A081528 -id:A081528 - cp's OEIS frontend

A027446 Triangle read by rows: square of the lower triangular mean matrix.

1, 3, 1, 11, 5, 2, 25, 13, 7, 3, 137, 77, 47, 27, 12, 147, 87, 57, 37, 22, 10, 1089, 669, 459, 319, 214, 130, 60, 2283, 1443, 1023, 743, 533, 365, 225, 105, 7129, 4609, 3349, 2509, 1879, 1375, 955, 595, 280, 7381, 4861, 3601, 2761, 2131, 1627, 1207, 847, 532, 252

Offset: 1

Olivier Gérard

Numerators of nonzero elements of A^2, written as rows using the least common denominator, where A[i,j] = 1/i if j <= i, 0 if j > i. [Edited by M. F. Hasler, Nov 05 2019]

			Triangle starts
     1
     3,    1
    11,    5,    2
    25,   13,    7,    3
   137,   77,   47,   27,   12
   147,   87,   57,   37,   22,   10
  1089,  669,  459,  319,  214,  130,  60
  2283, 1443, 1023,  743,  533,  365, 225, 105
  7129, 4609, 3349, 2509, 1879, 1375, 955, 595, 280
  ... - _Joerg Arndt_, Mar 29 2013

L. Bendersky, Sur la fonction gamma généralisée, Acta Math. 61 (1933), p. 263-322. See p. 295.

The row sums give A081528(n), n>=1.
The column sequences give A025529, A027457, A027458 for j=1..3.
The diagonal sequences give A002944, A027449, A027450.
Cf. A027447, A027448.

rows = 10;
M = MatrixPower[Table[If[j <= i, 1/i, 0], {i, 1, rows}, {j, 1, rows}], 2];
T = Table[M[[n]]*LCM @@ Denominator[M[[n]]], {n, 1, rows}];
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 05 2013, updated May 06 2022 *)

A027446_upto(n)={my(M=matrix(n, n, i, j, (j<=i)/i)^2); vector(n,r,M[r,1..r]*denominator(M[r,1..r]))} \\ M. F. Hasler, Nov 05 2019

The rational matrix A^2, where the matrix A has elements a[i,j] = 1/A002024(i,j), is equal to A119947(i,j)/A119948(i,j).

a(i,j) = lcm(seq(A119948(i,m),m=1..i))*A119947(i,j)/A119948(i,j), 1 <= j =< i and zero otherwise.

Edited by M. F. Hasler, Nov 05 2019

A097344 Numerators in binomial transform of 1/(n+1)^2.

Original entry on oeis.org

1, 5, 29, 103, 887, 1517, 18239, 63253, 332839, 118127, 2331085, 4222975, 100309579, 184649263, 1710440723, 6372905521, 202804884977, 381240382217, 13667257415003, 25872280345103, 49119954154463, 93501887462903, 4103348710010689, 7846225754967739, 75162749477272151

Offset: 0

Paul Barry, Aug 06 2004

Is this identical to A097345? - Aaron Gulliver, Jul 19 2007. The answer turns out to be No - see A134652.
If the putative formula a(n)=A081528(n) sum{k=0..n, binomial(n, k)/(k+1)^2} were true, then this sequence coincides with A097345 according to Mathar's notes. However, the term n=9 in the binomial transform of 1/(n+1)^2 has the denominator 5040=A081528(9)/4=A081528(10)/5. So the formula cannot be true. - M. F. Hasler, Jan 25 2008
a(n) is also the numerator of u(n+1) with u(n) = (1/n)*Sum_{k=1..n} (2^k-1)/k and we have the formula: polylog(2,x/(1-x)) = Sum_{n>=1} u(n)*x^n on the interval [-1/2, 1/2]. - Groux Roland, Feb 01 2009

			The first values of the binomial transform of 1/(n+1)^2 are 1, 5/4, 29/18, 103/48, 887/300, 1517/360, 18239/2940, 63253/6720, 332839/22680, 118127/5040, 2331085/60984, ...

