A231345 -id:A231345 - cp's OEIS frontend

A271344 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers multiplied by -1, interleaved with k-1 zeros, but T(n,1) = 1 and the first element of column k is in row k(k+1)/2.

1, 1, 1, -1, 1, 0, 1, -3, 1, 0, -1, 1, -5, 0, 1, 0, 0, 1, -7, -3, 1, 0, 0, -1, 1, -9, 0, 0, 1, 0, -5, 0, 1, -11, 0, 0, 1, 0, 0, -3, 1, -13, -7, 0, -1, 1, 0, 0, 0, 0, 1, -15, 0, 0, 0, 1, 0, -9, -5, 0, 1, -17, 0, 0, 0, 1, 0, 0, 0, -3, 1, -19, -11, 0, 0, -1, 1, 0, 0, -7, 0, 0, 1, -21, 0, 0, 0, 0, 1, 0, -13, 0, 0, 0

Offset: 1

Omar E. Pol, Apr 19 2016

Gives an identity for the deficiency of n. Alternating sum of row n equals the deficiency of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A033879(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
The number of nonzero elements of row n is A001227(n).
If T(n,k) is the second nonzero term in column k then T(n+1,k+1) = -1 is the first element of column k+1.

			Triangle begins:
  1;
  1;
  1,  -1;
  1,   0;
  1,  -3;
  1,   0,  -1;
  1,  -5,   0;
  1,   0,   0;
  1,  -7,  -3;
  1,   0,   0,  -1;
  1,  -9,   0,   0;
  1,   0,  -5,   0;
  1, -11,   0,   0;
  1,   0,   0,  -3;
  1, -13,  -7,   0,  -1;
  1,   0,   0,   0,   0;
  1, -15,   0,   0,   0;
  1,   0,  -9,  -5,   0;
  1, -17,   0,   0,   0;
  1,   0,   0,   0,  -3;
  1, -19, -11,   0,   0,  -1;
  1,   0,   0,  -7,   0,   0;
  1, -21,   0,   0,   0,   0;
  1,   0, -13,   0,   0,   0;
  1, -23,   0,   0,  -5,   0;
  1,   0    0,  -9,   0,   0;
  1, -25, -15,   0,   0,  -3;
  1,   0,   0,   0,   0,   0,  -1;
  ...
For n = 24 the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 so the deficiency of 24 is 24 - 12 - 8 - 6 - 4 - 3 - 2 - 1 = -12. On the other hand the 24th row of triangle is 1, 0, -13, 0, 0, 0, and the alternating row sum is 1 - 0 +(-13) - 0 + 0 - 0 = -12, equaling the deficiency of 24; A033879(24) = -12, so 24 is an abundant number (A005101).
For n = 27 the divisors of 27 are 1, 3, 9, 27 so the deficiency of 27 is 27 - 9 - 3 - 1 = 14. On the other hand the 27th row of triangle is 1, -25, -15, 0, 0, -3, and the alternating row sum is 1 -(-25) +(-15) - 0 + 0 -(-3) = 14, equalling the deficiency of 27; A033879(27) = 14, so 27 is a deficient number (A005100).
For n = 28 the divisors of 28 are 1, 2, 4, 7, 14, 28 so the deficiency of 28 is 28 - 14 - 7 - 4 - 2 - 1 = 0. On the other hand the 28th row of triangle is 1, 0, 0, 0, 0, 0, -1, and the alternating row sum is 1 - 0 + 0 - 0 + 0 - 0 +(-1) = 0, equaling the deficiency of 28; A033879(28) = 0, so 28 is a perfect number (A000396).

Column 1 is A000012. Column 2 is -1*A193356.

Cf. A000203, A000217, A000396, A001227, A003056, A033879, A033880, A005100, A005101, A196020, A231345, A237593.

T(n,k) = -1*A231345(n,k).

T(n,k) = -1*A196020(n,k), if k >= 2.

A239313 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros, except the first column which lists 0 together with the nonnegative integers, and the first element of column k is in row k*(k+1)/2.

