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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A213998 Numerators of the triangle of fractions read by rows: pf(n,0) = 1, pf(n,n) = 1/(n+1) and pf(n+1,k) = pf(n,k) + pf(n,k-1) with 0 < k < n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 11, 1, 1, 7, 13, 25, 1, 1, 9, 47, 77, 137, 1, 1, 11, 37, 57, 87, 49, 1, 1, 13, 107, 319, 459, 223, 363, 1, 1, 15, 73, 533, 743, 341, 481, 761, 1, 1, 17, 191, 275, 1879, 2509, 3349, 4609, 7129, 1, 1, 19, 121, 1207, 1627, 2131, 2761, 3601, 4861, 7381, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 03 2012

Keywords

Comments

T(n,0) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A188386(n-2) for n > 2;
T(n,n-3) = A124837(n-2) for n > 2;
T(n,n-2) = A027612(n-1) for n > 1;
T(n,n-1) = A001008(n) for n > 0;
T(n,n) = 1;
A214075(n,k) = floor(T(n,k) / A213999(n,k)).

Examples

			Start of triangle pf with corresponding triangles of numerators and denominators:
. 0:                            1
. 1:                         1    1/2
. 2:                     1     3/2    1/3
. 3:                  1    5/2    11/6    1/4
. 4:              1   7/2    13/3    25/12    1/5
. 5:           1    9/2   47/6    77/12   137/60   1/6
. 6:        1  11/2   37/3    57/4    87/10    49/20    1/7
. 7:     1  13/2  107/6  319/12  459/20   223/20  363/140   1/8
. 8:  1  15/2  73/3  533/12  743/15  341/10   481/35   761/280  1/9,
.
. 0:   numerators     1                          1    denominators
. 1:                1  1                        1  2       A213999
. 2:              1   3  1                     1 2  3
. 3:            1   5  11 1                   1 2 6  4
. 4:          1  7  13  25  1                1 2 3  12 5
. 5:        1  9  47  77 137  1             1 2 6 12  60 6
. 6:      1 11  37 57  87  49  1           1 2 3 4 10  20  7
. 7:    1 13 107 319 459 223 363 1        1 2 6 12 20 20 140 8
. 8:  1 15 73 533 743 341 481 761 1,     1 2 3 12 15 10 35 280 9.
		

Crossrefs

Cf. A005408, A188386 (columns).
Cf. A001008, A027612, A124837 (diagonals).
Cf. A213999 (denominators).

Programs

  • Haskell
    import Data.Ratio ((%), numerator, denominator, Ratio)
    a213998 n k = a213998_tabl !! n !! k
    a213998_row n = a213998_tabl !! n
    a213998_tabl = map (map numerator) $ iterate pf [1] where
       pf row = zipWith (+) ([0] ++ row) (row ++ [-1 % (x * (x + 1))])
                where x = denominator $ last row
  • Mathematica
    T[, 0] = 1; T[n, n_] := 1/(n + 1);
    T[n_, k_] := T[n, k] = T[n - 1, k] + T[n - 1, k - 1];
    Table[T[n, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 10 2021 *)

A081528 a(n) = n*lcm{1,2,...,n}.

Original entry on oeis.org

1, 4, 18, 48, 300, 360, 2940, 6720, 22680, 25200, 304920, 332640, 4684680, 5045040, 5405400, 11531520, 208288080, 220540320, 4423058640, 4655851200, 4888643760, 5121436320, 123147264240, 128501493120, 669278610000, 696049754400
Offset: 1

Views

Author

Amarnath Murthy, Mar 27 2003

Keywords

Comments

Denominators in binomial transform of 1/(n + 1)^2. - Paul Barry, Aug 06 2004
Construct a sequence S_n from n sequences b_1, b_2, ..., b_n of periods 1, 2, ..., n, respectively, say, b_1 = [1, 1, ...], b_2 = [1, 2, 1, 2, ...], ..., b_n = [1, 2, 3, ..., n, 1, 2, 3, ..., n, ...], by taking S_n = [b_1(1), b_2(1), ..., b_n(1), b_1(2), b_2(2), ..., b_n(2), ..., b_1(n), b_2(n), ..., b_n(n), ...] (by listing the b_i sequences in rows and taking each column in turn as the next n terms of S_n). Then a(n) is the period of sequence S_n. - Rick L. Shepherd, Aug 21 2006
This is a sequence that goes in strictly ascending order. The related sequence A003418 also goes in ascending order but has consecutive repeated terms. Since n increases, then so too does a(n) even when A003418(n) doesn't. - Alonso del Arte, Nov 25 2012

Examples

			a(2) = 4 because the least common multiple of 1 and 2 is 2, and 2 * 2 = 4.
a(3) = 18 because lcm(1,2,3) = 6, and 3 * 6 = 18.
a(4) = 48 because lcm(1, 2, 3, 4) = 12, and 4 * 12 = 48.
		

