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 33 results. Next

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387
Offset: 1

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Author

Keywords

Comments

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

Examples

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
  • H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

Programs

  • GAP
    List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018
  • Haskell
    import Data.Ratio ((%), numerator)
    a001008 = numerator . sum . map (1 %) . enumFromTo 1
    a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016
    
  • Maple
    A001008 := proc(n)
        add(1/k,k=1..n) ;
        numer(%) ;
    end proc:
    seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016
  • Mathematica
    Table[Numerator[HarmonicNumber[n]], {n, 30}]
    (* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
    Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
    Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)
  • PARI
    A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009
    
  • PARI
    H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019
    
  • Python
    from sympy import Integer
    [sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017
    
  • Sage
    def harmonic(a, b):  # See the F. Johansson link.
        if b - a == 1:
            return 1, a
        m = (a+b)//2
        p, q = harmonic(a,m)
        r, s = harmonic(m,b)
        return p*s+q*r, q*s
    def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
    [A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012
    

Formula

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Extensions

Edited by Max Alekseyev, Oct 21 2011
Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A103345 Numerator of Sum_{k=1..n} 1/k^6 = Zeta(6,n).

Original entry on oeis.org

1, 65, 47449, 3037465, 47463376609, 47464376609, 5584183099672241, 357389058474664049, 260537105518334091721, 52107472322919827957, 92311616995117182948130877, 92311647383100199924330877, 445570781131605573859221176881493, 445570839299219762020391212081493
Offset: 1

Views

Author

Wolfdieter Lang, Feb 15 2005

Keywords

Comments

For the rationals Zeta(k,n) for k = 1..10 and n = 1..20, see the W. Lang link.
a(n) gives the partial sum, Zeta(6,n), of Euler's (later Riemann's) Zeta(6). Zeta(k,n), k >= 2, is sometimes also called H(k,n) because for k = 1 these would be the harmonic numbers A001008/A002805. However, H(1,n) does not give partial sums of a convergent series.

Examples

			The first few fractions are 1, 65/64, 47449/46656, 3037465/2985984, 47463376609/46656000000, ... = A103345/A103346. - _Petros Hadjicostas_, May 10 2020
		

Crossrefs

Cf. A013664, A291456. For the denominators, see A103346.

Programs

Formula

a(n) = numerator(Sum_{k=1..n} 1/k^6) = numerator(A291456(n)/(n!)^6).
G.f. for rationals Zeta(6, n): polylogarithm(6, x)/(1-x).
Zeta(6, n) = (1/945)*Pi^6 - psi(5, n+1)/5!, see eq. 6.4.3 with m = 5, p. 260, of the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 03 2013

A099828 Numerator of the generalized harmonic number H(n,5) = Sum_{k=1..n} 1/k^5.

Original entry on oeis.org

1, 33, 8051, 257875, 806108207, 268736069, 4516906311683, 144545256245731, 105375212839937899, 105376229094957931, 16971048697474072945481, 16971114472329088045481, 6301272372663207205033976933
Offset: 1

Views

Author

Alexander Adamchuk, Oct 27 2004

Keywords

Comments

From Alexander Adamchuk, Nov 07 2006: (Start)
a(n) is prime for n = {23, 25, 85, 147, 167, ...}.
There is a Wolstenholme-like theorem: p divides a(p-1) for prime p and p^2 divides a(p-1) for prime p > 7.
Also, p^3 divides a(p-1) for prime p = 5; p divides a((p-1)/2) for prime p = 37; p divides a((p-1)/3) for prime p = 37; p divides a((p-1)/4) for prime p = 37; p divides a((p-1)/5) for prime p = 11; p^2 divides a((p-1)/6) for prime p = 37; p divides a((p+1)/4) for prime p = 83; p divides a((p+1)/5) for prime p = 29; and p divides a((p+1)/6) for prime p = 11. (End)
See the Wolfdieter Lang link for information about Zeta(k, n) = H(n, k) with the rationals for k = 1..10, g.f.s, and polygamma formulas. - Wolfdieter Lang, Dec 03 2013

Examples

			H(n,5) = {1, 33/32, 8051/7776, 257875/248832, ... } = A099828/A069052.
For example, a(2) = numerator(1 + 1/2^5) = numerator(33/32) = 33 and a(3) = numerator(1 + 1/2^5 + 1/3^5) = numerator(8051/7776) = 8051. [Edited by _Petros Hadjicostas_, May 10 2020]
		

Crossrefs

Denominators are A069052.
A099827 = H(n,5) multiplied by (n!)^5.

