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

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

Author

Paul Curtz, Apr 27 2012

Keywords

• sign
• frac
• tabl

Comments

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).

Crossrefs

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

Formula

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.