A060546 -id:A060546 - cp's OEIS frontend

A347171 Triangle read by rows where T(n,k) is the sum of Golay-Rudin-Shapiro terms GRS(j) (A020985) for j in the range 0 <= j < 2^n and having binary weight wt(j) = A000120(j) = k.

1, 1, 1, 1, 2, -1, 1, 3, -1, 1, 1, 4, 0, 0, -1, 1, 5, 2, -2, 1, 1, 1, 6, 5, -4, 3, -2, -1, 1, 7, 9, -5, 3, -3, 3, 1, 1, 8, 14, -4, 0, 0, 2, -4, -1, 1, 9, 20, 0, -6, 6, -4, 0, 5, 1, 1, 10, 27, 8, -14, 12, -10, 8, -3, -6, -1, 1, 11, 35, 21, -22, 14, -10, 10, -11, 7, 7, 1

Offset: 0

Kevin Ryde, Aug 21 2021

Doche and Mendès France form polynomials P_n(y) = Sum_{j=0..2^n-1} GRS(j) * y^wt(j) and here row n is the coefficients of P_n starting from the constant term, so P_n(y) = Sum_{k=0..n} T(n,k)*y^k. They conjecture that the number of real roots of P_n is A285869(n).

Row sum n is the sum of GRS terms from j = 0 to 2^n-1 inclusive, which Brillhart and Morton (Beispiel 6 page 129) show is A020986(2^n-1) = 2^ceiling(n/2) = A060546(n). The same follows by substituting y=1 in the P_n recurrence or the generating function.

			Triangle begins
        k=0 k=1 k=2 k=3 k=4 k=5 k=6 k=7
  n=0:   1
  n=1:   1,  1
  n=2:   1,  2, -1
  n=3:   1,  3, -1,  1
  n=4:   1,  4,  0,  0, -1
  n=5:   1,  5,  2, -2,  1,  1
  n=6:   1,  6,  5, -4,  3, -2, -1
  n=7:   1,  7,  9, -5,  3, -3,  3,  1
For T(5,3), those j in the range 0 <= j < 2^5 with wt(j) = 3 are
  j      =  7 11 13 14 19 21 22 25 26 28
  GRS(j) = +1 -1 -1 +1 -1 +1 -1 -1 -1 +1 total -2 = T(5,3)

Kevin Ryde, Table of n, a(n) for rows 0 to 100, flattened
John Brillhart and Patrick Morton, Über Summen von Rudin-Shapiroschen Koeffizienten, Illinois Journal of Mathematics, volume 22, issue 1, 1978, pages 126-148.
Christophe Doche and Michel Mendès France, An Exercise on the Average Number of Real Zeros of Random Real Polynomials, Finite and Infinite Combinatorics conference, Budapest, 2001, pages 1-14, see Rudin-Shapiro example page 9.

Cf. A020985 (GRS), A020986 (GRS partial sums), A000120 (binary weight), A285869.
Columns k=0..3: A000012, A001477, A000096, A275874.
Cf. A165326 (main diagonal), A248157 (second diagonal negated).
Cf. A060546 (row sums), A104969 (row sums squared terms).
Cf. A329301 (antidiagonal sums).
Cf. A104967 (rows reversed, up to signs).

my(M=Mod('x, 'x^2-(1-'y)*'x-2*'y)); row(n) = Vecrev(subst(lift(M^n),'x,'y+1));

T(n,k) = T(n-1,k) - T(n-1,k-1) + 2*T(n-2,k-1) for n>=2, and taking T(n,k)=0 if k<0 or k>n.
T(n,k) = (-1)^k * A104967(n,n-k).
Row polynomial P_n(y) = (1-y)*P_{n-1}(y) + 2*y*P_{n-2}(y) for n>=2. [Doche and Mendès France]
G.f.: (1 + 2*x*y)/(1 + x*(y-1) - 2*x^2*y).
Column g.f.: C_k(x) = 1/(1-x) for k=0 and C_k(x) = x^k * (2*x-1)^(k-1) / (1-x)^(k+1) for k>=1.

A381899 Irregular triangular array read by rows. T(n,k) is the number of length n words x on {0,1} such that I(x) + W(x)*(n-W(x)) = k, where I(x) is the number of inversions in x and W(x) is the number of 1's in x, n >= 0, 0 <= k <= floor(n^2/2).