Crossrefs

Programs

  • Derive
    a(n) := (n + 1)*LCM(VECTOR(k + 1, k, 0, n)) " Paul Barry, Aug 06 2004 "
    
  • Mathematica
    Table[n*LCM@@Range[n], {n, 30}] (* Harvey P. Dale, Oct 09 2012 *)
  • PARI
    l=vector(35); l[1]=1; print1("1, "); for(n=2,35, l[n]=lcm(l[n-1],n); print1(n*l[n],", ")) \\ Rick L. Shepherd, Aug 21 2006

Formula

a(n) = A003418(n) * n. - Martin Fuller, Jan 03 2006

Extensions

More terms from Paul Barry, Aug 06 2004
Entry revised by N. J. A. Sloane, Jan 15 2006

A115071 Numerator of 1^n/n + 2^n/(n-1) + 3^n/(n-2) +...+ (n-1)^n/2 + n^n/1.

Original entry on oeis.org

1, 9, 94, 3625, 18631, 1120581, 34793764, 5692787001, 29669041771, 30708223774261, 134127439064434, 302304605103335861, 2387352152511746837, 109134149200789179825, 24217460586461892638584
Offset: 1

Views

Author

Alexander Adamchuk, Jun 17 2006

Keywords

Comments

a(p-1) is divisible by p^3 for prime p>3. a(p^2-1) is divisible by p^6 for prime p>3. a(p^3-1) is divisible by p^9 for prime p>3.

Examples

			a(1) = numerator(1/1) = 1.
a(2) = numerator(1/2 + 4/1) = 9.
a(3) = numerator(1/3 + 8/2 + 27/1) = 94.
a(4) = numerator(1/4 + 16/3 + 81/2 + 256/1) = 3625.
		

Crossrefs

Cf. A001008, A027612, A120487 (denominator).

Programs

  • Mathematica
    Table[Numerator[Sum[i^n/(n+1-i),{i,1,n}]],{n,1,20}]

Formula

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

A124838 Denominators of third-order harmonic numbers (defined by Conway and Guy, 1996).

Original entry on oeis.org

1, 2, 6, 4, 20, 10, 70, 56, 504, 420, 4620, 3960, 3432, 6006, 90090, 80080, 1361360, 408408, 369512, 67184, 470288, 1293292, 29745716, 27457584, 228813200, 212469400, 5736673800, 5354228880, 155272637520, 291136195350, 273491577450
Offset: 1

Views

Author

Jonathan Vos Post, Nov 10 2006

Keywords

Comments

Numerators are A124837. All fractions reduced. Thanks to Jonathan Sondow for verifying these calculations. He suggests that the equivalent definition in terms of first order harmonic numbers may be computationally simpler. We are happy with the description of A027612 Numerator of 1/n + 2/(n-1) + 3/(n-2) +...+ (n-1)/2 + n, but baffled by the description of A027611.

Examples

			a(1) = 1 = denominator of 1/1.
a(2) = 2 = denominator of 1/1 + 5/2 = 7/2.
a(3) = 6 = denominator of 7/2 + 13/3 = 47/6.
a(4) = 4 = denominator of 47/6 + 77/12 = 57/4.
a(5) = 20 = denominator of 57/4 + 87/10 = 549/20.
a(6) = 10 = denominator of 549/20 + 223/20 = 341/10
a(7) = 70 = denominator of 341/10 + 481/35 = 3349/70.
a(8) = 1260 = denominator of 3349/70 + 4609/280 = 88327/1260.
a(9) = 45 = denominator of 88327/1260 + 4861/252 = 3844/45.
a(10) = 504 = denominator of 3844/45 + 55991/2520 = 54251/504, or, untelescoping:
a(10) = 504 = denominator of 1/1 + 5/2 + 13/3 + 77/12 + 87/10 + 223/20 + 481/35 + 4609/252 + 4861/252 + 55991/2520 = 54251/504.
		

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, pp. 143 and 258-259, 1996.

Crossrefs

Programs

  • Haskell
    a124838 n = a213999 (n + 2) (n - 1) -- Reinhard Zumkeller, Jul 03 2012
  • Mathematica
    Table[Denominator[(n+2)!/2!/n!*Sum[1/k,{k,3,n+2}]],{n,1,40}] (* Alexander Adamchuk, Nov 11 2006 *)

Formula

A124837(n)/A124838(n) = Sum_{i=1..n} A027612(n)/A027611(n+1).
a(n) = denominator(Sum_{m=1..n} Sum_{L=1..m} Sum_{k=1..L} 1/k).
a(n) = denominator(((n+2)!/(2!*n!)) * Sum_{k=3..n+2} 1/k). - Alexander Adamchuk, Nov 11 2006
a(n) = A213999(n+2,n-1). - Reinhard Zumkeller, Jul 03 2012

Extensions

Corrected and extended by Alexander Adamchuk, Nov 11 2006

A363000 a(n) = numerator(R(n, n, 1)), where R are the rational polynomials R(n, k, x) = Sum_{u=0..k} ( Sum_{j=0..u} x^j * binomial(u, j) * (j + 1)^n ) / (u + 1).

