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Revision History for A081733 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Triangle read by rows, T(n,k) = 2^(n-k)*[x^k] Euler_polynomial(n, x), for n >= 0, k >= 0.
(history; published version)
#63 by N. J. A. Sloane at Mon Jan 08 01:41:30 EST 2018
STATUS

proposed

approved

#62 by Jon E. Schoenfield at Mon Jan 08 01:38:53 EST 2018
STATUS

editing

proposed

#61 by Jon E. Schoenfield at Mon Jan 08 01:38:49 EST 2018
NAME

Triangle read by rows, T(n,k) = 2^(n-k)*[x^k] Euler_polynomial(n, x), for n >= 0, k >= 0.

EXAMPLE

1,

-1, 1,

0, -2, 1,

2, 0, -3, 1,

0, 8, 0, -4, 1.

STATUS

approved

editing

#60 by N. J. A. Sloane at Mon Jan 08 01:36:21 EST 2018
EXTENSIONS

New name by _from _Peter Luschny_, Jul 18 2012

Discussion
Mon Jan 08
01:36
OEIS Server: https://oeis.org/edit/global/2730
#59 by Bruno Berselli at Fri Oct 09 18:19:41 EDT 2015
STATUS

reviewed

approved

#58 by Michel Marcus at Sat Oct 03 18:28:33 EDT 2015
STATUS

proposed

reviewed

#57 by Tom Copeland at Sat Oct 03 16:49:56 EDT 2015
STATUS

editing

proposed

Discussion
Sat Oct 03
17:35
Tom Copeland: I detailed the relationship to the seq.
#56 by Tom Copeland at Sat Oct 03 16:49:30 EDT 2015
FORMULA

E.g.f.: [2/(e^(2t)+1)] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x - 2/[e^(-2D)+1], i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x). Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x - 1 - D + 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, and P_n(0) are the coefficients (mod signs/shift) of these sequences. The polynomials PI_n(x) of A119468 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Note that 2/[e^(2t)+1] = 2 Sum_{n >= 0} Eta(-n) (-2t)^n/n!], where Eta(s) is the Dirichlet eta function, and b_n = 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers, so P_n(x) = (b. + x)^n, as an Appell polynomial. - Tom Copeland, Sep 27 2015

CROSSREFS
STATUS

proposed

editing

#55 by Jon E. Schoenfield at Fri Oct 02 17:26:01 EDT 2015
STATUS

editing

proposed

Discussion
Sat Oct 03
14:43
Peter Luschny: Please add the referenced sequences to the CROSSREFS.
14:52
Peter Luschny: The comment "2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers" belongs to A155585; I feel it is misplaced here as it does not add to the sequence at hand.
#54 by Jon E. Schoenfield at Fri Oct 02 17:25:58 EDT 2015
FORMULA

E.g.f.: [2/(e^(2t)+1)] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x - 2/[e^(-2D)+1], i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x). Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x - 1 - D + 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, and P_n(0) are the coefficients (mod signs/shift) of these sequences. The polynomials PI_n(x) of A119468 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Note that 2/[e^(2t)+1] = 2 sum[Sum_{n >= 0, } Eta(-n) (-2t)^n/n!], where Eta(s) is the Dirichlet eta function, and 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers. - Tom Copeland, Sep 27 2015

STATUS

proposed

editing