Chai Wah Wu, Table of n, a(n) for n = 0..500
R. J. Mathar, Notes on an attempt to prove that A097344 and A097345 are identical

Cf. A097345, A134652.

f:=n->numer(add( binomial(n,k)/(k+1)^2, k=0..n));

Table[HypergeometricPFQ[{1, 1, -n}, {2, 2}, -1] // Numerator, {n, 0, 24}] (* Jean-François Alcover, Oct 14 2013 *)

a(n):=if n<0 then 1 else 1/((n+1)^2)*((n)*(3*n+1)*a(n-1)-2*(n-1)*(n)*a(n-2)+1);
makelist(num(a(n),n,0,10); /* Vladimir Kruchinin, Jun 01 2016 */

A097344(n)=numerator(sum(k=0,n,binomial(n,k)/(k+1)^2)) \\ M. F. Hasler, Jan 25 2008

from fractions import Fraction
A097344_list, tlist = [1], [Fraction(1,1)]
for i in range(1,100):
    for j in range(len(tlist)):
        tlist[j] *= Fraction(i,i-j)
    tlist += [Fraction(1,(i+1)**2)]
    A097344_list.append(sum(tlist).numerator) # Chai Wah Wu, Jun 04 2015

def A097344_list(size):
    R, L = [1], [1]
    inc = sqr = 1
    for i in range(1, size):
        for j in range(i):
            L[j] *= i / (i - j)
        inc += 2; sqr += inc
        L.extend(1 / sqr)
        R.append(sum(L).numerator())
    return R
A097344_list(50) # (after Chai Wah Wu) Peter Luschny, Jun 05 2016

a(n) = numerator(b(n)), b(n) = 1/((n+1)^2)*((n)*(3*n+1)*b(n-1)-2*(n-1)*(n)*b(n-2)+1). - Vladimir Kruchinin, May 31 2016

Edited and corrected by Daniel Glasscock (glasscock(AT)rice.edu), Jan 04 2008 and M. F. Hasler, Jan 25 2008

Moved comment on numerators of a logarithmic g.f. over to A097345 - R. J. Mathar, Mar 04 2010

A119948 Triangle of denominators in the square of the matrix with A[i,j] = 1/i for j <= i, 0 otherwise.

Original entry on oeis.org

1, 4, 4, 18, 18, 9, 48, 48, 48, 16, 300, 300, 300, 100, 25, 120, 120, 120, 360, 180, 36, 980, 980, 980, 2940, 1470, 294, 49, 2240, 2240, 2240, 6720, 6720, 1344, 448, 64, 22680, 22680, 22680, 22680, 22680, 4536

Offset: 1

Wolfdieter Lang, Jul 20 2006

The triangle of the corresponding numerators is A119947. The rationals appear in lowest terms.

The least common multiple (LCM) of row i gives [1, 4, 18, 48, 300, 360, 2940, 6720, 22680, ...], which coincides with A081528.

			The first rows of the table are:
  [1];
  [4, 4];
  [18, 18, 9];
  [48, 48, 48, 16];
  [300, 300, 300, 100, 25];
  [120, 120, 120, 360, 180, 36]; ...

Row sums give A119950. Row sums of the triangle of rationals always give 1.

A119948_upto(n)={my(M=matrix(n, n, i, j, (j<=i)/i)^2); vector(n, r, apply(denominator, M[r, 1..r]))} \\ M. F. Hasler, Nov 05 2019

T(i,j) = denominator((A^2)[i,j]), where the lower triangular matrix A has elements a[i,j] = 1/i if j <= i, 0 if j > i.

Edited by M. F. Hasler, Nov 05 2019

A081530 a(n) = running sum of the first n harmonic numbers, multiplied by the LCM of 1..n.

Original entry on oeis.org

1, 5, 26, 77, 522, 669, 5772, 13827, 48610, 55991, 699612, 785633, 11359222, 12530955, 13726712, 29889983, 550271934, 593094837, 12094689300, 12932216325, 13780828710, 14640022575, 356714770680, 376932115005, 1986818142426

Offset: 1

Amarnath Murthy, Mar 27 2003

nonn

Consider triangle in A081525. Write terms in k-th row with denominator = LCM of terms in that row. Sequence gives sum of numerators of terms in n-th row.

			(1), 2*(1 + 3/2), 6*(1 + 3/2 + 11/6), 12*(1 + 3/2 + 11/6 + 25/12).

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A081525, A081526, A081527, A081528, A081529.

Cf. A001705, A025527.

