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.

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A126196 Numbers k such that gcd(A001008(k), A001008(floor(k/2))) > 1.

Original entry on oeis.org

7, 546, 1092, 1755, 3510, 4896, 52447, 670668
Offset: 1

Views

Author

Max Alekseyev and Tanya Khovanova, Mar 07 2007, corrected Mar 10 2007

Keywords

Comments

Note a connection to the Wieferich primes A001220: a(2) = (A001220(1) - 1)/2, a(3) = A001220(1) - 1, a(4) = (A001220(2) - 1)/2, a(5) = A001220(2) - 1. [Comment regarding a(2) added by Kevin J. Gomez, Jul 11 2017]
a(9) > 840000. - Giovanni Resta, May 13 2016

Crossrefs

The corresponding GCDs are given by A126197.

Programs

  • Mathematica
    Select[Range[5000], GCD @@ Numerator@ HarmonicNumber@{#, Floor[#/2]} > 1 &] (* Giovanni Resta, May 13 2016 *)
  • PARI
    a001008(n)=numerator(sum(i=1, n, 1/i))
    for(n=1, 1e6, if(gcd(a001008(n), a001008(n/2)) > 1, print1(n, ", "))) \\ Felix Fröhlich, Aug 08 2014

Extensions

a(8) from Giovanni Resta, May 13 2016

A117731 Numerator of the fraction n*Sum_{k=1..n} 1/(n+k).

Original entry on oeis.org

1, 7, 37, 533, 1627, 18107, 237371, 95549, 1632341, 155685007, 156188887, 3602044091, 18051406831, 54260455193, 225175759291, 13981692518567, 14000078506967, 98115155543129, 3634060848592973, 3637485804655193
Offset: 1

Views

Author

Alexander Adamchuk, Apr 14 2006

Keywords

Comments

a(n) almost always equals A082687(n), but differs for n in A125740.
p divides a((p-1)/3) for primes p in A002476, that is, primes of form 6*n + 1. - Alexander Adamchuk, Jul 16 2006

Examples

			The first few fractions are 1/2, 7/6, 37/20, 533/210, 1627/504, 18107/4620, 237371/51480, ... = A117731/A296519.
For n=2, the n X n Hilbert matrix is
  1 1/2
  1/2 1/3
Thus, a(2) = numerator(1 + 1/2 + 1/2 + 1/3) = numerator(7/3) = 7.
The n X n Hilbert matrix begins as follows:
    1 1/2 1/3 1/4  1/5  1/6  1/7  1/8 ...
  1/2 1/3 1/4 1/5  1/6  1/7  1/8  1/9 ...
  1/3 1/4 1/5 1/6  1/7  1/8  1/9 1/10 ...
  1/4 1/5 1/6 1/7  1/8  1/9 1/10 1/11 ...
  1/5 1/6 1/7 1/8  1/9 1/10 1/11 1/12 ...
  1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
  ...
		

Crossrefs

Programs

  • Magma
    [Numerator(n*(HarmonicNumber(2*n) -HarmonicNumber(n))): n in [1..40]]; // G. C. Greubel, Jul 24 2023
    
  • Mathematica
    Numerator[Table[n Sum[1/(n + k), {k, n}], {n, 1, 100}]]
    Numerator[Table[Sum[Sum[1/(i + j - 1), {i, n}], {j, n}], {n, 30}]] (* Alexander Adamchuk, Apr 23 2006 *)
    Table[n (HarmonicNumber[2 n] - HarmonicNumber[n]), {n, 20}] // Numerator (* Eric W. Weisstein, Dec 14 2017 *)
  • PARI
    a(n) = numerator(n*sum(k=1, n, 1/(n+k))); \\ Michel Marcus, Dec 14 2017
    
  • SageMath
    [numerator(n*(harmonic_number(2*n,1) - harmonic_number(n,1))) for n in range(1,41)] # G. C. Greubel, Jul 24 2023

Formula

a(n) = numerator(n*Sum_{k=1..n} 1/(n+k)).
a(n) = numerator(n*(Psi(2*n+1) - Psi(n+1))).
a(n) = numerator(n*Sum_{k=1..2*n} (-1)^(k+1)/k).
a(n) = numerator(n*A058313(2*n)/A058312(2*n)).
a(n) = numerator(Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1)), which is the numerator of the sum of all matrix elements of n X n Hilbert Matrix M(i,j) = 1/(i+j-1), (i,j = 1..n). The denominator is A117664(n). - Alexander Adamchuk, Apr 23 2006