Programs

  • Mathematica
    Numerator[Table[Sum[1/k^5, {k, 1, n}], {n, 1, 20}]]
    Numerator[Table[HarmonicNumber[n, 5], {n, 1, 20}]]
    Table[Numerator[Sum[1/k^5,{k,1,n}]],{n,1,100}] (* Alexander Adamchuk, Nov 07 2006 *)
  • PARI
    a(n) = numerator(sum(k=1, n, 1/k^5)); \\ Michel Marcus, May 10 2020

Formula

a(n) = numerator(Sum_{k=1..n} 1/k^5) = numerator(HarmonicNumber[n, 5]).

A120296 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^4.

Original entry on oeis.org

1, 15, 1231, 19615, 12280111, 4090037, 9824498837, 157151464517, 38193952437631, 7637983935923, 111835788321880643, 111830093529238643, 3194097388508809394723, 3194009594644356242723, 15970381078317764649391
Offset: 1

Views

Author

Alexander Adamchuk, Jul 10 2006

Keywords

Comments

p divides a(p-1) for prime p > 2 - similar to Wolstenholme's theorem for A007406(n) (= numerator of Sum_{k=1..n} 1/k^2) and for A007410(n) (= numerator of Sum_{k=1..n} 1/k^4).
Lim_{n -> infinity} a(n)/A334585(n) = A267315 = (7/8)*A013662. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 15/16, 1231/1296, 19615/20736, 12280111/12960000, 4090037/4320000, 9824498837/10372320000, ... = A120296/A334585. - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Cf. A007406, A007410, A013662, A119682, A267315, A334585 (denominators).

Programs

  • Mathematica
    Numerator[Table[Sum[(-1)^(k+1)/k^4,{k,1,n}],{n,1,20}]]
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^4)); \\ Michel Marcus, May 07 2020

Formula

a(n) = numerator(Sum_{k=1..n} (-1)^(k+1)/k^4).

Extensions

Name edited by Petros Hadjicostas, May 07 2020

A007480 a(n) = denominator of sum_{k=1..n} k^(-4).

Original entry on oeis.org

1, 16, 1296, 20736, 12960000, 12960000, 31116960000, 497871360000, 40327580160000, 40327580160000, 590436101122560000, 590436101122560000, 16863445484161436160000, 16863445484161436160000, 16863445484161436160000, 269815127746582978560000
Offset: 1

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Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A007410.

Programs

  • Mathematica
    Denominator[Accumulate[1/Range[20]^4]] (* Harvey P. Dale, Dec 25 2013 *)

Formula

a(n) = denominator of (Pi^4/90 - zeta(4, n+1)). - Arkadiusz Wesolowski, Nov 23 2012
Denominators of coefficients in expansion of PolyLog(4, x)/(1 - x). - Ilya Gutkovskiy, Apr 10 2017

A136675 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^3.

Original entry on oeis.org

1, 7, 197, 1549, 195353, 194353, 66879079, 533875007, 14436577189, 14420574181, 19209787242911, 19197460851911, 42198121495296467, 6025866788581781, 6027847576222613, 48209723660000029, 236907853607882606477
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

a(n) is prime for n in A136683.
Lim_{n -> infinity} a(n)/A334582(n) = A197070. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 7/8, 197/216, 1549/1728, 195353/216000, 194353/216000, 66879079/74088000, 533875007/592704000, ... = a(n)/A334582(n). - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Programs

  • Maple
    map(numer,ListTools:-PartialSums([seq((-1)^(k+1)/k^3, k=1..100)])); # Robert Israel, Nov 09 2023
  • Mathematica
    (* Program #1 *) Table[Numerator[Sum[(-1)^(k+1)/k^3, {k,1,n}]], {n,1,50}]
    (* Program #2 *) Numerator[Accumulate[Table[(-1)^(k+1) 1/k^3, {k,50}]]] (* Harvey P. Dale, Feb 12 2013 *)
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^3)); \\ Michel Marcus, May 07 2020

A136677 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^6.

