators in an asymptotic expansion for the cubic recurrence sequence A123851.

Original entry on oeis.org

1, 3, -15, 113, -5397, 84813, -3267755, 74391561, -15633072909, 465681118929, -31041303829713, 1145088996404679, -185348722911971841, 8165727090278785521, -778296382754673737187, 39898888480559205453945, -35033447016186321707305533
Offset: 0

Views

Author

Keywords

Comments

A cubic analog of the asymptotic expansion A116603 of Somos's quadratic recurrence sequence A052129. Denominators are A123854.

Examples

			A123851(n) ~ c^(3^n)*n^(-1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) - 5397/(2048*n^4) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Cf. A052129, A112302, A116603, A123851, A123852, A123854 (denominators).

Programs

  • Maple
    f:=proc(t,x) exp(sum(ln(1+m*x)/t^m,m=1..infinity)); end; for j from 0 to 29 do numer(coeff(series(f(3,x),x=0,30),x,j)); od;
  • PARI
    {a(n) = local(A); if(n < 0, 0, A = 1 + O(x) ; for( k = 1, n, A = truncate(A) + x * O(x^k); A += x^k * polcoeff( 3/4 * (subst(1/A, x, x^2/(1-x^2))^2/(1-x^2) - 1/subst(A, x, x^2)^(2/3)), 2*k ) ); numerator( polcoeff( A, n ) ) ) } /* Michael Somos, Aug 23 2007 */

A384332 Expansion of Product_{k>=1} (1 + k*x)^((2/3)^k).

Original entry on oeis.org

1, 6, 3, 20, -207, 2538, -36381, 599760, -11210229, 234779146, -5455240455, 139445920452, -3892724842549, 117916363928070, -3854035833235839, 135241405277665656, -5072575747811807052, 202559732310632082120, -8581116791103001216108
Offset: 0

Views

Author

Seiichi Manyama, May 26 2025

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = -2*A[x] + 3*A[x/(1+x)]^(2/3) * (1+x)^2 + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Vaclav Kotesovec, May 27 2025 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(exp(3*sum(k=1, N, (-1)^(k-1)*sum(j=0, k, 2^j*j!*stirling(k, j, 2))*x^k/k)))

Formula

G.f. A(x) satisfies A(x) = (1+x)^2 * A(x/(1+x))^(2/3).
G.f.: exp(3 * Sum_{k>=1} (-1)^(k-1) * A004123(k+1) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A384324.
G.f.: B(x)^6, where B(x) is the g.f. of A384344.
a(n) ~ (-1)^(n+1) * (n-1)! / log(3/2)^(n+1). - Vaclav Kotesovec, May 27 2025

A384333 Expansion of Product_{k>=1} (1 + k*x)^((3/4)^k).

Original entry on oeis.org

1, 12, 30, 76, -819, 15120, -320568, 7719984, -208986462, 6300545128, -209806494828, 7660698340008, -304718887446110, 13127557400200944, -609336227455254936, 30330991088734345200, -1612305658103085757467, 91179423240593288760396, -5467060640706775435713298
Offset: 0

Views

Author

Seiichi Manyama, May 26 2025

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = -3*A[x] + 4*A[x/(1+x)]^(3/4) * (1+x)^3 + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Vaclav Kotesovec, May 27 2025 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(exp(4*sum(k=1, N, (-1)^(k-1)*sum(j=0, k, 3^j*j!*stirling(k, j, 2))*x^k/k)))

Formula

G.f. A(x) satisfies A(x) = (1+x)^3 * A(x/(1+x))^(3/4).
G.f.: exp(4 * Sum_{k>=1} (-1)^(k-1) * A032033(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A384325.
G.f.: B(x)^12, where B(x) is the g.f. of A381890.
a(n) ~ (-1)^(n+1) * (n-1)! / log(4/3)^(n+1). - Vaclav Kotesovec, May 27 2025

A384334 Expansion of Product_{k>=1} (1 + k*x)^((4/5)^k).

Original entry on oeis.org

1, 20, 110, 340, -1995, 53904, -1534600, 49159600, -1758057650, 69662897000, -3037327435860, 144787947993000, -7502235351828450, 420296374337607600, -25335189019626256200, 1636008982452733508400, -112721505676611504401025, 8256863266451569604835900
Offset: 0

Views

Author

Seiichi Manyama, May 26 2025

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = -4*A[x] + 5*A[x/(1+x)]^(4/5) * (1+x)^4 + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Vaclav Kotesovec, May 27 2025 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(exp(5*sum(k=1, N, (-1)^(k-1)*sum(j=0, k, 4^j*j!*stirling(k, j, 2))*x^k/k)))

Formula

G.f. A(x) satisfies A(x) = (1+x)^4 * A(x/(1+x))^(4/5).
G.f.: exp(5 * Sum_{k>=1} (-1)^(k-1) * A094417(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A384326.
G.f.: B(x)^20, where B(x) is the g.f. of A384345.
a(n) ~ (-1)^(n+1) * (n-1)! / log(5/4)^(n+1). - Vaclav Kotesovec, May 27 2025
Showing 1-10 of 10 results.