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.

A051179 a(n) = 2^(2^n) - 1.

Original entry on oeis.org

1, 3, 15, 255, 65535, 4294967295, 18446744073709551615, 340282366920938463463374607431768211455, 115792089237316195423570985008687907853269984665640564039457584007913129639935
Offset: 0

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Author

Alan DeKok (aland(AT)ox.org)

Keywords

Comments

In a tree with binary nodes (0, 1 children only), the maximum number of unique child nodes at level n.
Number of binary trees (each vertex has 0, or 1 left, or 1 right, or 2 children) such that all leaves are at level n. Example: a(1) = 3 because we have (i) root with a left child, (ii) root with a right child and (iii) root with two children. a(n) = A000215(n) - 2. - Emeric Deutsch, Jan 20 2004
Similarly, this is also the number of full balanced binary trees of height n. (There is an obvious 1-to-1 correspondence between the two sets of trees.) - David Hobby (hobbyd(AT)newpaltz.edu), May 02 2010
Partial products of A000215.
The first 5 terms n (only) have the property that phi(n)=(n+1)/2, where phi(n) = A000010(n) is Euler's totient function. - Lekraj Beedassy, Feb 12 2007
If A003558(n) is of the form 2^n and A179480(n+1) is even, then (2^(A003558(n)) - 1) is in A051179. Example: A003558(25) = 8 with A179480(25) = 4, even. Then (2^8 - 1) = 255. - Gary W. Adamson, Aug 20 2012
For any odd positive a(0), the sequence defined by a(n) = a(n-1) * (a(n-1) + 2) gives a constructive proof that there exist integers with at least n distinct prime factors, e.g., a(n), since omega(a(n)) >= n. As a corollary, this gives a constructive proof of Euclid's theorem stating that there are infinitely many primes. - Daniel Forgues, Mar 07 2017
From Sergey Pavlov, Apr 24 2017: (Start)
I conjecture that, for n > 7, omega(a(n)) > omega(a(n-1)) > n.
It seems that the largest prime divisor p(n+1) of a(n+1) is always bigger than the largest prime divisor of a(n): p(n+1) > p(n). For 3 < n < 8, p(n+1) > 100 * p(n).
(End)
It appears that a(n) is the integer whose bits indicate the possible subset sums of the first n powers of two. For another example, see the calculation for primes at A368491 - Yigit Oktar, Mar 20 2025

Examples

			15 = 3*5; 255 = 3*5*17; 65535 = 3*5*17*257; ... - _Daniel Forgues_, Mar 07 2017
		

References

  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 4.

Crossrefs

Programs

Formula

a(n) = A000215(n) - 2.
a(n) = (a(n-1) + 1)^2 - 1, a(0) = 1. [ or a(n) = a(n-1)(a(n-1) + 2) ].
1 = 2/3 + 4/15 + 16/255 + 256/65535 + ... = Sum_{n>=0} A001146(n)/a(n+1) with partial sums: 2/3, 14/15, 254/255, 65534/65535, ... - Gary W. Adamson, Jun 15 2003
a(n) = b(n-1) where b(1)=1, b(n) = Product_{k=1..n-1} (b(k) + 2). - Benoit Cloitre, Sep 13 2003
A136308(n) = A007088(a(n)). - Jason Kimberley, Dec 19 2012
A000215(n) = a(n+1) / a(n). - Daniel Forgues, Mar 07 2017
Sum_{n>=0} 1/a(n) = A048649. - Amiram Eldar, Oct 27 2020

A051158 Decimal expansion of Sum_{n >= 0} 1/(2^2^n+1).

Original entry on oeis.org

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

Views

Author

Robert Lozyniak (11(AT)onna.com)

Keywords

Examples

			0.59606317211782167942...
		

Crossrefs

Terms in continued fraction: A159243. - Enrique Pérez Herrero, Nov 17 2009

Programs

  • Mathematica
    RealDigits[Sum[1/(2^2^n + 1), {n, 0, 10}], 10, 111][[1]] (* Robert G. Wilson v, Jul 03 2014 *)
  • PARI
    /* divisionless routine from fxtbook */
    s2(y, N=7)=
    { local(in, y2, A); /* as powerseries correct to order = 2^N-1 */
        in = 1; /* 1+y+y^2+y^3+...+y^(2^k-1) */
        A = y; for(k=2, N, in *= (1+y); y *= y; A += y*(in + A); );
        return( A ); }
    a=0.5*s2(0.5) /* computation of the constant 0.596063172117821... */
    /* Joerg Arndt, Apr 15 2010 */
    
  • PARI
    suminf(n=0, 1/(2^2^n+1)) \\ Michel Marcus, May 15 2020

Formula

Equals (1/2) * Sum_{k>=1} A000120(k)/2^k (S. Audinarayana Moorthy, 1974). - Amiram Eldar, May 15 2020
Equals 1 - Sum_{n>=1} A007814(n)/2^n = 2/3 - Sum_{n>=1} A007814(n)/4^n = 3/5 - Sum_{n>=1} A007814(n)/16^n. - Amiram Eldar, Nov 06 2020

A384251 The number of integers k from 1 to n such that the greatest divisor of k that is an infinitary divisor of n is odd.