Original entry on oeis.org

1, 63, 45991, 2942695, 45982595359, 5109066151, 601081707598999, 38469080386820311, 252396118308232060471, 252395862211967012407, 447134922152359540530757327, 447134770212444455649757327, 2158234586764514215343657417779543, 308319185132349039219686748825649
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

p divides a(p-1) for prime p > 2. a(n) is prime for n = {19, 47, 164, ...} = A136686.
Lim_{n -> infinity} a(n)/A334605(n) = A275703 = (31/32)*A013664. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 63/64, 45991/46656, 2942695/2985984, 45982595359/46656000000, 5109066151/5184000000, ... = a(n)/A334605(n). - _Petros Hadjicostas_, May 07 2020
		

Crossrefs

Programs

  • Mathematica
    Table[ Numerator[ Sum[ (-1)^(k+1)/k^6, {k,1,n} ] ], {n,1,30} ]
    Accumulate[Table[(-1)^(k+1)/k^6,{k,20}]]//Numerator (* Harvey P. Dale, Aug 21 2023 *)

A136676 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^5.

Original entry on oeis.org

1, 31, 7565, 241837, 755989457, 755889457, 12705011703799, 406547611705943, 98792790681344149, 98791774426324117, 15910615688635928566967, 15910549913780913466967, 5907492176026179821253778331
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

a(n) is prime for n in A136685.
Lim_{n -> infinity} a(n)/A334604(n) = A267316 = (15/16)*A013663. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 31/32, 7565/7776, 241837/248832, 755989457/777600000, 755889457/777600000, ... = a(n)/A334604(n). - _Petros Hadjicostas_, May 07 2020
		

Crossrefs

Programs

  • Mathematica
    Table[ Numerator[ Sum[ (-1)^(k+1)/k^5, {k,1,n} ] ], {n,1,30} ]
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^5)); \\ Michel Marcus, May 07 2020

A119722 Numerator of generalized harmonic number H(p-1,p)= Sum[ 1/k^p, {k,1,p-1}] divided by p^3 for prime p>3.

Original entry on oeis.org

2063, 2743174627, 19563315706517008974432827112201617, 2597378078067393746941400113704449589199274012223316613, 777478358612529699991463948563778410644748121498526065585976638854277886379480749840301120148933
Offset: 3

Views

Author

Alexander Adamchuk, Jun 13 2006

Keywords

Comments

Generalized harmonic number is H(n,m)= Sum[ 1/k^m, {k,1,n} ]. The numerator of generalized harmonic number H(p-1,p) is divisible by p^3 for prime p>3.

Examples

			Prime[3] = 5.
a(3) = numerator[ 1 + 1/2^5 + 1/3^5 + 1/4^5 ] / 5^3 = 257875/125 = 2063.
Prime[4] = 7
a(4) = numerator[ 1 + 1/2^7 + 1/3^7 + 1/4^7 + 1/5^7 + 1/6^7 ] / 7^3 = 2743174627.
		

Crossrefs

Programs

  • Mathematica
    Numerator[Table[Sum[1/k^Prime[n],{k,1,Prime[n]-1}],{n,3,9}]]/Table[Prime[n]^3,{n,3,9}]

Formula

a(n) = numerator[ Sum[ 1/k^Prime[n], {k,1,Prime[n]-1} ]] / Prime[n]^3 for n>2.

A136681 Numbers k such that A058313(k) is prime.

Original entry on oeis.org

3, 4, 5, 6, 9, 10, 13, 16, 17, 18, 37, 43, 58, 121, 124, 126, 137, 203, 247, 283, 285, 286, 289, 317, 424, 508, 751, 790, 937, 958, 1066, 1097, 1151, 1166, 1194, 1199, 1235, 1414, 1418, 1460, 1498, 1573, 2090, 2122, 2691, 2718, 3030, 3426, 3600, 3653, 3737
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A058313(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,317} ]
  • PARI
    isok(n) = isprime(numerator(sum(k=1, n, (-1)^(k+1)/k))); \\ Michel Marcus, Mar 14 2019

Extensions

a(25)-a(30) from James R. Buddenhagen, Sep 22 2015
a(31)-a(51) from Amiram Eldar, Mar 14 2019
Showing 1-10 of 33 results. Next