Original entry on oeis.org

0, 0, 1, 1, 2, 0, 3, 3, 4, 0, 1, 5, 5, 0, 6, 0, 0, 7, 7, 3, 8, 0, 0, 1, 9, 9, 0, 0, 10, 0, 5, 0, 11, 11, 0, 0, 12, 0, 0, 3, 13, 13, 7, 0, 1, 14, 0, 0, 0, 0, 15, 15, 0, 0, 0, 16, 0, 9, 5, 0, 17, 17, 0, 0, 0, 18, 0, 0, 0, 3, 19, 19, 11, 0, 0, 1, 20, 0, 0, 7, 0, 0

Offset: 1

Omar E. Pol, Mar 15 2014

Alternating row sums give the Chowla's function, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A048050(n).
Row n has length A003056(n) hence column k starts in row A000217(k).
Column 1 gives 0 together with A001477.
Column 2 is A193356.
The number of positive terms in row n is A001227(n), if n >= 3. - Omar E. Pol, Apr 18 2016

			Triangle begins (row n = 1..24):
0;
0;
1,   1;
2,   0;
3,   3;
4,   0,  1;
5,   5,  0;
6,   0,  0;
7,   7,  3;
8,   0,  0,  1;
9,   9,  0,  0;
10,  0,  5,  0;
11, 11,  0,  0;
12,  0,  0,  3;
13, 13,  7,  0,  1;
14,  0,  0,  0,  0;
15, 15,  0,  0,  0;
16,  0,  9,  5,  0;
17, 17,  0,  0,  0;
18,  0,  0,  0,  3;
19, 19, 11,  0,  0,  1;
20,  0,  0,  7,  0,  0;
21, 21,  0,  0,  0,  0;
22,  0, 13,  0,  0,  0;
...
For n = 15 the divisors of 15 are 1, 3, 5, 15 therefore the sum of divisors of 15 except 1 and 15 is 3 + 5 = 8. On the other hand the 15th row of triangle is 13, 13, 7, 0, 1, hence the alternating row sum is 13 - 13 + 7 - 0 + 1 = 8, equalling the sum of divisors of 15 except 1 and 15.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of divisors of n, except 1 and n. Example: the sum of divisors of 24 except 1 and 24 is 2 + 3 + 4 + 6 + 8 + 12 = 35, and the alternating sum of the 24th row of triangle is 22 - 0 + 13 - 0 + 0 - 0 = 35.

Cf. A000203, A000217, A001065, A001227, A001477, A193356, A196020, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112.

T(n,k) = A196020(n,k), if k >= 2. - Omar E. Pol, Apr 18 2016

A347737 Zero together with the partial sums of A238005.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 5, 7, 9, 11, 13, 15, 16, 20, 23, 25, 28, 31, 33, 37, 41, 45, 48, 52, 54, 59, 64, 67, 72, 78, 81, 86, 89, 94, 100, 106, 110, 116, 122, 126, 132, 138, 141, 148, 155, 162, 168, 174, 179, 186, 193, 198, 204, 212, 218, 226, 234, 240, 248, 256, 260

Offset: 0

Omar E. Pol, Sep 11 2021

nonn

a(n) is also the total number of ones in the first n rows of A347579, n >= 1.

a(n) is also the total number of zeros in the first n rows of the triangles A196020, A211343, A231345, A236106, A237048 (simpler), A239662, A261699, A271344, A272026, A280850, A285574, A285891, A285914, A286013, A296508 (and possibly others), n >= 1.

Cf. A001227, A003056, A006463, A060831, A237593, A238005, A347579.

Accumulate@Table[Length@Select[Select[IntegerPartitions@n,DuplicateFreeQ],Differences@MinMax@#=={Length@#}&],{n,60}] (* Giorgos Kalogeropoulos, Sep 12 2021 *)

from math import isqrt
def A347737(n): return (r:=isqrt((n+1<<3)+1)-1>>1)*(6*n+4-r*(r+3))//6-((t:=isqrt(m:=n>>1))+(s:=isqrt(n)))*(t-s)-(sum(n//k for k in range(1,s+1))-sum(m//k for k in range(1,t+1))<<1) # Chai Wah Wu, Jun 07 2025

a(n) = A006463(n+1) - A060831(n).

A380231 Alternating row sums of triangle A237591.

