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.

Showing 1-3 of 3 results.

A212705 a(n) is the difference between numbers of nonnegative multiples of 2*n+1 with even and odd digit sum in base 2*n in interval [0, (2*n)^8).

Original entry on oeis.org

54, 3220, 38794, 237832, 995710, 3256540, 8954258, 21645200, 47366982, 95758500, 181475866, 325939096, 559444366, 923676652, 1474657570, 2286163232, 3453646934, 5098701492, 7374096042, 10469422120, 14617383838
Offset: 1

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Formula

a(n) = 2/(2*n+1)*sum{i=1..n}tan^8(pi*i/(2*n+1)).
a(n) = 2/315*n*(1088*n^6+3808*n^5+3920*n^4+280*n^3-868*n^2+322n-45).
G.f.: 2*x*(27+1394*x+7273*x^2+7308*x^3+1373*x^4+34*x^5-x^6)/(1-x)^8. [Bruno Berselli, May 24 2012]

A212706 a(n) is the difference between numbers of nonnegative multiples of 2*n+1 with even and odd digit sum in base 2*n in interval [0, (2*n)^9).

Original entry on oeis.org

81, 5825, 73745, 461313, 1951057, 6418369, 17712657, 42921473, 94087249, 190446273, 361259537, 649305089, 1115101521, 1841932225, 2941740049, 4561961985, 6893373521, 10179012289, 14724250641, 20908086785, 29195724113, 40152508353, 54459292177, 72929296897
Offset: 1

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Programs

  • Magma
    [1+n/315*(4352*n^6+15232*n^5+12992*n^4-5600*n^3- 5152*n^2+5488*n-2112): n in [1..25]]; // Vincenzo Librandi, Dec 02 2015
  • Mathematica
    Table[1 + n/315 (4352 n^6 + 15232 n^5 + 12992 n^4 - 5600 n^3 - 5152 n^2 + 5488 n - 2112), {n, 30}] (* Vincenzo Librandi, Dec 02 2015 *)
    LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{81,5825,73745,461313,1951057,6418369,17712657,42921473},30] (* Harvey P. Dale, Aug 05 2025 *)
  • PARI
    Vec(x*(81+5177*x+29413*x^2+29917*x^3+4883*x^4+171*x^5-9*x^6-x^7)/(1-x)^8 + O(x^40)) \\ Colin Barker, Dec 01 2015
    

Formula

a(n) = 2/(2*n+1) * Sum_{i=1..n} tan^9(Pi*i/(2*n+1)) * sin(2*Pi*i/(2*n+1)).
a(n) = 1+n/315*(4352*n^6 + 15232*n^5 + 12992*n^4 - 5600*n^3 - 5152*n^2 + 5488*n - 2112).
G.f.: x*(81+5177*x+29413*x^2+29917*x^3+4883*x^4+171*x^5-9*x^6-x^7) / (1-x)^8. - Colin Barker, Dec 01 2015

Extensions

Typo in data fixed by Colin Barker, Dec 01 2015

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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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.
Showing 1-3 of 3 results.