A011934 -id:A011934 - cp's OEIS frontend

A059342 Triangle giving denominators of coefficients of Euler polynomials, highest powers first.

1, 1, 2, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 2, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 4, 1, 2, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 0

N. J. A. Sloane, Jan 27 2001

			1; x-1/2; x^2-x; x^3-3*x^2/2+1/4; ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 48, [14b].

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A059341. See also A004172 A004173 A004174 A004175 A011934 A020523 A020524 A020525 A020526 A020547 A020548 A058940.

for n from 0 to 30 do for k from n to 0 by -1 do printf(`%d,`,denom(coeff(euler(n,x), x, k))) od:od:

Denominator[Table[Reverse[CoefficientList[Series[EulerE[n, x], {x, 0, 20}], x]], {n, 0, 10}]] (* G. C. Greubel, Jan 07 2017 *)

More terms from James Sellers, Jan 29 2001

A215097 a(n) = n^3 - a(n-2) for n >= 2 and a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 8, 26, 56, 99, 160, 244, 352, 485, 648, 846, 1080, 1351, 1664, 2024, 2432, 2889, 3400, 3970, 4600, 5291, 6048, 6876, 7776, 8749, 9800, 10934, 12152, 13455, 14848, 16336, 17920, 19601, 21384, 23274, 25272, 27379, 29600, 31940, 34400, 36981, 39688, 42526

Offset: 0

Alex Ratushnyak, Aug 03 2012

Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-7,4,-1).

Cf. A000217 (n^2 - a(n-1)).
Cf. A125577 (n^2 - a(n-1) with a(0)=1).
Cf. A011934 (n^3 - a(n-1)).
Cf. A153026 (n^3 - a(n-1) with a(1)=0).
Cf. A194274 (n^2 - a(n-2)).
Cf. A187093 (n^2 - a(n-2) with a(0)=a(1)=1, a(-1)=0).
Cf. A107386 ((n-2)^2 - a(n-1) with a(0)=0, a(1)=a(2)=1, a(3)=2).
Cf. A206481 ((n-1)^3 - a(n-2)).

RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == n^3 - a[n - 2]}, a[n], {n, 0, 43}] (* Bruno Berselli, Aug 07 2012 *)

prpr = 0
prev = 1
for n in range(2,77):
    print(prpr, end=',')
    curr = n*n*n - prpr
    prpr = prev
    prev = curr

G.f.: (x+4*x^2+x^3)/((-1+x)^4*(1+x^2)). - David Scambler, Aug 06 2012

a(n) = (n*(n^2-3)-(1-(-1)^n)*i^(n+1))/2, where i=sqrt(-1). - Bruno Berselli, Aug 07 2012

A382973 a(n) = 4n^3 - 6n^2 + 6*n - 2 + (-1)^n.

Original entry on oeis.org

1, 19, 69, 183, 377, 683, 1117, 1711, 2481, 3459, 4661, 6119, 7849, 9883, 12237, 14943, 18017, 21491, 25381, 29719, 34521, 39819, 45629, 51983, 58897, 66403, 74517, 83271, 92681, 102779, 113581, 125119, 137409, 150483, 164357, 179063, 194617, 211051, 228381, 246639

Offset: 1

Nicolay Avilov, Jun 02 2025

a(n) is the number of 1 X 1 X 1 black cubes in a cube (2*n - 1) X (2*n - 1) X (2*n - 1), which is made up of 1 X 1 X 1 black cubes and 1 X 1 X 1 white cubes. In this case, any 1 X 1 X 1 cube is either completely black or completely white. The black and white cubes are arranged as follows: if the central cube has coordinates (0, 0, 0), then all cubes with coordinates (0, y, z), (x, 0, z) and (x, y, 0) are black. Then the cubes adjacent to the black ones are painted white so that a triangular layer with all white cubes is obtained. There will be eight such layers. In the next step, all cubes adjacent to the white ones are painted black so that a triangular layer of black cubes is formed, and so on, alternating layers of black cubes and layers of white cubes until all the cubes are painted (see the link "Illustration of coloring a cube").

			a(2) = 3^3 - 8*1 = 19;
a(3) = 5^3 - 8*7 = 69.

Nicolay Avilov, Illustration of a cube coloring.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A000578, A011934.

LinearRecurrence[{3, -2, -2, 3, -1}, {1, 19, 69, 183, 377}, 20] (* Hugo Pfoertner, Jun 12 2025 *)

a(n) = (2n - 1)^3 - 8*A011934(n-1).

G.f.: x*(1 + 16*x + 14*x^2 + 16*x^3 + x^4)/((1 - x)^4*(1 + x)). - Stefano Spezia, Jun 12 2025

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A059342 Triangle giving denominators of coefficients of Euler polynomials, highest powers first.

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A215097 a(n) = n^3 - a(n-2) for n >= 2 and a(0)=0, a(1)=1.

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A382973 a(n) = 4n^3 - 6n^2 + 6*n - 2 + (-1)^n.

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cp's OEIS Frontend

A059342 Triangle giving denominators of coefficients of Euler polynomials, highest powers first.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A215097 a(n) = n^3 - a(n-2) for n >= 2 and a(0)=0, a(1)=1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

A382973 a(n) = 4*n^3 - 6*n^2 + 6*n - 2 + (-1)^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A382973 a(n) = 4n^3 - 6n^2 + 6*n - 2 + (-1)^n.