A085737 -id:A085737 - cp's OEIS frontend

A110841 a(n) is the number of prime factors, with multiplicity, of abs(A014509(n)).

0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7, 7, 2, 2, 4, 3, 3, 7, 1, 6, 4, 5, 14, 4, 9, 5, 10, 3, 11, 2, 5, 3, 7, 11, 5, 3, 4, 15, 6, 5, 19, 10, 6, 13, 15, 5, 10, 5, 5, 6, 7, 5, 15, 7, 5, 2, 13, 4, 3, 10, 5, 9, 7, 5, 4, 9, 5, 4, 1, 7, 4, 4, 5, 3, 11, 13, 10, 5, 5, 7, 6

Offset: 0

Jonathan Vos Post, Sep 16 2005

			a(10) = 2 because A014509(10) = 529 = 23^2.
a(8) = a(19) = 1 since A014509(8) and A014509(19) are prime.

Sean A. Irvine, Table of n, a(n) for n = 0..91

Cf. A001222, A073276, A075178, A076547, A076549, A079294, A083687, A084217, A085092, A085737, A089170, A089644, A089655.

a(n) = my(b=bernfrac(2*n), c=floor(abs(b))*sign(b)); if (c==0, 0, bigomega(c)); \\ Michel Marcus, Mar 29 2020

a(n) = A001222(abs(A014509(n))).

More terms from Michel Marcus, Mar 29 2020
a(51)-a(65) from Jinyuan Wang, Apr 02 2020
More terms from Sean A. Irvine, Jul 29 2024

A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

Original entry on oeis.org

1, 1, -3, 1, -5, 5, 1, -7, 25, -5, 1, -9, 23, -35, 49, 1, -11, 73, -27, 112, -49, 1, -13, 53, -77, 629, -91, 58, 1, -15, 145, -130, 1399, -451, 753, -58, 1, -17, 95, -135, 2699, -2301, 8573, -869, 341, 1, -19, 241

Offset: 0

Paul Curtz, Apr 27 2012

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
Hence
                    1    1/2   1/6      0
            1    -1/2   -1/3  -1/6  -1/30
      1   -3/2    1/6    1/6  2/15   1/15
  1 -5/2   5/3      0  -1/30 -1/15 -8/105.
The second row is A051716/A051717.
The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
B(0)=    1 = 1                 Redbernou1li
B(1)= -1/2 = 1  -3/2
B(2)=  1/6 = 1  -5/2  5/3
B(3)=    0 = 1  -7/2 25/6  -5/3
B(4)=-1/30 = 1  -9/2 23/3 -35/6  49/30
B(5)=    0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
For the main diagonal, see A165142.
Denominator b(n) will be submitted.
This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
1
1  -3/2
1  -5/2 10/6
1  -7/2 25/6 -10/6
1  -9/2 46/6 -35/6  49/30
1 -11/2 73/6 -81/6 224/30 -49/30.
For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
Triangle Checkbernou1 with the same denominator A080326 for every row is
1/1
(2    -3)/2
(6   -15  +10)/6
(6   -21  +25  -10)/6
(30 -135 +230 -175  +49)/30
(30 -165 +365 -405 +224 -49)/30;
Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.

CB(0,x) = 1,
CB(1,x) = 1 - 3*x/2,
CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.

A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

Original entry on oeis.org

1, 1, 3, 1, 2, 13, 0, 1, 5, 3, -1, -1, 2, 29, 119, 0, -1, -1, 1, 31, 5, 1, 1, -1, -8, -1, 43, 253, 0, 1, 1, 4, -4, -1, 41, 7, -1, -1, -1, 4, 8, 4, -1, 29, 239, 0, -1, -1, -8, -4, 4, 8, 1, 31, 9, 5, 5, 7, -4, -116, -32, -116, -4, 7, 71, 665, 0

Offset: 0

Paul Curtz, Feb 03 2015

The difference table of Bernoulli(n,2) or B(n,2) = A164558(n)/A027642(n) is defined by placing the fractions in the upper row and calculating further rows as the differences of their preceding row:
1,       3/2, 13/6,     3, 119/30, ...
1/2,     2/3,  5/6, 29/30, ...
1/6,     1/6, 2/15, ...
0,     -1/30, ...
-1/30, ...
etc.
The first column is A164555(n)/A027642(n).
In particular, the sums of the antidiagonals
1                              = 1
1/2 + 3/2                      = 2
1/6 + 2/3 + 13/6               = 3
0   + 1/6 +  5/6 + 3           = 4
etc. are the positive natural numbers. (This is rewritten for Bernoulli(n,3) in A157809).
We also have for Bernoulli(.,2)
B(0,2) = 1
B(0,2) + 2*B(1,2) = 4
B(0,2) + 3*B(1,2) + 3*B(2,2) = 12
B(0,2) + 4*B(1,2) + 6*B(2,2) + 4*B(3,2) = 32
etc. with right hand sides provided by A001787.
More generally sum_{s=0..t-1} binomial(t,s)*Bernoulli(s,q) gives A027471(t) for q=3, A002697 for q=4 etc, by reading A104002 downwards the q-th column.

Cf. A027641, A027642, A074909, A085737, A085738, A104002, A157809, A157920, A157930, A157945, A157946, A157965, A164555, A164558, A190339, A158302, A181131 (numerators and denominators of the main diagonal).

nmax = 11; A164558 = Table[BernoulliB[n,2], {n, 0, nmax}]; D164558 = Table[ Differences[A164558, n], {n, 0, nmax}]; Table[ D164558[[n-k+1, k+1]] // Numerator, {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2015 *)

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A110841 a(n) is the number of prime factors, with multiplicity, of abs(A014509(n)).

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A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

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A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

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