A091042 -id:A091042 - cp's OEIS frontend

A214458 Let S_3(n) denote difference between multiples of 3 in interval [0,n) with even and odd binary digit sums. Then a(n)=(-1)^A000120(n)(S_3(n)-3S_3(floor(n/4))).

0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jul 18 2012

In 1969, D. J. Newman (see the reference) proved L. Moser's conjecture that difference between numbers of multiples of 3 with even and odd binary digit sums in interval [0,x] is always positive. This fact is known as Moser-Newman phenomenon.

Theorem: The sequence is periodic with period of length 24.

J. Coquet, A summation formula related to the binary digits, Inventiones Mathematicae 73 (1983), pp. 107-115.
D. J. Newman, On the number of binary digits in a multiple of three, Proc. Amer. Math. Soc. 21 (1969) 719-721.
Vladimir Shevelev, Two algorithms for evaluation of the Newman digit sum, and a new proof of Coquet's theorem, arXiv:0709.0885 [math.NT], 2007-2012.

Cf. A091042, A212500.

Recursion for evaluation S_3(n): S_3(n)=3*S_3(floor(n/4))+(-1)^A000120(n)*a(n). As a corollary, we have |S_3(n)-3*S_3(n/4)|<=2.

A377657 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..k} tan(jPi/(1 + 2k))^(2*n).

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 9, 10, 4, 0, 27, 90, 21, 5, 0, 81, 850, 371, 36, 6, 0, 243, 8050, 7077, 1044, 55, 7, 0, 729, 76250, 135779, 33300, 2365, 78, 8, 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9, 0, 6561, 6841250, 50028755, 34420356, 5476405, 312390, 8295, 136, 10

Offset: 0

Peter Luschny, Nov 10 2024

Based on an observation made by Fredrik Johansson about A376777, which is the main diagonal of this array.

			Array begins
  [0] 1,    2,       3,         4,           5,            6, ... A000027
  [1] 0,    3,      10,        21,          36,           55, ... A014105
  [2] 0,    9,      90,       371,        1044,         2365, ... A377858
  [3] 0,   27,     850,      7077,       33300,       113311, ... A376778
  [4] 0,   81,    8050,    135779,     1070244,      5476405, ...
  [5] 0,  243,   76250,   2606261,    34420356,    264893255, ...
  [6] 0,  729,  722250,  50028755,  1107069876,  12813875437, ...
  [7] 0, 2187, 6841250, 960335173, 35607151476, 619859803695, ...
  .
Seen as a triangle T(n, k) = A(n-k, k):
  [0] 1;
  [1] 0,    2;
  [2] 0,    3,      3;
  [3] 0,    9,     10,       4;
  [4] 0,   27,     90,      21,       5;
  [5] 0,   81,    850,     371,      36,      6;
  [6] 0,  243,   8050,    7077,    1044,     55,    7;
  [7] 0,  729,  76250,  135779,   33300,   2365,   78,   8;
  [8] 0, 2187, 722250, 2606261, 1070244, 113311, 4654, 105, 9;

Rows: A000027, A014105, A377858, A376778.
Columns: A376478.
Cf. A376777 (main diagonal), A377658 (antidiagonal sums).
Cf. A091042.

A := (n, k) -> add(tan(j*Pi/(1 + 2*k))^(2*n), j = 0..k):
seq(print(seq(round(evalf(A(n, k), 32)), k = 0..6)), n = 0..7);

A(n, k) = {trace(matcompanion(sum(m=0, k, x^m*binomial(2*k+1, 2*(k-m))*(-1)^(m+1)))^n)+(n==0) } \\ Thomas Scheuerle, Nov 11 2024