Original entry on oeis.org

1, 1, 3, 3, 5, 3, 7, 4, 9, 5, 11, 9, 13, 7, 15, 15, 17, 9, 19, 15, 21, 11, 23, 12, 25, 13, 27, 21, 29, 15, 31, 16, 33, 17, 35, 27, 37, 19, 39, 20, 41, 21, 43, 33, 45, 23, 47, 45, 49, 25, 51, 39, 53, 27, 55, 28, 57, 29, 59, 45, 61, 31, 63, 48, 65, 33, 67, 51, 69
Offset: 1

Views

Author

Amiram Eldar, May 23 2025

Keywords

Crossrefs

Analogous sequences: A026741, A384055.
The number of integers k from 1 to n such that the greatest divisor of k that is an infinitary divisor of n is: A384247(1), A384249 (squarefree), A384250 (powerful), this sequence (odd), A384252 (power of 2).

Programs

  • Mathematica
    a[n_] := n * If[OddQ[n], 1, 1 - 1/2^(2^(IntegerExponent[IntegerExponent[n, 2], 2]))]; Array[a, 100]
  • PARI
    a(n) = n * if(n%2, 1, (1 - 1/(1 << (1 << valuation(valuation(n, 2), 2)))));

Formula

Multiplicative with a(2^e) = 2^e * (1 - 1/2^A006519(e)), and a(p^e) = p^e if p is an odd prime.
a(n) = A384247(n)/A384252(n).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = (3 - A048649)/2 = 0.79803158605891083971... .

A048164 a(0)=1, a(n+1)=1+(2^(2^n)+1)*a(n).

Original entry on oeis.org

1, 4, 21, 358, 92007, 6029862760, 25898063359598159721, 477734946799221833229035410333259818858, 162564778457687820218065957445498785826947155451688293007128627114802460256107
Offset: 0

Views

Author

N. J. A. Sloane, E. M. Rains

Keywords

Comments

a(n) = height of lattice of orthogonal arrays with 2^2^n runs.

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[0]==1,a[n]==1+(2^(2^(n-1))+1)a[n-1]},a,{n,10}] (* Harvey P. Dale, Dec 13 2013 *)

Formula

a(n) converges to nearest integer to c*(2^(2^n)-1), where c = 1.403936827882178... (see A048649).

A346192 Decimal expansion of Sum_{k>=0} 1/(2^(2^k)*(2^(2^k) - 1)).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jul 09 2021

Keywords

Comments

This constant is transcendental (Schwarz, 1967).

Examples

			0.58751531886028517686812633660668413283432738560133...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sum[1/((2^(2^n) - 1)*2^(2^n)), {n, 0, 10}], 10, 100][[1]]

Formula

Equals Sum_{k>=2} 1/A087046(k).

A048650 Continued fraction for Sum_{m>=0} 1/(2^2^m - 1).

Original entry on oeis.org

1, 2, 2, 9, 1, 3, 5, 1, 2, 1, 1, 1, 1, 8, 2, 1, 1, 2, 1, 12, 19, 24, 1, 18, 12, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 8, 1, 1, 5, 2, 5, 8, 1, 4, 2, 5, 1, 1, 8, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 12, 18, 1, 7, 2, 1, 1, 2, 4, 1, 5, 4, 2, 1, 1, 1, 1, 1, 4, 64, 14, 1, 6, 3, 1, 6
Offset: 0

Views

Author

Keywords

Examples

			1.40393682788217832057620607413720935453...
1.403936827882178320576206074... = 1 + 1/(2 + 1/(2 + 1/(9 + 1/(1 + ...)))). - _Harry J. Smith_, May 03 2009
		

Crossrefs

Cf. A048164, A048649 (decimal expansion).

Programs

  • PARI
    { allocatemem(932245000); default(realprecision, 21000); x=suminf(m=0, 1/(2^2^m - 1)); x=contfrac(x); for (n=1, 20000, write("b048650.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 07 2009

Extensions

Deleted old PARI program. - Harry J. Smith, May 20 2009
Offset changed by Andrew Howroyd, Aug 03 2024

A346190 Decimal expansion of Sum_{k>=0} 1/(2^(2^(2*k+1)) - 1).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jul 09 2021

Keywords

Comments

This constant is transcendental (Schwarz, 1967).

Examples

			0.50781250046566128730773925781250000000587747175411...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sum[1/(2^(2^(2*n + 1) - 1)), {n, 0, 10}], 10, 100][[1]]

Formula

Equals A076214 - A346191.

A346191 Decimal expansion of Sum_{k>=0} 1/(2^(2^(2*k)) - 1).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jul 09 2021

Keywords

Comments

This constant is transcendental (Schwarz, 1967).

Examples

			1.12503051757812500010842021724855044340074528008699...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sum[1/(2^(2^(2*n) - 1)), {n, 0, 10}], 10, 100][[1]]

Formula

Equals A076214 - A346190.

A386261 a(n) = A001511(A001511(n)), where A001511 is the ruler function.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 17 2025

Keywords

Comments

The first occurrence of k = 1, 2, ... is at n = 2^(2^(k-1) - 1) = A058891(k).
The asymptotic density of the occurrences of k = 1, 2, ... in this sequence is 2^(2^(k-1))/(2^(2^k)-1) = 2/3, 4/15, 16/255, 256/65535, 65536/4294967295, ...

Crossrefs

Programs

  • Mathematica
    f[n_] := IntegerExponent[n, 2] + 1; a[n_] := f[f[n]]; Array[a, 100]
  • PARI
    a(n) = valuation(valuation(n, 2) + 1, 2) + 1;

Formula

a(n) >= 1, with equality if and only if n is in A003159.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{m>=0} 1/(2^(2^m) - 1) = 1.4039368... (A048649).
Showing 1-9 of 9 results.