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.

Previous Showing 11-18 of 18 results.

A030450 Related to number of elements in the free band (idempotent semigroup) on n generators.

Original entry on oeis.org

1, 1, 4, 144, 331776, 2751882854400, 272622932796264897576960000, 3641839910835401567626683591527643364677019238400000000
Offset: 0

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Author

Marcel Jackson (marcel_j(AT)hilbert.maths.utas.edu.au)

Keywords

Comments

Continued square root 2 = sqrt(1 + sqrt(1 + sqrt(4 + sqrt(144 + ...)))) = sqrt(1 + sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + ...))))) [S. Ramanujan]. - Michael Somos, Dec 03 2017

References

  • John M. Howie, Fundamentals of Semigroup Theory, Oxford University Press 1995, p. 123.

Crossrefs

A052129(n) = n*a(n-1) if n > 0.

Programs

Formula

Binomial transform is A005345. - Michael Somos, Oct 22 2006
a(n) = (n*a(n-1))^2 if n > 0. a(0)=1. - Michael Somos, Oct 22 2006
a(n) = Product_{i=1..n} (n-i+1)^(2^i).
Sum_{n>=1} 1/a(n) = A258621. - Amiram Eldar, Nov 19 2020

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 */

A088679 a(n) = a(n-1)^2 * n / (n-1), n>1, a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 6, 48, 2880, 9953280, 115579079884800, 15266884236590834264309760000, 262212473580148912869121218589990322256745385164800000000
Offset: 0

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Author

Michael Somos, Oct 05 2003

Keywords

Comments

Unreduced numerators of: f(1) = 1, f(n) = f(n-1) + f(n-1)/(n-1). - Daniel Suteu, Jul 29 2016

Examples

			x + 2*x^2 + 6*x^3 + 48*x^4 + 2880*x^5 + 9953280*x^6 + ...
		

Crossrefs

Cf. A052129.

Programs

  • Mathematica
    Join[{0},RecurrenceTable[{a[1]==1,a[n]==a[n-1]^2 n/(n-1)},a,{n,10}]] (* Harvey P. Dale, Jan 16 2015 *)
  • PARI
    {a(n) = if( n<2, n>0, a(n-1)^2 * n / (n-1))}

Formula

a(n) is asymptotic to c^(2^n)*(1-1/n+2/n^2-6/n^3+25/n^4-137/n^5+...) where c=1.28906475773... and coefficient of n^-k is (-1)^k*A084784(k).
a(0) = 0, a(1) = 1, a(n) = n * Product i=1..(n-1) a(i) for n > 1. - Gerald McGarvey, Jun 11 2004 Corrected by Jaroslav Krizek, Oct 16 2009
a(n)^2 = n * A052129(n). Michael Somos, May 13 2012
a(n+1)/A052129(n) = n+1. - Daniel Suteu, Jul 29 2016

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

Views

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.

A097635 Triangle read by rows: T(n,k) = number of unique-valued sequences of length k, n >= 1, 1 <= k <= 2n-3, in the symmetric group S_n.

Original entry on oeis.org

1, 2, 6, 18, 12, 24, 456, 5664, 20640, 576, 120, 13560, 1395840
Offset: 1

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Author

Aleksandar Blazhevski-Cane (CaneB(AT)mt.net.mk), Aug 17 2004

Keywords

Comments

Definition: Let G(*) be a semigroup. A finite sequence u_1, u_2,... u_n of elements from G is called unique-valued with value v and length n provided v = u_1 * u_2 *... * u_n and v != u_p(1) * u_p(2) *... * u_p(n) for any non-identity permutation p of the indices {1, 2,... n}.
In other words, the only way to obtain the unique value v from the elements u_1, u_2,... u_n is by multiplying them in that particular order; any other order always gives a value different from v.
When the length of the sequence is 2, the meaning of "unique-valued" is equivalent to "the two elements do not commute under *".
I proved that the maximal possible length of a unique-valued sequence in the monoid M_n(*) of all n X n matrices (with entries in some ring with 1) is exactly 2n-1 (that 2n-1 is an upper bound follows from the Amitsur-Levitzki theorem), providing a positive example that this limit is reached.
I also proved that the maximal possible length of unique-valued sequences in S_n is 2n-3 (using n-simplex and again using the Amitsur-Levitzki theorem), but didn't find examples that this limit is really reached. My computer said "yes" for n=2 to 5, but even 6 is too large to compute.
The numbers of unique-valued sequences in S_n of the maximal length 2n-3 form a sequence 1, 2, 12, 576, which seems to coincide with A002860, the number of Latin squares of order n.

Examples

			Triangle begins:
    1
    2
    6    18      12
   24   456    5664 20640 576
  120 13560 1395840 ?
		

Crossrefs

Possibly related to A002860 (the number of Latin Squares) or A052129.

Formula

a(n*(n-1)/2) = n!.

Extensions

Entry revised Dec 31 2005

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

Views

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.

A275611 a(0) = 1, a(n) = n * a(n-1)^2 + a(n-1).

Original entry on oeis.org

1, 2, 10, 310, 384710, 740009305210, 3285682630785061608169810, 75569972451698504356522006689642008796426176222510, 45686565890803766858880247710072390769807010129716258796255559717444312778982342222557219570421823310
Offset: 0

Views

Author

Daniel Suteu, Dec 24 2016

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n - 1] (n*a[n - 1] + 1); a[0] = 1; Array[a, 9, 0] (* Robert G. Wilson v, Dec 24 2016 *)
  • PARI
    a(n) = if(n==0, 1, n * a(n-1)^2 + a(n-1));

Formula

a(n) ~ c^(2^n) / n, where c = 2.49513423157979814033214501746049499374370018515926861932281537537499551261588753466... - Vaclav Kotesovec, Jan 28 2019
a(n) = A295391(n)*a(n-1), n >= 1. - N. J. A. Sloane, Sep 03 2024

A347289 Number of independent sets in the binomial tree of order n.

Original entry on oeis.org

2, 3, 8, 60, 3456, 11612160, 132090377011200, 17175244766164688547348480000, 291347192866832125410134687322211469174161539072000000000, 84034354923469245337680441503007090893711465882978424632224243601869256327175152475648504794972160000000000000000
Offset: 0

Views

Author

Kevin Ryde, Sep 26 2021

Keywords

Comments

The binomial tree of order 0 is a single root vertex and order n>=1 is an order n-1 with another order n-1 joined as a subtree of the root.
Going by induction with this construction shows the number of sets where the root is in the set is with(n) = A052129(n), and the number where the root is not in the set is without(n) = A088679(n+1).
Among the total a(n), the respective proportions are without(n)/a(n) = (n+1)/(n+2) and with(n)/a(n) = 1/(n+2).
Also, a(n-1) is the number of maximum independent sets in binomial tree order n.
Tree n can be constructed from tree n-1 by adding a new child under each vertex. Each independent set in tree n-1 corresponds one-to-one with a maximum independent set in tree n by putting each new child in or out of the set opposite to its parent.

Examples

			For n=5, the product formula is a(5) = 7 * 5 * 4^2 * 3^4 * 2^8 = 11612160.
		

Crossrefs

Programs

  • PARI
    a(n) = my(P=1); for(k=2,n,P=sqr(P)*k); (n+2)*P;

Formula

a(n) = (n+2) * Product_{k=2..n} k^(2^(n-k)).
a(n) = A052129(n) + A088679(n+1).
a(n) = a(n-1)^2 - A052129(n-1)^2.
Previous Showing 11-18 of 18 results.