A365968 -id:A365968 - cp's OEIS frontend

A367076 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). T(n,k) = -Sum_{i=0..k} A365968(n,i).

0, 1, 0, 3, 4, 3, 0, 6, 10, 12, 12, 12, 10, 6, 0, 10, 18, 24, 28, 32, 34, 34, 32, 34, 34, 32, 28, 24, 18, 10, 0, 15, 28, 39, 48, 57, 64, 69, 72, 79, 84, 87, 88, 89, 88, 85, 80, 85, 88, 89, 88, 87, 84, 79, 72, 69, 64, 57, 48, 39, 28, 15, 0, 21, 40, 57, 72, 87

Offset: 0

John Tyler Rascoe, Nov 05 2023

			Triangle begins:
    k=0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
n=0:  0;
n=1:  1,  0;
n=2:  3,  4,  3,  0;
n=3:  6, 10, 12, 12, 12, 10,  6,  0;
n=4; 10, 18, 24, 28, 32, 34, 34, 32, 34, 34, 32, 28, 24, 18, 10,  0;

John Tyler Rascoe, Rows n = 0..12, flattened
Wikipedia, Blancmange curve.
Index entries for sequences related to binary expansion of n

Cf. A000217 (column k=0), A028552 (column k=1), A192021 (row sums).

Cf. A004074, A249071, A274575, A277914, A296062, A365968.

nmax=10; row[n_]:=Join[CoefficientList[Series[1/(1-x)*Sum[ i/(1+x^2^(i-1))*Product[1+x^2^j,{j,0,i-2}],{i,n}],{x,0,2^n-1}],x],{0}]; Array[row,6,0] (* Stefano Spezia, Dec 23 2023 *)

def row_gen(n):
    x = 0
    for k in range(2**n):
        b = bin(k)[2:].zfill(n)
        x += sum((-1)**(int(b[n-i])+1)*i for i in range(1,n+1))
        yield(-x)
def A367076_row_n(n): return(list(row_gen(n)))

T(n,k) = Sum_{i=0..n} abs(k + 1 - (2^i) * round((k+1)/2^i)) * i.

G.f. for n-th row: 1/(1-x) * Sum_{i=1..n} (i/(1+x^2^(i-1)) * Product_{j=0..i-2} 1 + x^2^j).

A377170 Sum of the nonnegative terms in the n-th row of A365968.

Original entry on oeis.org

0, 1, 4, 12, 36, 98, 250, 616, 1484, 3508, 8140, 18620, 42164, 94632, 210518, 464840, 1020556, 2229014, 4843316, 10476164, 22576728, 48489154, 103790370, 221484824, 471427432, 1001027226, 2120503144, 4482083616, 9455815160, 19913405076, 41862056992, 87857540836

Offset: 0

John Tyler Rascoe, Oct 18 2024

By symmetry -a(n) is the sum of the nonpositive terms in the n-th row of A365968.

			The 4th row of A365968 is: [-10, -8, -6, -4, -4, -2, 0, 2, -2, 0, 2, 4, 4, 6, 8, 10], so a(4) = 2 + 2 + 4 + 4 + 6 + 8 + 10 = 36.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000079, A365968, A367076, A384868.

b:= proc(n, s) option remember; `if`(n=0, s,
      b(n-1, abs(s-n))+b(n-1, s+n))
    end:
a:= n-> b(n, 0)/2:
seq(a(n), n=0..31);  # Alois P. Heinz, Jun 13 2025

a(n) = (1/2) * Sum_{k=0..2^n-1} abs(A365968(n,k)).

a(n) = (1/2) * Sum_{i=0..2^n-1} abs(A384868(i+2^n-1)). - Alois P. Heinz, Jun 13 2025

A384868 a(n) = Sum_{i=1...|b|} i*(-1)^b_i where b is the lexicographically n-th binary string.

Original entry on oeis.org

0, 1, -1, 3, -1, 1, -3, 6, 0, 2, -4, 4, -2, 0, -6, 10, 2, 4, -4, 6, -2, 0, -8, 8, 0, 2, -6, 4, -4, -2, -10, 15, 5, 7, -3, 9, -1, 1, -9, 11, 1, 3, -7, 5, -5, -3, -13, 13, 3, 5, -5, 7, -3, -1, -11, 9, -1, 1, -9, 3, -7, -5, -15, 21, 9, 11, -1, 13, 1, 3, -9, 15, 3, 5, -7, 7, -5, -3, -15, 17

Offset: 0

Christopher Purcell, Jun 11 2025

The first binary string is the empty string and is indexed n=0.