Original entry on oeis.org

1, 2, 1, 2, 1, 4, 3, 4, 5, 4, 3, 6, 5, 4, 7, 8, 7, 8, 7, 10, 9, 8, 7, 10, 11, 10, 9, 12, 11, 14, 13, 14, 13, 12, 15, 16, 15, 14, 13, 16, 15, 18, 17, 16, 19, 18, 17, 20, 21, 22, 21, 20, 19, 22, 21, 24, 23, 22, 21, 24, 23, 22, 25, 26, 25, 28, 27, 26, 25, 28, 27, 32, 31, 30, 29, 28, 31, 30, 29

Offset: 1

Omar E. Pol, Jan 17 2025

nonn

Consider the symmetric Dyck path in the first quadrant of the square grid described in the n-th row of A237593. Let C = (A240542(n), A240542(n)) be the middle point of the Dyck path.
a(n) is also the coordinate on the x axis of the point (a(n),n) and also the coordinate on the y axis of the point (n,a(n)) such that the middle point of the line segment [(a(n),n),(n,a(n))] coincides with the middle point C of the symmetric Dyck path.
The three line segments [(a(n),n),C], [(n,a(n)),C] and [(n,n),C] have the same length.
For n > 2 the points (n,n), C and (a(n),n) are the vertices of a virtual isosceles right triangle.
For n > 2 the points (n,n), C and (n,a(n)) are the vertices of a virtual isosceles right triangle.
For n > 2 the points (a(n),n), (n,n) and (n,a(n)) are the vertices of a virtual isosceles right triangle.

			For n = 14 the 14th row of A237591 is [8, 3, 1, 2] hence the alternating row sum is 8 - 3 + 1 - 2 = 4, so a(14) = 4.
On the other hand the 14th row of A237593 is the 14th row of A237591 together with the 14 th row of A237591 in reverse order as follows: [8, 3, 1, 2, 2, 1, 3, 8].
Then with the terms of the 14th row of A237593 we can draw a Dyck path in the first quadrant of the square grid as shown below:
.
         (y axis)
          .
          .
          .    (4,14)              (14,14)
          ._ _ _ . _ _ _ _            .
          .               |
          .               |
          .               |_
          .                 |
          .                 |_ _
          .                C    |_ _ _
          .                           |
          .                           |
          .                           |
          .                           |
          .                           . (14,4)
          .                           |
          .                           |
          . . . . . . . . . . . . . . | . . . (x axis)
        (0,0)
.
In the example the point C is the point (9,9).
The three line segments [(4,14),(9,9)], [(14,4),(9,9)] and [(14,14),(9,9)] have the same length.
The points (14,14), (9,9) and (4,14) are the vertices of a virtual isosceles right triangle.
The points (14,14), (9,9) and (14,4) are the vertices of a virtual isosceles right triangle.
The points (4,14), (14,14) and (14,4) are the vertices of a virtual isosceles right triangle.

Other alternating row sums (ARS) related to the Dyck paths of A237593 and the stepped pyramid described in A245092 are as follows:
ARS of A237593 give A000004.
ARS of A196020 give A000203.
ARS of A252117 give A000203.
ARS of A271343 give A000593.
ARS of A231347 give A001065.
ARS of A236112 give A004125.
ARS of A236104 give A024916.
ARS of A249120 give A024916.
ARS of A271344 give A033879.
ARS of A231345 give A033880.
ARS of A239313 give A048050.
ARS of A237048 give A067742.
ARS of A236106 give A074400.
ARS of A235794 give A120444.
ARS of A266537 give A146076.
ARS of A236540 give A153485.
ARS of A262612 give A175254.
ARS of A353690 give A175254.
ARS of A239446 give A235796.
ARS of A239662 give A239050.
ARS of A235791 give A240542.
ARS of A272026 give A272027.
ARS of A211343 give A336305.
Cf. A221529, A237270, A237271, A237591, A262626, A322141.

row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
a(n) = my(orow = concat(row235791(n), 0)); vecsum(vector(#orow-1, i, (-1)^(i+1)*(orow[i] - orow[i+1]))); \\ Michel Marcus, Apr 13 2025

a(n) = 2*A240542(n) - n.
a(n) = n - 2*A322141(n).
a(n) = A240542(n) - A322141(n).

cp's OEIS Frontend

A271344 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers multiplied by -1, interleaved with k-1 zeros, but T(n,1) = 1 and the first element of column k is in row k(k+1)/2.

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A239313 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros, except the first column which lists 0 together with the nonnegative integers, and the first element of column k is in row k*(k+1)/2.

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A347737 Zero together with the partial sums of A238005.

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A380231 Alternating row sums of triangle A237591.

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