Original entry on oeis.org

1, 2, 2, 1, 1, 2, 0, 2, 2, 2, 2, 0, 0, 2, 3, 3, 4, 1, 1, 2, 0, 0, 0, 2, 2, 4, 4, 6, 4, 4, 2, 2, 2, 0, 0, 0, 0, 2, 2, 2, 4, 5, 7, 6, 9, 7, 7, 5, 4, 1, 1, 2, 0, 0, 0, 0, 0, 2, 2, 2, 2, 4, 4, 8, 6, 10, 12, 14, 12, 14, 10, 10, 6, 4, 2, 2

Offset: 0

Geoffrey Critzer, Mar 09 2025

Sum_{k>=0} T(n,k)*2^k = A132186(n).
Sum_{k>=0} T(n,k)*3^k = A053846(n).
Sum_{k>=0} T(n,k)*q^k = the number of idempotent n X n matrices over GF(q).
It appears that if n is even the n-th row converges to 2,0,0,...,21,13,9,5,4,1,1 which is A226622 reversed, and if n is odd the sequence is twice A226635.
From Alois P. Heinz, Mar 09 2025: (Start)
Sum_{k>=0} k * T(n,k) = 3*A001788(n-1) for n>=1.
Sum_{k>=0} (-1)^k * T(n,k) = A060546(n). (End)

			Triangle T(n,k) begins:
  1;
  2;
  2, 1, 1;
  2, 0, 2, 2, 2;
  2, 0, 0, 2, 3, 3, 4, 1, 1;
  2, 0, 0, 0, 2, 2, 4, 4, 6, 4, 4, 2, 2;
  ...
T(4,5) = 3 because we have: {0, 1, 0, 0}, {0, 1, 0, 1}, {1, 1, 0, 1}.

Alois P. Heinz, Rows n = 0..50, flattened

Row sums give A000079.

Cf. A001788, A060546, A226635, A226622, A132186, A053846, A083906.

b:= proc(i, j) option remember; expand(`if`(i+j=0, 1,
     `if`(i=0, 0, b(i-1, j))+`if`(j=0, 0, b(i, j-1)*z^i)))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(
         expand(add(b(n-j, j)*z^(j*(n-j)), j=0..n))):
seq(T(n), n=0..10);  # Alois P. Heinz, Mar 09 2025

nn = 7; B[n_] := FunctionExpand[QFactorial[n, q]]*q^Binomial[n, 2];e[z_] := Sum[z^n/B[n], {n, 0, nn}]; Map[CoefficientList[#, q] &, Table[B[n], {n, 0, nn}] CoefficientList[Series[e[z]^2, {z, 0, nn}],z]]

Sum_{n>=0} Sum_{k>=0} T(n,k)*q^k*x^n/(n_q!*q^binomial(n,2)) = e(x)^2 where e(x) = Sum_{n>=0} x^n/(n_q!*q^binomial(n,2)) where n_q! = Product{i=1..n} (q^n-1)/(q-1).

A387025 Start with the list of positive integers L_1 = (1, 2, ...); for n = 1, 2, ..., let m be the least integer > n such that L_n(n) divides L_n(m); L_{n+1}(k) = L_n(k) for any k <> m, L_{n+1}(m) = L_n(m)/L_n(n); a(n) = L_n(n).

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 7, 8, 9, 2, 11, 6, 13, 2, 15, 1, 17, 2, 19, 10, 21, 2, 23, 2, 25, 1, 27, 28, 29, 2, 31, 16, 33, 2, 35, 18, 37, 2, 39, 2, 41, 1, 43, 44, 45, 2, 47, 3, 49, 1, 17, 52, 53, 2, 55, 1, 57, 2, 59, 30, 61, 2, 63, 32, 65, 2, 67, 2, 69, 1, 71, 4, 73, 2

Offset: 1

Rémy Sigrist, Aug 13 2025

nonn

Applying the same procedure to the powers of two yields A060546.

Applying the same procedure to the factorial numbers yields A006882.

			The first terms are:
  n   a(n)  L_n
  --  ----  ------------------------------------------------------
   1     1  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...
   2     2  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...
   3     3  1, 2, 3, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...
   4     2  1, 2, 3, 2, 5, 2, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...
   5     5  1, 2, 3, 2, 5, 1, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...
   6     1  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11, 12, 13, 14, 15, ...
   7     7  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11, 12, 13, 14, 15, ...
   8     8  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11, 12, 13,  2, 15, ...
   9     9  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11, 12, 13,  2, 15, ...
  10     2  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11, 12, 13,  2, 15, ...
  11    11  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11,  6, 13,  2, 15, ...
  12     6  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11,  6, 13,  2, 15, ...
  13    13  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11,  6, 13,  2, 15, ...
  14     2  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11,  6, 13,  2, 15, ...
  15    15  1, 2, 3, 2, 5, 1, 7, 8, 9,  2, 11,  6, 13,  2, 15, ...

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A006882, A060546.

