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 1-8 of 8 results.

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

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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

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

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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 */

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

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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

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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

A116603 Coefficients in asymptotic expansion of sequence A052129.

Original entry on oeis.org

1, 2, -1, 4, -21, 138, -1091, 10088, -106918, 1279220, -17070418, 251560472, -4059954946, 71250808916, -1351381762990, 27552372478592, -601021307680207, 13969016314470386, -344653640328891233, 8997206549370634644, -247772400254700937149, 7178881153198162073002
Offset: 0

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Author

Michael Somos, Feb 18 2006

Keywords

Examples

			G.f. = 1 + 2*x - x^2 + 4*x^3 - 21*x^4 + 138*x^5 - 1091*x^6 + 10088*x^7 + ...
		

References

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

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = -A[x] + 2/A[x/(1+x)]^(-1/2)*(1+x) + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jul 28 2011, updated Jan 12 2018 *)
  • PARI
    {a(n) = my(A); if( n<0, 0, A=1; for( k=1, n, A = truncate( A + O(x^k)) + x * O(x^k); A = -A + 2 / subst(A^(-1/2), x, x/(1 + x)) * (1 + x);); polcoeff(A, n))};

Formula

a(0) = 1; thereafter, a(n) = (1/n)*Sum_{j=1..n} (-1)^(j-1)*2*b(j)*a(n-j), where b(j) = A000670(j) [Nemes]. - N. J. A. Sloane, Sep 11 2017
G.f. A(x) satisfies (1 + x)^2 = A(x)^2 / A(x/(1 + x)).
A003504(n+1) ~ C^(2^n) * (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where C = 1.04783144757... (see A115632).
A052129(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).
From Seiichi Manyama, May 26 2025: (Start)
G.f.: Product_{k>=1} (1 + k*x)^(1/2^k).
G.f.: exp(2 * Sum_{k>=1} (-1)^(k-1) * A000670(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A084785. (End)
a(n) ~ (-1)^(n+1) * (n-1)! / log(2)^(n+1). - Vaclav Kotesovec, May 27 2025

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

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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

A164334 Quartic recurrence sequence a(0) = 1, a(n) = n*a(n-1)^4.

Original entry on oeis.org

1, 1, 2, 48, 21233664, 1016411962239204484414785454080
Offset: 0

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Author

David Willingham (D.Willingham(AT)wmin.ac.uk), Aug 13 2009

Keywords

Comments

Number of different orderings for n-input trees in a Free Quaternary Decision Diagram.
The next term has 121 digits. - Harvey P. Dale, Dec 19 2016

Crossrefs

Quartic extension of A052129 and A123851.

Programs

  • Mathematica
    nxt[{n_,a_}]:={n+1,(n+1)a^4}; NestList[nxt,{0,1},5][[All,2]] (* Harvey P. Dale, Dec 19 2016 *)
  • PARI
    a(n) = if (n==0, 1, n*a(n-1)^4); \\ Michel Marcus, Sep 14 2017

Formula

a(0) = 1, a(n) = n*a(n-1)^4.

A164335 Quintic recurrence sequence a(0) = 1, a(n) = n*a(n-1)^5.

Original entry on oeis.org

1, 1, 2, 96, 32614907904, 184523119031305377426211669050277696887837070322565120
Offset: 0

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Author

David Willingham (D.Willingham(AT)wmin.ac.uk), Aug 13 2009

Keywords

Comments

Number of different orderings for n-input trees in a Free Quinary Decision Diagram.
a(7) onward have more than 1000 digits. - G. C. Greubel, Sep 14 2017

Crossrefs

Quintic extension of A052129, A123851 and A164334.

Programs

  • Mathematica
    nxt[{n_, a_}] := {n + 1, (n + 1) a^5}; NestList[nxt, {0, 1}, 5][[All, 2]] (* G. C. Greubel, Sep 14 2017 *)
  • PARI
    a(n) = if (n==0, 1, n*a(n-1)^5); \\ Michel Marcus, Sep 14 2017

Formula

a(0) = 1, a(n) = n*a(n-1)^5.
Showing 1-8 of 8 results.