			The lexicographically 8th binary string is 001; therefore, a(8) = 1 + 2 - 3 = 0.
Sequence can be written as triangle T(n,k) with row lengths 2^n:
   0;
   1, -1;
   3, -1, 1, -3;
   6,  0, 2, -4, 4, -2, 0, -6;
  10,  2, 4, -4, 6, -2, 0, -8, 8, 0, 2, -6, 4, -4, -2, -10;
  ...

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Cf. A000079, A000217, A000225, A007931, A100575, A231204, A365968, A377170.

a(n) = my(b=[d|d<-binary(n+1)[^1]]); sum(i=1, #b, i*(-1)^b[i]); \\ Michel Marcus, Jun 11 2025

from math import comb
def A384868(n): return comb(len(s:=bin(n+1)[3:])+1,2)-(sum(i for i,j in enumerate(s,1) if j=='1')<<1) # Chai Wah Wu, Jun 13 2025

def a384868(n): return sum(i if b == '0' else -i for i, b in enumerate(bin(n + 1)[3:], 1)) # David Radcliffe, Jun 15 2025

From Alois P. Heinz, Jun 13 2025: (Start)
a(A000225(n)) = A000217(n).
a(2*(2^n-1)) = (-1)*A000217(n).
Sum_{i=0..2^n-1} a(i+2^n-1) = 0.
Sum_{i=0..2^n-1} i*a(i+2^n-1) = (-1)*A100575(n+1).
Sum_{i=0..2^n-1} abs(a(i+2^n-1)) = 2*A377170(n). (End)

A373202 Product over all sums of {n, +/-(n-1), +/-(n-2), ..., +/-1}, where "+/-" means here we use all possible combination of signs, this means for n > 0 that we have 2^(n-1) factors.

Original entry on oeis.org

1, 1, 3, 0, 0, -28733079375, -821329806742140930609375, 0, 0

Offset: 0

Thomas Scheuerle, May 28 2024

sign

a(9) is roughly 10^250.72 and is too big for the data section.
Consider the polynomial recurrence: P_{n}(x) = P_{n-1}(x + n)*P_{n-1}(x - n), with P_{1}(x) = 1 + x and P_{0}(x) = 1, then |a(n)| = P_{n}(0).
Each nonzero term divides all nonzero terms with higher index. This can be understood by looking at the tree structure in A365968 as the factors of a(n) are obtained in row n (second half of row). Each factor is only dependent by inheritance from parent and grandparent in this tree. Each row contains either all even numbers from 0 to A000217(n) or all odd numbers from 0 to A000217(n). In the even case 0 will be a factor and thus a(n) = 0. In the odd case some factors are duplicates, but by inheritance we will keep the duplicates from parent rows included.

			a(4) = (4+3+2+1)*(4+3+2-1)*(4+3-2+1)*(4+3-2-1)*(4-3+2+1)*(4-3+2-1)*(4-3-2+1)*(4-3-2-1) = 10*8*6*4*4*2*0*(-2) = 0.

Cf. A365968 (row n gives factors of a(n)).

Cf. A001787 (result if we would use summation only).

a[n_]:=Product[Sum[(n+1-m)(2*Part[IntegerDigits[2^(n-1)+k,2],m]-1),{m,n}],{k,0,2^(n-1)-1}]; Array[a,10,0] (* Stefano Spezia, May 29 2024 *)

a(n) = prod(k=0, 2^(n-1)-1, sum(m=1, n, (n+1-m)*(-1+2*(digits(2^(n-1)+k, 2)[m]))))

cp's OEIS Frontend

A367076 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). T(n,k) = -Sum_{i=0..k} A365968(n,i).

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A377170 Sum of the nonnegative terms in the n-th row of A365968.

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A384868 a(n) = Sum_{i=1...|b|} i*(-1)^b_i where b is the lexicographically n-th binary string.

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A373202 Product over all sums of {n, +/-(n-1), +/-(n-2), ..., +/-1}, where "+/-" means here we use all possible combination of signs, this means for n > 0 that we have 2^(n-1) factors.

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Author

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Programs

Mathematica

PARI