Extensions

Various sections edited by Petros Hadjicostas and Michel Marcus, May 07 2020

A075829 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)*x + c(n))/(d(n)*x + a(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

1, 0, 1, 1, 5, 13, 23, 101, 307, 641, 893, 7303, 9613, 97249, 122989, 19793, 48595, 681971, 818107, 13093585, 77107553, 66022193, 76603673, 1529091919, 1752184789, 7690078169, 8719737569, 23184641107, 3721854001, 96460418429
Offset: 1

Views

Author

Benoit Cloitre, Oct 14 2002

Keywords

Comments

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]
Difference between the denominator and the numerator of the (n-1)-th alternating harmonic number Sum_{k=1..n-1} (-1)^(k+1)*1/k = A058313(n-1)/A058312(n-1). - Alexander Adamchuk, Jul 22 2006
From Petros Hadjicostas, May 06 2020: (Start)
Inspired by Michael Somos's result below, we established the following formulas (valid for n >= 2). All the denominators in the first three formulas are equal to A334958(n).
b(n) = A024167(n)/gcd(A024167(n-1), A024167(n)).
c(n) = A024168(n)/gcd(A024168(n-1), A024168(n)).
d(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).
b(n) + c(n) = n*(d(n) + a(n)).
u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)). (End)

Crossrefs

Cf. A075827 (= b), A075828 (= c), A075830 (= d).

Programs

  • Mathematica
    Denominator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]]-Numerator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]] (* Alexander Adamchuk, Jul 22 2006 *)
  • PARI
    u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
    a(n) = polcoeff(denominator(u(n)), 0, x);

Formula

a(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)). - Michael Somos, Oct 29 2002
From Alexander Adamchuk, Jul 22 2006: (Start)
a(n) = A058312(n-1) - A058313(n-1) for n > 1 with a(1) = 1.
a(n) = denominator(Sum_{k=1..n-1} (-1)^(k+1)*1/k) - numerator(Sum_{k=1..n-1}(-1)^(k+1)*1/k). (End)

Extensions

Name edited by Petros Hadjicostas, May 06 2020

A121594 Numbers k such that k does not divide the denominator of the k-th alternating Harmonic number.

Original entry on oeis.org

15, 28, 75, 77, 104, 187, 196, 203, 210, 222, 228, 235, 238, 328, 345, 375, 551, 620, 847, 888, 1036, 1107, 1204, 1349, 1352, 1372, 1391, 1430, 1457, 1469, 1470, 1498, 1666, 1687, 1855, 1875, 2133, 2301, 2425, 2440, 2556, 2678, 2948, 3179, 3337, 3477
Offset: 1

Views

Author

Alexander Adamchuk, Aug 09 2006

Keywords

Comments

Indices k such that A119788(k) is not equal to 1.
Also indices k such that numerators of k*H'(k) = A119787(k) and H'(k) = A058313(k) are different (H'(k) is the alternating harmonic number H'(k) = Sum_{j=1..k} (-1)^(j+1)*1/j). The ratio of numerators A119787(k)/A058313(k) for k = 1..400 is given in A119788(k). A121595(k) = A119788(a(k)) is the compressed version of A119788(k) (all 1 entries are excluded).

Crossrefs

Cf. A058312 = Denominator of the n-th alternating harmonic number, Sum_{k=1..n} (-1)^(k+1)/k. A074791 = numbers k such that k does not divide the denominator of the k-th Harmonic number.

Programs

  • Mathematica
    Do[H=Sum[(-1)^(i+1)*1/i, {i, 1, n}]; a=Numerator[n*H]; b=Numerator[H]; If[ !Equal[a,b],Print[{n,a/b}]],{n,1,6000}]
    f=0;Do[f=f+(-1)^(n+1)/n;If[ !IntegerQ[Denominator[f]/n],Print[n]],{n,1,100}] (* Alexander Adamchuk, Jan 02 2007 *)

A128672 Numbers m such that m^k does not divide the denominator of the m-th generalized harmonic number H(m,k) nor the denominator of the m-th alternating generalized harmonic number H'(m,k), for k = 2.