Original entry on oeis.org

1, 5, 19, 188, 1249, 125744, 283517, 303923456, 138604561, 599865008128, 118159023973, 7078040993755136, 155792758736921, 146303841678548271104, 294014633772018349, 64670474732430319157248, 752324747622089633569, 3224753626003393505960919040, 2507759850059601711479669
Offset: 0

Views

Author

Peter Luschny, May 12 2023

Keywords

Comments

R(n, n, 0) are the (0-based) harmonic numbers, R(n, n, -1) are the Bernoulli numbers, and R(n, n, 1) is this sequence in its rational form.

Examples

			a(n) are the numerators of the terms on the main diagonal of the triangle:
[0] 1;
[1] 1,  5/2;
[2] 1,  7/2,  19/2;
[3] 1, 11/2, 121/6,  188/3;
[4] 1, 19/2,  95/2,  369/2,   1249/2;
[5] 1, 35/2, 721/6, 1748/3, 35164/15, 125744/15;
[6] 1, 67/2, 639/2, 3877/2,  18533/2,   76317/2, 283517/2;
		

Crossrefs

Cf. A363001 (denominators), A362999 (odd-indexed denominators), A362998.

Programs

  • Maple
    # For better context we put A362998, A362999, A363000, and A363001 together here.
    R := (n, k, x) -> add(add(x^j*binomial(u, j)*(j+1)^n, j=0..u)/(u + 1), u=0..k):
    ### x = 1 -> this sequence
     for n from 0 to 7 do [n], seq(R(n, k, 1), k = 0..n) od;
     seq(R(n, n, 1), n = 0..9);
     A363000 := n -> numer(R(n, n, 1)): seq(A363000(n), n = 0..10);
     A363001 := n -> denom(R(n, n, 1)): seq(A363001(n), n = 0..20);
     A362999 := n -> denom(R(2*n+1, 2*n+1, 1)): seq(A362999(n), n = 0..11);
     A362998 := n -> add(R(2*n, k, 1), k = 0..2*n): seq(A362998(n), n = 0..9);
    ### x = -1 -> Bernoulli(n, 1)
    # for n from 0 to 9 do [n], seq(R(n, k,-1), k = 0..n) od;
    # seq(R(n, n, -1), n = 0..12); seq(bernoulli(n, 1), n = 0..12);
    ### x = 0 -> Harmonic numbers
    # for n from 0 to 9 do [n], seq(R(n, k, 0), k = 0..n) od;
    # seq(R(n, n, 0), n = 0..9); seq(harmonic(n+1), n = 0..9);

Formula

Sum_{k=0..n} R(n, k, 0) = Sum_{j=0..n} (n-j+1)/(j+1) = (n+2)*Harmonic(n+1)-n-1.
Sum_{k=0..n} R(n, k,-1) = (n + 2 - 0^n) * Bernoulli(n, 1).
Sum_{k=0..2*n} R(2*n, k, 1) = A362998(n).
2*R(n, 1, 1) = A062709(n).

A124880 Primes in A124837.

Original entry on oeis.org

7, 47, 42131, 23763863, 192066102203, 5733412167187, 34745876421709, 185813891783454008069, 171312804637561107990389, 29207630124216024960052176833, 6300447575454970515437116064749
Offset: 1

Views

Author

Alexander Adamchuk, Nov 11 2006

Keywords

Examples

			A124837(n) begins {1, 7, 47, 57, 459, 341, 3349, 3601, 42131, 44441, ...}.
Thus a(1) = 7, a(2) = 47, a(3) = 42131.
		

Crossrefs

A124837 are the numerators of third-order harmonic numbers H(n, (3)).
Corresponding numbers n such that A124837(n) is prime are listed in A124881.

Programs

  • Mathematica
    s=3/2;Do[s=s+1/n;f=Numerator[n*(n-1)/2*(s-3/2)]; If[PrimeQ[f],Print[{n-2,f}]],{n,3,125}]

Extensions

Crossrefs edited by Michel Marcus, Jul 14 2018

A124881 Numbers k such that A124837(k) is prime.