H:=n->add(1/i,i=1..n):seq((n+1)*ilcm(seq(j,j=1..n))*(H(n+1)-1),n=1..30); # C. Ronaldo

Table[Sum[HarmonicNumber[k], {k, n}] LCM @@ Range[n], {n, 36}] (* Wouter Meeussen *)

a(n) = lcm(1..n)*(n+1)*(H(n+1)-1), where H(n) is the n-th harmonic number. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 19 2004

Equal to A001705(n) / A025527(n). - Martin Fuller, Jan 03 2006

More terms from Wouter Meeussen, Apr 13 2003

A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

Original entry on oeis.org

0, 1, 0, 2, 3, 0, 1, 4, 0, 2, 5, 1, 0, 3, 0, 1, 6, 2, 4, 0, 1, 0, 3, 2, 7, 0, 5, 1, 0, 2, 4, 1, 3, 0, 0, 1, 6, 2, 0, 3, 5, 1, 4, 8, 0, 2, 0, 1, 0, 3, 2, 1, 0, 4, 7, 0, 5, 1, 6, 2, 3, 0, 1, 0, 2, 4, 0, 1, 3, 0, 2, 5, 1, 0, 0, 3, 4, 1, 6, 2, 0, 1, 0, 7, 2, 3, 0

Offset: 1

Jwalin Bhatt, Aug 10 2025

The pdf of the sequence is the probability to see i as the remainder when you divide 1,2,3... with a random number from [1,10].
The respective probabilities are:
  p(0) = 7381/25200 = 0.292896...
  p(1) = 4861/25200 = 0.192896...
  p(2) = 3601/25200 = 0.142896...
  p(3) = 2761/25200 = 0.109563...
  p(4) = 2131/25200 = 0.084563...
  p(5) = 1627/25200 = 0.064563...
  p(6) = 1207/25200 = 0.047896...
  p(7) = 121/3600   = 0.033611...
  p(8) = 19/900     = 0.021111...
  p(9) = 1/100      = 0.01
The sequence repeats after 25200 terms. If the random number is picked from [1,n] the period of the sequence is given by n*lcm{1,2,...,n} (A081528).
The arithmetic mean of this sequence is 2.25.
For a sequence with length n, the n-th root of the product of nonzero terms is (2**(5333/12600))*(3**(559/3150))*(5**(1627/25200))*(7**(121/3600)) which is 1.9302557640832...

			Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms.
We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
| n | (c(0)+1)/p(0) | (c(1)+1)/p(1) | (c(2)+1)/p(2) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     3.414     |       -       |       -       |   0    |
| 2 |     6.828     |     5.181     |       -       |   1    |
| 3 |     6.828     |    10.362     |     6.998     |   0    |
| 4 |    10.242     |    10.362     |     6.998     |   2    |

Jwalin Bhatt, Table of n, a(n) for n = 1..25200

Cf. A081528.

samplePDF[numCoeffs_] := Module[{coeffs, counts},
  coeffs = {}; counts = ConstantArray[0, 10];
  Do[
    minTime = Infinity;
    Do[
      time = 10*(counts[[i]] + 1)/(HarmonicNumber[10] - HarmonicNumber[i - 1]);
      If[time < minTime,minIndex = i;minTime = time],{i, 1, 10}];
    counts[[minIndex]] += 1;
    coeffs = Append[coeffs, minIndex - 1],
    {numCoeffs}
   ];
  coeffs
 ]
A386950 = samplePDF[120]

from sympy import harmonic
def sample_pdf(num_coeffs):
  coeffs, counts = [], [0]*10
  for _ in range(num_coeffs):
    min_time = float('inf')
    for i, count in enumerate(counts):
      time = 10*(count+1) / (harmonic(10)-harmonic(i))
      if time < min_time:
        min_index, min_time = i, time
    counts[min_index] += 1
    coeffs.append(min_index)
  return coeffs
A386950 = sample_pdf(120)

cp's OEIS Frontend

A027446 Triangle read by rows: square of the lower triangular mean matrix.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A097344 Numerators in binomial transform of 1/(n+1)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Python

Sage

Formula

Extensions

A119948 Triangle of denominators in the square of the matrix with A[i,j] = 1/i for j <= i, 0 otherwise.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

Extensions

A081530 a(n) = running sum of the first n harmonic numbers, multiplied by the LCM of 1..n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python