A249120 -id:A249120 - cp's OEIS frontend

A252117 Irregular triangle read by row: T(n,k), n>=1, k>=1, in which column k lists the numbers of A000716 multiplied by A000330(k), and the first element of column k is in row A000217(k).

1, 3, 9, 5, 22, 15, 51, 45, 108, 110, 14, 221, 255, 42, 429, 540, 126, 810, 1105, 308, 1479, 2145, 714, 30, 2640, 4050, 1512, 90, 4599, 7395, 3094, 270, 7868, 13200, 6006, 660, 13209, 22995, 11340, 1530, 21843, 39340, 20706, 3240, 55, 35581, 66045, 36960, 6630, 165, 57222, 109215, 64386, 12870, 495

Offset: 1

Omar E. Pol, Dec 14 2014

Gives an identity for sigma(n). Alternating sum of row n equals A000203(n), the sum of the divisors of n.
Row n has length A003056(n) hence column k starts in row A000217(k).
Column 1 is A000716, but here the offset is 1 not 0.
The 1st element of column k is A000330(k).
The 2nd element of column k is A059270(k).
The 3rd element of column k is A220443(k).
The partial sums of column k give the k-th column of A249120.
This triangle has been constructed after Mircea Merca's formula for A000203.
From Omar E. Pol, May 05 2022: (Start)
In the Honda-Yoda paper see "3. String theory and Riemann hypothesis". The coefficients that are mentioned in 3.11 are the first 16 terms of A000716, the coefficients that are mentioned in 3.12 are the first 5 terms of A000330, and the coefficients that are mentioned in 3.13 are the first 16 terms of A000203.
Another triangle with the same row lengths and whose alternating row sums give A000203 is A196020. (End)

			Triangle begins:
       1;
       3;
       9,      5;
      22,     15;
      51,     45;
     108,    110,     14;
     221,    255,     42;
     429,    540,    126;
     810,   1105,    308;
    1479,   2145,    714,     30;
    2640,   4050,   1512,     90;
    4599,   7395,   3094,    270;
    7868,  13200,   6006,    660;
   13209,  22995,  11340,   1530;
   21843,  39340,  20706,   3240,    55;
   35581,  66045,  36960,   6630,   165;
   57222, 109215,  64386,  12870,   495;
   90882, 177905, 110152,  24300,  1210;
  142769, 286110, 184926,  44370,  2805;
  221910, 454410, 305802,  79200,  5940;
  341649, 713845, 498134, 137970, 12155, 91;
...
For n = 6 the divisors of 6 are 1, 2, 3, 6, so the sum of the divisors of 6 is 1 + 2 + 3 + 6 = 12. On the other hand, the 6th row of the triangle is 108, 110, 14, so the alternating row sum is 108 - 110 + 14 = 12, equaling the sum of the divisors of 6.
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of the divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 21843, 39340, 20706, 3240, 55, so the alternating row sum is 21843 - 39340 + 20706 - 3240 + 55 = 24, equaling the sum of the divisors of 15.

Masazumi Honda and Takuya Yoda, String theory, N = 4 SYM and Riemann hypothesis, arXiv:2203.17091 [hep-th], (2022), pp. 5-6.
Index entries for sequences related to sigma(n)

Cf. A000203, A000217, A000330, A000716, A003056, A059270, A196020, A220443, A238442, A249120.

A003056(n) = (sqrtint(8*n+1)-1)\2;
A000330(n) = n*(n+1)*(2*n+1)/6;
A000217(n) = n*(n+1)/2;
A000716(n) = polcoef(1/eta('x+O('x^(n+1)))^3, n, x);
T(n, k) = A000330(k)*A000716(n-A000217(k));
row(n) = vector(A003056(n), k, T(n,k)); \\ Michel Marcus, Oct 29 2022

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).

A353690 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers of A353689 multiplied by A000330(k), and the first element of column k is in row A000217(k).

Original entry on oeis.org

1, 5, 18, 5, 53, 25, 139, 90, 333, 265, 14, 748, 695, 70, 1592, 1665, 252, 3246, 3740, 742, 6379, 7960, 1946, 30, 12152, 16230, 4662, 150, 22524, 31895, 10472, 540, 40764, 60760, 22288, 1590, 72213, 112620, 45444, 4170, 125505, 203820, 89306, 9990, 55, 214378, 361065, 170128, 22440, 275

Offset: 1

Omar E. Pol, May 04 2022

The alternating sum of the n-th row equals A175254(n), the volume of the stepped pyramid with n levels described in A245092, also the n-th term of the convolution of A000203 and A000027.
Column k is the partial sums of the k-th column of the triangle A249120.
Another triangle with the same row lengths and whose alternating row sums give A175254 is A262612.

			Triangle begins:
        1;
        5;
       18,       5;
       53,      25;
      139,      90;
      333,     265,      14;
      748,     695,      70;
     1592,    1665,     252;
     3246,    3740,     742;
     6379,    7960,    1946,     30;
    12152,   16230,    4662,    150;
    22524,   31895,   10472,    540;
    40764,   60760,   22288,   1590;
    72213,  112620,   45444,   4170;
   125505,  203820,   89306,   9990,    55;
   214378,  361065,  170128,  22440,   275;
   360473,  627525,  315336,  47760,   990;
   597450, 1071890,  570696,  97380,  2915;
   977196, 1802365, 1010982, 191370,  7645;
  1578852, 2987250, 1757070, 364560, 18315;
  2522157, 4885980, 3001292, 675720, 41140, 91;
  ...
For n = 6 we have that A175254(6) is equal to [1] + [1 + 3] + [1 + 3 + 4] + [1 + 3 + 4 + 7] + [1 + 3 + 4 + 7 + 6] + [1 + 3 + 4 + 7 + 6 + 12] = 1 + 4 + 8 + 15 + 21 + 33 = 82. On the other hand the alternating sum of the 6th row of the triangle is 333 - 265 + 14 = 82, equaling A175254(6).

Column 1 is A353689.
Row n has length A003056(n).
Column k starts in row A000217(k).
The first element in column k is A000330(k).
Alternating row sums give A175254.
Cf. A000203, A000716, A196020, A210843, A236104, A237593, A245092, A249120, A252117, A262612.

A175254(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).

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

A252117 Irregular triangle read by row: T(n,k), n>=1, k>=1, in which column k lists the numbers of A000716 multiplied by A000330(k), and the first element of column k is in row A000217(k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A353690 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers of A353689 multiplied by A000330(k), and the first element of column k is in row A000217(k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A380231 Alternating row sums of triangle A237591.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula