cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A160143 a(n) = Numerator((-1)^n*Euler(2*n)*(2*n+1)/(4^(2*n+1)-2^(2*n+1))), where Euler(n) = A122045(n).

Original entry on oeis.org

1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 1111113564712575, 1595024111042171723, 387863354088927172625, 110350957750914345093747
Offset: 0

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Author

Peter Luschny, May 03 2009

Keywords

Comments

Resembles the coefficients of the series for x/cos(x).
The first difference with sequence A009843 (expansion of x/cos(x)) occurs at a(10). An explanation can be found in the similarity of the numerators of (2*n+1)/(2^(2*n+1)-1) and the odd numbers 2n+1 (cf. A160144).
Similarly, A156769 resembles A036279 (from the expansion of tan(x)).

Crossrefs

Programs

  • Maple
    a := n -> (-1)^iquo(n,2)*euler(n)*(n+1)/(4^(n+1)-2^(n+1));
    seq(numer(a(2*n)),n=0..13);

A212822 Triangle of coefficients of polynomials concerning Newman-like phenomenon of multiples of b+1 in even base b in interval [0,b^n) (see comment).

Original entry on oeis.org

1, 2, -1, 1, 3, -1, 2, 6, -8, 3, 2, 10, 10, -10, 3, 4, 20, 10, -50, 46, -15, 17, 119, 245, 35, -217, 161, -45, 34, 238, 406, -350, -644, 1372, -1056, 315, 62, 558, 1722, 1638, -1092, -1008, 1828, -1188, 315, 124, 1116, 3138, 1134, -5838, 1134, 9452, -14724, 10134, -2835
Offset: 2

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Keywords

Comments

In 1969, D. J. Newman (see the reference) proved that difference between numbers of multiples of 3 with even and odd binary digit sums in interval [0,x] is always positive. This fact now is known as Newman phenomenon.
Consider difference between numbers of multiples of b+1 with even and odd digit sums in even base b in interval [0, b^n). It is a polynomial in b P_n(b) of degree n-1 and multiple of b, if n is even, and n-2, if n is odd, such that all polynomials Q_n(b):=A156769(n/2)*P_n(b)/b, if n is even, and Q_n(b):=A156769((n-1)/2)*P_n(b), if n is odd, presumably have integer coefficients and are of degree n-2. The sequence is triangle of coefficients of polynomials Q_n(b).
The r-th row contains r-1 entries.
Since, evidently, P_n(1)=1, then the row sums form sequence A156769 repeated.

Examples

			Triangle begins (r is the number of row or the number of polynomial; coefficients of b^k, k=r-2-i, i=0,1,..., r-2)
r/i.|..0......1......2.....3.....4......5......6.....7
======================================================
.2..|..1
.3..|..2.....-1
.4..|..1......3.....-1
.5..|..2......6.....-8.....3
.6..|..2.....10.....10...-10.....3
.7..|..4.....20.....10...-50....46....-15
.8..|.17....119....245....35..-217....161....-45
.9..|.34....238....406..-350..-644...1372..-1056....315
For example, if r=4, the polynomial
P_4(b)=b*(b^2+3*b-1)/A156769(4/2)=b/3*(b^2+3*b-1) (b==0 mod 2)
gives difference between multiples of b+1 with even and odd digit sums in  base b in interval [0, b^4). Note also that P_2(b)=b. Therefore, setting in the formula n=r=3, again for P_4(b) we have P_4(b)=b*C(b+1,2)-C(b,3)=b/3*(b^2+3*b-1).
		

Crossrefs

Programs

  • Mathematica
    A156769[n_] := Denominator[(2^(2*n-2)/Factorial[2*n-1])]; poly[1, b_] := 1; poly[2, b_] := b; poly[n_, b_] :=  poly[n, b] = If[OddQ[n], (-1)^((n - 1)/2) (FunctionExpand[Binomial[b - 1, n - 1]] - Sum[(-1)^(k/2) FunctionExpand[Binomial[b + 1, n - k - 1]] poly[k + 1, b], {k, 0, n - 2, 2}]), (-1)^((n - 2)/2) (FunctionExpand[Binomial[b, n - 1]] - Sum[(-1)^((k - 1)/2) FunctionExpand[Binomial[b + 1, n - k - 1]] poly[k + 1, b], {k, 1, n - 2, 2}])]; Table[If[EvenQ[z], Most[Reverse[CoefficientList[poly[z, b] A156769[z/2], b]]], Reverse[CoefficientList[poly[z, b] A156769[(z - 1)/2], b]]], {z, 2, 12}]

Formula

If n>=2 is even, then P_(n+1)(b) = (-1)^((n-2)/2)*(C(b+1,n)-C(b-1,n))-sum{i=2,4,...,n-2}(-1)^((n+i)/2)*C(b+1, n-i)*P_(i+1)(b), where P_n(b)=b*Q_n(b)/A156769(n/2);
if n>=3 is odd, then P_(n+1)(b) = (-1)^((n-1)/2)*(C(b,n)-b*C(b+1,n-1))+sum{i=3,5,...,n-2}(-1)^((n+i)/2)*C(b+1, n-i)*P_(i+1)(b), where
P_n(b) = Q_n(b)/A156769((n-1)/2).
P_n(b) = 2/(b+1)*Sum_{j=1..b/2}(tan(j*Pi/(b+1)))^n, if n is even, and
P_n(b) = 2/(b+1)*Sum_{j=1..b/2}(tan(j*Pi/(b+1)))^n*sin(j*Pi/(b+1)), if n is odd.
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