{ for (n = 1, #a = vector(74, n, n), print1 (a[n]", "); forstep (k = ceil((n+1)/a[n])*a[n], #a, a[n], if (a[k] % a[n]==0, a[k] /= a[n]; break;););); }

a(p) = p for any prime number p.

a(2*p) = 1 or 2 for any prime number p.

A060549 a(n) is the number of distinct patterns (modulo geometric D3-operations) with strict median-reflective (palindrome) symmetry (i.e., having no other symmetry) which can be formed by an equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.

Original entry on oeis.org

0, 1, 2, 2, 6, 6, 12, 14, 28, 28, 60, 60, 120, 124, 248, 248, 504, 504, 1008, 1016, 2032, 2032, 4080, 4080, 8160, 8176, 16352, 16352, 32736, 32736, 65472, 65504, 131008, 131008, 262080, 262080, 524160, 524224, 1048448, 1048448, 2097024, 2097024

Offset: 1

André Barbé (Andre.Barbe(AT)esat.kuleuven.ac.be), Apr 03 2001

Harry J. Smith, Table of n, a(n) for n=1..500
A. Barbé, Symmetric patterns in the cellular automaton that generates Pascal's triangle modulo 2, Discr. Appl. Math. 105(2000), 1-38.
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (0,2,0,0,0,2,0,-4).

Cf. A060546, A060548, A008619, A008615.

a(n) = { 2^ceil(n/2) - 2^(floor((n + 3)/6) + (n%6==1)) } \\ Harry J. Smith, Jul 07 2009

a(n) = 2^ceiling(n/2) - 2^(floor((n+3)/6) + d(n)), with d(n)=1 if n mod 6=1 else d(n)=0.
a(n) = A060546(n) - A060548(n) = 2^A008619(n-1) - 2^A008615(n+1), for n >= 1.
G.f.: x^2*(2*x^4 + 2*x^3 + 2*x + 1) / ((2*x^2-1)*(2*x^6-1)). - Colin Barker, Aug 29 2013

A060551 a(n) is the number of nonsymmetric patterns (no reflective, nor rotational symmetry) which may be formed by an equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.

Original entry on oeis.org

0, 0, 0, 6, 12, 42, 84, 210, 420, 924, 1860, 3900, 7800, 15996, 31992, 64728, 129528, 260568, 521136, 1045464, 2090928, 4187952, 8376240, 16764720, 33529440, 67084080, 134168160, 268385376, 536772192, 1073642592, 2147285184, 4294769760, 8589539520, 17179472064

Offset: 1

André Barbé (Andre.Barbe(AT)esat.kuleuven.ac.be), Apr 03 2001

Harry J. Smith, Table of n, a(n) for n=1..500
A. Barbé, Symmetric patterns in the cellular automaton that generates Pascal's triangle modulo 2, Discr. Appl. Math. 105(2000), 1-38.
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-4,-4,10,-4,-4,4,8,8,-16).

Cf. A000079, A060546, A060547, A060548, A008619, A008611, A008615.

LinearRecurrence[{2,2,-2,-4,-4,10,-4,-4,4,8,8,-16},{0,0,0,6,12,42,84,210,420,924,1860,3900},40] (* Harvey P. Dale, Feb 01 2015 *)

a(n) = { 2^n-3*2^ceil(n/2)-2^(floor(n/3)+(n%3)%2)+3*2^(floor((n+3)/6)+(n%6==1)) } \\ Harry J. Smith, Jul 07 2009

a(n) = 2^n - 3*2^ceiling(n/2) - 2^(floor(n/3)+(n mod 3)mod 2) + 3*2^(floor((n+3)/6) + d(n)), with d(n)=1 if n mod 6=1 else d(n)=0.
a(n) = A000079(n) - 3*A060546(n) - A060547(n) + 3*A060548(n).
a(n) = A000079(n) - 3*2^A008619(n-1) - 2^A008611(n-1) + 3*2^A008615(n+1), for n >= 1.
G.f.: -6*x^4*(2*x^6 + 2*x^5 - x^4 + 2*x^3 - x^2 - 1) / ((2*x-1)*(2*x^2-1)*(2*x^3-1)*(2*x^6-1)). - Colin Barker, Aug 29 2013
a(n) = 6*A060552(n). - Andrew Howroyd, Dec 24 2024

More terms from Colin Barker, Aug 29 2013

A060553 a(n) is the number of distinct (modulo geometric D3-operations) patterns which can be formed by an equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.

Original entry on oeis.org

2, 2, 4, 6, 10, 16, 32, 52, 104, 192, 376, 720, 1440, 2800, 5600, 11072, 22112, 43968, 87936, 175296, 350592, 700160, 1400192, 2798336, 5596672, 11188992, 22377984, 44747776, 89495040, 178973696, 357947392, 715860992, 1431721984, 2863378432, 5726754816

Offset: 1

André Barbé (Andre.Barbe(AT)esat.kuleuven.ac.be), Apr 03 2001

Harry J. Smith, Table of n, a(n) for n=1..500
A. Barbé, Symmetric patterns in the cellular automaton that generates Pascal's triangle modulo 2, Discr. Appl. Math. 105(2000), 1-38.
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-4,-4,8).