Original entry on oeis.org

20, 42, 100, 110, 156, 272, 294, 342, 500, 506, 812, 930, 1210, 1332, 1640, 1806, 2028, 2058, 2162, 2500, 2756, 3422, 3660, 4422, 4624, 4970, 5256, 6162, 6498, 6806, 7832, 9312, 10100, 10506, 11026, 11342, 11638, 11772, 12500, 12656, 13310, 14406, 16002, 17030
Offset: 1

Views

Author

Alexander Adamchuk, Mar 20 2007

Keywords

Comments

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.
Sequence contains all geometric progressions of the form (p-1)*p^k for k > 0 and some primes p > 3. Note the factorization of initial terms of {a(n)} = {4*5, 6*7, 4*5^2, 10*11, 12*13, 16*17, 6*7^2, 18*19, 4*5^3, 22*23, 28*29, 30*31, 10*11^2, 36*37, 40*41, 42*43, 12*13^2, 6*7^3, 46*47, 4*5^4, 52*53, 58*59, 60*61, 66*67, 16*17^2, 70*71, 72*73, 78*79, 18*19^2, 82*83, ...}. The smallest term that does not fit this pattern is 11026 = ((149-1)/2) * 149.

Crossrefs

Similar sequences for generalized harmonic numbers with different k: A125581 (k=1), A128673 (k=3), A128674 (k=4), A128675 (k=5); A128676 (k=6).
For the least numbers k > 0 such that k^n does not divide the denominator of H(k,n) nor the denominator of H'(k,n), see A128670. See also A128671(n) = A128670(prime(n)).

Programs

  • Mathematica
    k=2; f=0; g=0; Do[ f=f+1/n^k; g=g+(-1)^(n+1)*1/n^k; kf=Denominator[f]; kg=Denominator[g]; If[ !IntegerQ[kf/n^k] && !IntegerQ[kg/n^k], Print[n] ], {n,1,7000} ]

Extensions

Edited and extended by Max Alekseyev, May 07 2010

A128673 Numbers m such that m^k does not divide the denominator of the m-th generalized harmonic number H(m,k) nor the denominator of the m-th alternating generalized harmonic number H'(m,k), for k = 3.

Original entry on oeis.org

94556602, 141834903, 189113204, 283669806, 450820422
Offset: 1

Views

Author

Alexander Adamchuk, Apr 18 2007

Keywords

Comments

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.
Note that {a(n)} contains the following geometric progressions: ((16843-1)/3)*16843^m found by Max Alekseyev, ((16843-1)/2)*16843^m found by Max Alekseyev, ((16843-1)*2/3)*16843^m, (16843-1)*16843^m, 20826*21647^m found by Max Alekseyev, ((2124679-1)/3)*2124679^m, ((2124679-1)/2)*2124679^m, ((2124679-1)*2/3)*2124679^m, (2124679-1)*2124679^m. Here {16843, 2124679} = A088164 are the only two currently known Wolstenholme Primes: primes p such that {2p-1} choose {p-1} == 1 mod p^4. See more details in Comments at A128672 and A125581.

Crossrefs

Programs

  • Mathematica
    k=3; f=0; g=0; Do[ f=f+1/n^k; g=g+(-1)^(n+1)*1/n^k; kf=Denominator[f]; kg=Denominator[g]; If[ !IntegerQ[kf/n^k] && !IntegerQ[kg/n^k], Print[n] ], {n, 1, 450820422} ]

A128676 Numbers m such that m^k does not divide the denominator of the m-th generalized harmonic number H(m,k) nor the denominator of the m-th alternating generalized harmonic number H'(m,k), for k = 6.

Original entry on oeis.org

20, 100, 110, 156, 161, 272, 342, 345, 500, 506, 812, 930, 1210, 1332, 1640, 1806, 2028, 2162, 2500, 2756, 3051, 3422, 3660, 3703, 4422, 4624, 4970, 5256, 6162, 6498, 6806, 7832, 7935, 9312, 9605, 10100, 10506, 11342, 11638, 11772, 12500, 12656, 13310
Offset: 1

Views

Author

Alexander Adamchuk, Mar 20 2007

Keywords

Comments

Generalized harmonic numbers are defined as H(m,k) = Sum_{j=1..m} 1/j^k. Alternating generalized harmonic numbers are defined as H'(m,k) = Sum_{j=1..m} (-1)^(j+1)/j^k.
Sequence contains all terms of geometric progressions of the form (p-1)*p^k, k > 0, for some primes p >= 5, such as 4*5^k, 7*23^k, 15*23^k, 27*113^k, etc. Note the factorization of initial terms of {a(n)} = {4*5, 4*5^2, 10*11, 12*13, 7*23, 16*17, 18*19, 15*23, 4*5^3, 22*23, 28*29, 30*31, 10*11^2, 36*37, 40*41, 42*43, 12*13^2, 46*47, 4*5^4, 52*53, 27*113, 58*59, 60*61, 7*23^2, ...}. See more details in Comments at A128672 and A125581.