Row n of A091042(n, k) = binomial(2*n+1, 2*k) gives the polynomial Pe(n, x), with zeros in -tan(Pi/2*n+1)^2, -tan(2*Pi/2*n+1)^2, ..., -tan(n*Pi/2*n+1)^2. Let Pm(n, k, x) be the polynomial with zeros in (-tan(Pi/2*n+1)^2)^k, (-tan(2*Pi/2*n+1)^2)^k, ..., (-tan(n*Pi/2*n+1)^2)^k, then A(k, n) is the coefficient of X^(n-1) in the polynomial Pm(n, k, x). A way to do this calculation without evaluation of irrational numbers is to obtain the companion matrix M of the polynomial Pe(n, x), then A(k, n) = tr(M^k) (the trace of M^k). - Thomas Scheuerle, Nov 11 2024

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

Vladimir Shevelev and Peter J. C. Moses, May 28 2012

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.

			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).

D. J. Newman, On the number of binary digits in a multiple of three, Proc. Amer. Math. Soc. 21 (1969) 719-721.
Vladimir Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m, arXiv:0710.3177v2 [math.NT], 2007.

Cf. A156769, A038754, A084990, A091042, A212500, A212592, A212593, A212594, A212668, A212669, A212670, A212705, A212706.

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}]

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.

A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2n,2k) + binomial(2n+1,2k))/2.

Original entry on oeis.org

1, 1, 2, 1, 8, 3, 1, 18, 25, 4, 1, 32, 98, 56, 5, 1, 50, 270, 336, 105, 6, 1, 72, 605, 1320, 891, 176, 7, 1, 98, 1183, 4004, 4719, 2002, 273, 8, 1, 128, 2100, 10192, 18590, 13728, 4004, 400, 9, 1, 162, 3468, 22848, 59670, 68068, 34476, 7344, 561, 10, 1, 200, 5415

Offset: 0

Philippe Deléham, Nov 25 2013

Sum_{k=0..n}T(n,k)*x^k = A164111(n), A000012(n), A002001(n), A001653(n+1), A001019(n), A166965(n) for x =-1, 0, 1, 2, 4, 9 respectively.

			Triangle begins:
1
1, 2
1, 8, 3
1, 18, 25, 4
1, 32, 98, 56, 5
1, 50, 270, 336, 105, 6
1, 72, 605, 1320, 891, 176, 7
1, 98, 1183, 4004, 4719, 2002, 273, 8
1, 128, 2100, 10192, 18590, 13728, 4004, 400, 9

Cf. A086645, A091042
Cf. Columns : A000012, A001105, A180324 ; Diagonals: A000027, A131423
Cf. T(2*n,n): A228329, Row sums : A002001

T := (n,k) -> binomial(2*n, 2*k)*(2*n+1-k)/(2*n+1-2*k);
seq(seq(T(n,k), k=0..n), n=0..9); # Peter Luschny, Nov 26 2013

Flatten[Table[(Binomial[2n,2k]+Binomial[2n+1,2k])/2,{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jul 05 2015 *)

G.f.: (1-x)/(1-2*x*(1+y)+x^2*(1-y)^2).
T(n,k) = 2*T(n-1,k)+2*T(n-1,k-1)+2*T(n-2,k-1)-T(n-2,k)-T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
T(n,k) = (A086645(n,k) + A091042(n,k))/2.
T(n,k) = binomial(2*n,2*k)*(2*n+1-k)/(2*n+1-2*k). - Peter Luschny, Nov 26 2013

cp's OEIS Frontend

A214458 Let S_3(n) denote difference between multiples of 3 in interval [0,n) with even and odd binary digit sums. Then a(n)=(-1)^A000120(n)*(S_3(n)-3*S_3(floor(n/4))).

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A377657 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..k} tan(j*Pi/(1 + 2*k))^(2*n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2*n,2*k) + binomial(2*n+1,2*k))/2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A214458 Let S_3(n) denote difference between multiples of 3 in interval [0,n) with even and odd binary digit sums. Then a(n)=(-1)^A000120(n)(S_3(n)-3S_3(floor(n/4))).

A377657 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..k} tan(jPi/(1 + 2k))^(2*n).

A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2n,2k) + binomial(2n+1,2k))/2.