Original entry on oeis.org

2, 3, 9, 15, 25, 27, 33, 45, 55, 67, 70, 93, 94, 97, 112, 113, 125, 137, 189, 193, 212, 232, 262, 273, 281, 381, 453, 528, 670, 677, 742, 743, 827, 996, 1257, 1349, 1402, 1645, 1683, 2110, 2217, 2408, 2480, 2623, 3208, 3517, 3637, 3665, 4571, 4730
Offset: 1

Views

Author

Alexander Adamchuk, Nov 11 2006

Keywords

Examples

			A124837(n) begins {1, 7, 47, 57, 459, 341, 3349, 3601, 42131, 44441, ...}.
Thus a(1) = 2, a(2) = 3, a(3) = 9.
		

Crossrefs

A124837 are the numerators of third-order harmonic numbers H(n, (3)).
Corresponding primes in A124837 are listed in A124880.

Programs

  • Mathematica
    s=3/2;Do[s=s+1/n;f=Numerator[n*(n-1)/2*(s-3/2)]; If[PrimeQ[f],Print[{n-2,f}]],{n,3,1000}]

Extensions

More terms from Stefan Steinerberger, May 09 2007
Crossrefs edited by Michel Marcus, Jul 14 2018

A120286 Numerator of 1/n^2 + 2/(n-1)^2 + 3/(n-2)^2 +...+ (n-1)/2^2 + n.

Original entry on oeis.org

1, 9, 65, 725, 3899, 28763, 419017, 864669, 7981633, 3586319, 200763407, 2649665993, 34899471137, 176508049513, 356606957297, 12234391348253, 209672027529221, 4012917216669239, 15350275129353301, 15443118015171841
Offset: 1

Views

Author

Alexander Adamchuk, Jul 07 2006

Keywords

Comments

p^2 divides a(p-1) for prime p>2. p divides a(p-2) for prime p>3.

Crossrefs

Programs

  • Mathematica
    Numerator[Table[Sum[Sum[1/i^2,{i,1,k}],{k,1,n}],{n,1,30}]]
    Table[-EulerGamma + HarmonicNumber[1 + n, 2] + n*HarmonicNumber[1 + n, 2] - PolyGamma[0, 2 + n], {n, 1, 20}] // Numerator (* Vaclav Kotesovec, May 02 2024 *)
  • Python
    from fractions import Fraction
    def A120286(n): return sum(Fraction(n-i+1,i**2) for i in range(1,n+1)).numerator # Chai Wah Wu, May 01 2024

Formula

a(n) = numerator[Sum[Sum[1/i^2,{i,1,k}],{k,1,n}]].

A160048 Numerator of the Harary number for the path graph P_n.

Original entry on oeis.org

0, 2, 5, 26, 77, 87, 223, 962, 4609, 4861, 55991, 58301, 785633, 811373, 835397, 3431678, 29889983, 30570663, 197698279, 201578155, 41054655, 13920029, 325333835, 990874363, 25128807667, 25472027467, 232222818803, 235091155703
Offset: 1

Views

Author

Eric W. Weisstein, Apr 30 2009

Keywords

Examples

			0, 2, 5, 26/3, 77/6, 87/5, 223/10, 962/35, 4609/140, 4861/126, ...
		

Crossrefs

Programs

  • PARI
    harmonic(n)=sum(k=1,n,1/k);
    a(n)=numerator(2*n*(harmonic(n)-1)); \\ Andrew Howroyd, Oct 31 2017

A354859 a(n) is the numerator of 1/prime(n) + 2/prime(n-1) + 3/prime(n-2) + ... + (n-1)/3 + n/2.

Original entry on oeis.org

1, 4, 71, 124, 11111, 92402, 257189, 43708274, 2344987003, 1855369084, 2729707472269, 11281836318542, 1705853427969059, 120757830191824486, 1124815045783478971, 118422287191742563724, 1291008724583331399881, 113743044027018860349034, 70575236921156825443680027, 8002471039307070610187173702
Offset: 1

Views

Author

Ilya Gutkovskiy, Jun 09 2022

Keywords

Comments

Numerator of a second order prime harmonic number.

Examples

			1/2, 4/3, 71/30, 124/35, 11111/2310, 92402/15015, 257189/34034, ...
		

Crossrefs

Cf. A024451, A027612, A120272, A354860 (denominators).

Programs

  • Mathematica
    Table[Sum[(n - k + 1)/Prime[k], {k, 1, n}], {n, 1, 20}] // Numerator
  • Python
    from fractions import Fraction
    from sympy import prime, primerange
    def a(n): return sum(Fraction(n-i, p) for i, p in enumerate(primerange(1, prime(n)+1))).numerator
    print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Jun 09 2022

Formula

a(n) is the numerator of Sum_{j=1..n} Sum_{i=1..j} 1/prime(i).
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