Crossrefs

Programs

  • Mathematica
    k=6; f=0; g=0; Do[ f=f+1/n^k; g=g+(-1)^(n+1)*1/n^k; kf=Denominator[f]; kg=Denominator[g]; If[ !IntegerQ[kf/n^k] && !IntegerQ[kg/n^k], Print[n] ], {n,1,3703} ]

Extensions

Edited and extended by Max Alekseyev, May 08 2010

A075830 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) = (b(n)*x + c(n))/(a(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

Original entry on oeis.org

0, 1, 1, 5, 7, 47, 37, 319, 533, 1879, 1627, 20417, 18107, 263111, 237371, 52279, 95549, 1768477, 1632341, 33464927, 155685007, 166770367, 156188887, 3825136961, 3602044091, 19081066231, 18051406831, 57128792093, 7751493599
Offset: 1

Views

Author

Benoit Cloitre, Oct 14 2002

Keywords

Comments

For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n)-n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by Petros Hadjicostas, May 18 2020]
From Petros Hadjicostas, May 05 2020: (Start)
Given x > 0, u(n) = (A075827(n)*x + A075828(n))/(a(n)*x + A075829(n)) = (b(n)*x + c(n))/(a(n)*x + d(n)) with gcd(gcd(b(n), c(n)), gcd(a(n), d(n))) = 1 for each n >= 1.
Conjecture 1: Define the sequences (A(n): n >= 1) and (B(n): n >= 1) by A(n+1) = n^2/A(n) + 1 for n >= 2 with A(1) = infinity and A(2) = 1, and B(n+1) = n^2/B(n) + 1 for n >= 3 with B(1) = 0, B(2) = infinity, and B(3) = 1. Then a(n) = denominator(A(n)), b(n) = numerator(A(n)), c(n) = numerator(B(n)), and d(n) = denominator(B(n)) (assuming infinity = 1/0). Also, gcd(a(n), d(n)) = 1.
In 2002, Michael Somos claimed that d(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)) for n >= 2. In 2006, N. J. A. Sloane claimed that a(n) = A058313(n-1) for n >= 2 while Alexander Adamchuk claimed that d(n) = A058312(n-1) - A058313(n-1) for n >= 2.
Conjecture 2: a(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).
Conjecture 3: b(p) = a(p+1) for p = 1 or prime. In general, it seems that b(n) = A048671(n)*a(n+1) for all n for which A048671(n) < n.
Conjecture 4: c(n) = n*(a(n) + d(n)) - b(n) for n >= 1. (End)
All conjectures are proved in the link below except for the second part of Conjecture 3. - Petros Hadjicostas, May 21 2020

Crossrefs

Apart from the leading term, same as A058313.
Cf. A075827 (= b), A075828 (= c), A075829 (= d).

Programs

  • PARI
    u(n)=if(n<2,x,(n-1)^2/u(n-1)+1);
    a(n)=polcoeff(denominator(u(n)),1,x);

Extensions

Name edited by Petros Hadjicostas, May 04 2020

A334958 GCD of consecutive terms of the factorial times the alternating harmonic series.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 48, 144, 1440, 1440, 17280, 17280, 241920, 18144000, 145152000, 145152000, 2612736000, 2612736000, 10450944000, 219469824000, 4828336128000, 4828336128000, 115880067072000, 579400335360000, 15064408719360000, 135579678474240000, 26573616980951040000, 26573616980951040000
Offset: 1

Views

Author

Petros Hadjicostas, May 17 2020

Keywords

Comments

For n = 1..14, we have a(n) = A025527(n), but a(15) = 18144000 <> 3628800 = A025527(15).
It appears that A025527(n) | a(n) for all n >= 1 and A025527(n) = a(n) for infinitely many n. In addition, it seems that a(n)/a(n-1) = A048671(n) for infinitely many n >= 2. However, I have not established these claims.
This sequence appears in formulas for sequences A075827, A075828, A075829, and A075830 (the first one of which was established in 2002 by Michael Somos).
Conjecture: a(n) = n! * Product_{p <= n} p^min(0, v_p(H'(n))), where the product ranges over primes p, H'(n) = Sum_{k=1..n} (-1)^(k+1)/k, and v_p(r) is the p-adic valuation of rational r (checked for n < 1100).

