The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A341091 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A341091

Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .
(history; published version)

#105 by Michael De Vlieger at Wed Jul 10 15:54:28 EDT 2024

STATUS

proposed

approved

#104 by Michel Marcus at Wed Jul 10 11:59:50 EDT 2024

STATUS

editing

proposed

#103 by Michel Marcus at Wed Jul 10 11:59:45 EDT 2024

COMMENTS

This These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).

STATUS

approved

editing

#102 by N. J. A. Sloane at Sun May 19 12:00:02 EDT 2024

STATUS

proposed

approved

#101 by Thomas Scheuerle at Mon May 13 10:59:45 EDT 2024

STATUS

editing

proposed

Discussion

Mon May 13

12:22

Thomas Scheuerle: Finally I am done. There was a huge amount of work. Now I can and will use this to solve a problem ( but not here , you will see later in some other sequence some day where I will need this operator).

#100 by Thomas Scheuerle at Mon May 13 10:56:02 EDT 2024

FORMULA

Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).

#99 by Thomas Scheuerle at Mon May 13 10:49:34 EDT 2024

FORMULA

Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j).

#98 by Thomas Scheuerle at Mon May 13 09:57:04 EDT 2024

FORMULA

How this sequence will be acting act as an operator onto a Dirichlet series may be developed by the formulas below:

#97 by Thomas Scheuerle at Mon May 13 09:56:01 EDT 2024

FORMULA

How this sequence will be acting as an operator onto a Dirichlet series may be developed by the formulas below:

Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.

#96 by Thomas Scheuerle at Fri May 10 16:54:07 EDT 2024

FORMULA

Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.

STATUS

proposed

editing