Cf. A000079, A060547, A060546, A008611, A008619.

a(n) = { (2^(n-1) + 2^(floor(n/3) + (n%3)%2))/3 + 2^floor((n-1)/2) } \\ Harry J. Smith, Jul 07 2009

a(n) = (2^(n-1)+2^(floor(n/3) + (n mod 3)mod 2))/3 + 2^floor((n-1)/2).
a(n) = (A000079(n-1) + A060547(n))/3 + A060546(n)/2.
a(n) = (A000079(n-1) + 2^A008611(n-1))/3 + 2^(A008619(n-1) - 1), for n >= 1.
G.f.: -2*x*(4*x^5 + x^4 - x^3 - 2*x^2 - x + 1) / ((2*x-1)*(2*x^2-1)*(2*x^3-1)). - Colin Barker, Aug 29 2013

More terms from Colin Barker, Aug 29 2013

A260211 Irregular triangle read by rows, T(n,k) is the decimal number conversion from an n-bit symmetric binary table arranged in ascending order for n > 1.

Original entry on oeis.org

0, 1, 0, 3, 0, 2, 5, 7, 0, 6, 9, 15, 0, 4, 10, 14, 17, 21, 27, 31, 0, 12, 18, 30, 33, 45, 51, 63, 0, 8, 20, 28, 34, 42, 54, 62, 65, 73, 85, 93, 99, 107, 119, 127, 0, 24, 36, 60, 66, 90, 102, 126, 129, 153, 165, 189, 195, 219, 231, 255

Offset: 1

Kival Ngaokrajang, Jul 19 2015

The sequence of row lengths is A060546(n).

Column 2 is A164073. See illustration.

			The irregular triangle begins:
n\k 0  1  2  3  4  5  6  7 ...
1   0  1
2   0  3
3   0  2  5  7
4   0  6  9 15
5   0  4 10 14 17 21 27 31
6   0 12 18 30 33 45 51 63
...

Kival Ngaokrajang, Illustration of initial terms

Cf. A060546, A164073.

Flatten[Table[Select[Range[0,2^k -1], #==IntegerReverse[#,2,k]&],{k,1,8}]] (* Ed Pegg Jr, May 03 2021 *)

A092434 Number of words X=x(1)x(2)x(3)...x(n) of length n in three digits {0,1,2} that are invariant under the mapping X -> Y, where y(i)=((AD)^(i-1))x(1) and where (AD) denotes the absolute difference (AD)x(i)=abs(x(i+1)-x(i)) (in other words, y(i) is the i-th element in the diagonal of leading entries in the table of absolute differences of {x(1), x(2),...,x(n)}).

Original entry on oeis.org

3, 4, 10, 12, 28, 32, 72, 80, 176, 192, 416, 448, 960, 1024

Offset: 1

John W. Layman, Mar 23 2004

In the two digits {0,1} the corresponding sequence is 2,2,4,4,8,8,16,16,32,32,64,64,... which appears to be A060546.

			The table of absolute differences of {2,1,1,0} is
2
1.1
1.0.1
0.1.1.0
with the diagonal of leading absolute differences again forming the word (2110).
Thus (2110) is one of the twelve words in the digits {0,1,2} that are counted in calculating a(4).

Cf. A060546.

It is conjectured that a(n)=(n+2)*2^((n-1) div 2).

cp's OEIS Frontend

A347171 Triangle read by rows where T(n,k) is the sum of Golay-Rudin-Shapiro terms GRS(j) (A020985) for j in the range 0 <= j < 2^n and having binary weight wt(j) = A000120(j) = k.

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A381899 Irregular triangular array read by rows. T(n,k) is the number of length n words x on {0,1} such that I(x) + W(x)*(n-W(x)) = k, where I(x) is the number of inversions in x and W(x) is the number of 1's in x, n >= 0, 0 <= k <= floor(n^2/2).

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A387025 Start with the list of positive integers L_1 = (1, 2, ...); for n = 1, 2, ..., let m be the least integer > n such that L_n(n) divides L_n(m); L_{n+1}(k) = L_n(k) for any k <> m, L_{n+1}(m) = L_n(m)/L_n(n); a(n) = L_n(n).

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A260211 Irregular triangle read by rows, T(n,k) is the decimal number conversion from an n-bit symmetric binary table arranged in ascending order for n > 1.

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Mathematica

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