Examples

			A024167(4) = 4!*(1 - 1/2 + 1/3 - 1/4) = 14, A024167(5) = 5!*(1 - 1/2 + 1/3 - 1/4 + 1/5) = 94, A024168(4) = 4!*(1/2 - 1/3 + 1/4) = 10, and A024168(5) = 5!*(1/2 - 1/3 + 1/4 - 1/5) = 26. Then a(4) = gcd(14, 94) = gcd(10, 26) = gcd(14, 4!) = gcd(10, 4!) = gcd(14, 10) = 2.
		

Crossrefs

Cf. A056612 (similar sequence for the harmonic series).

Programs

  • Maple
    b:= proc(n) b(n):= (-(-1)^n/n +`if`(n=1, 0, b(n-1))) end:
    a:= n-> (f-> igcd(b(n)*f, f))(n!):
    seq(a(n), n=1..30);  # Alois P. Heinz, May 18 2020
  • Mathematica
    b[n_] := b[n] = -(-1)^n/n + If[n == 1, 0, b[n-1]];
    a[n_] := GCD[b[n] #, #]&[n!];
    Array[a, 30] (* Jean-François Alcover, Oct 27 2020, after Alois P. Heinz *)
  • SageMath
    def A():
        a, b, n = 1, 1, 2
        while True:
            yield gcd(a, b)
            b, a = a, a + b * n * n
            n += 1
    a = A(); print([next(a) for  in range(29)]) # _Peter Luschny, May 19 2020

Formula

a(n) = gcd(A024167(n+1), A024167(n)) = gcd(A024168(n+1), A024168(n)) = gcd(A024167(n), n!) = gcd(A024168(n), n!) = gcd(A024167(n), A024168(n)).

A117664 Denominator of the sum of all elements in the n X n Hilbert matrix M(i,j) = 1/(i+j-1), where i,j = 1..n.

Original entry on oeis.org

1, 3, 10, 105, 252, 2310, 25740, 9009, 136136, 11639628, 10581480, 223092870, 1029659400, 2868336900, 11090902680, 644658718275, 606737617200, 4011209802600, 140603459396400, 133573286426580, 5215718803323600
Offset: 1

Views

Author

Alexander Adamchuk, Apr 11 2006

Keywords

Comments

Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) = A117731(n) / A117664(n) = 2n * H'(2n) = 2n * A058313(2n) / A058312(2n), where H'(2n) is 2n-th alternating sign Harmonic Number. H'(2n) = H(2n) - H(n), where H(n) is n-th Harmonic Number. - Alexander Adamchuk, Apr 23 2006

Examples

			For n=2, the 2 X 2 Hilbert matrix is [1, 1/2; 1/2, 1/3], so a(2) = denominator(1 + 1/2 + 1/2 + 1/3) = denominator(7/3) = 3.
The n X n Hilbert matrix begins:
    1 1/2 1/3 1/4  1/5  1/6  1/7  1/8 ...
  1/2 1/3 1/4 1/5  1/6  1/7  1/8  1/9 ...
  1/3 1/4 1/5 1/6  1/7  1/8  1/9 1/10 ...
  1/4 1/5 1/6 1/7  1/8  1/9 1/10 1/11 ...
  1/5 1/6 1/7 1/8  1/9 1/10 1/11 1/12 ...
  1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
  ...
		

Crossrefs

Programs

  • Mathematica
    Table[Denominator[Sum[1/(i + j - 1), {i, n}, {j, n}]], {n, 30}]

Formula

a(n) = A111876(n-1)/n.
a(n) = denominator( Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) ). Numerator is A117731(n). - Alexander Adamchuk, Apr 23 2006
a(n) = denominator( Sum_{k=1..n} (2*k)/(n+k) ). - Peter Bala, Oct 10 2021
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