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A360461
T(n,k) is the sum of all the k-th smallest divisors of all positive integers <= n. Irregular triangle read by rows (n>=1, k>=1).
1
1, 2, 2, 3, 5, 4, 7, 4, 5, 12, 4, 6, 14, 7, 6, 7, 21, 7, 6, 8, 23, 11, 14, 9, 26, 20, 14, 10, 28, 25, 24, 11, 39, 25, 24, 12, 41, 28, 28, 6, 12, 13, 54, 28, 28, 6, 12, 14, 56, 35, 42, 6, 12, 15, 59, 40, 57, 6, 12, 16, 61, 44, 65, 22, 12, 17, 78, 44, 65, 22, 12, 18, 80, 47, 71, 31, 30, 19, 99, 47, 71, 31, 30
OFFSET
1,2
COMMENTS
Also, looking at all the partitions into equal-sized parts of all positive integers <= n, T(n,k) is the total number of parts in the partitions with the k-th largest parts.
Column k lists the partial sums of the column k of A027750.
The rows where the length row increases to a record gives A002182.
LINKS
Michael De Vlieger, Log log scatterplot of row n, n = 1..5040, showing T(n, k) with trajectory of k in a color function where black indicates k = 1, red k = 2, ..., magenta k = 60, ignoring zeros.
EXAMPLE
Triangle begins:
1;
2, 2;
3, 5;
4, 7, 4;
5, 12, 4;
6, 14, 7, 6;
7, 21, 7, 6;
8, 23, 11, 14;
9, 26, 20, 14;
10, 28, 25, 24;
11, 39, 25, 24;
12, 41, 28, 28, 6, 12;
...
For n = 6 the divisors, in increasing order, of all positive integers <= 6 are as follows:
-----------------------------
n\k | 1 2 3 4
-----------------------------
1 | 1
2 | 1, 2
3 | 1, 3
4 | 1, 2, 4
5 | 1, 5
6 | 1, 2, 3, 6
.
The sum of the first divisors (k = 1) is equal to 1+1+1+1+1+1 = 6, so T(6,1) = 6.
The sum of the second divisors (k = 2) is equal to 2+3+2+5+2 = 14, so T(6,2) = 14.
The sum of the third divisors (k = 3) is equal to 4+3 = 7, so T(6,3) = 7.
The sum of the fourth divisors (k = 4) is equal to 6, so T(6,4) = 6.
So the 6th row of the triangle is [6, 14, 7, 6].
Also, for n = 6 the partitions into equal parts, with the sizes of the parts in decreasing order, of all positive integers <= 6 are as follows:
----------------------------------------------------
n\k | 1 2 3 4
----------------------------------------------------
1 | [1]
2 | [2], [1,1]
3 | [3], [1,1,1]
4 | [4], [2,2], [1,1,1,1]
5 | [5], [1,1,1,1,1]
6 | [6], [3,3], [2,2,2], [1,1,1,1,1,1]
.
The total number of parts in the 1st partitions (k = 1) is 6, so T(6,1) = 6.
The total number of parts in the 2nd partitions (k = 2) is 14, so T(6,2) = 14.
The total number of parts in the 3rd partitions (k = 3) is 7, so T(6,3) = 7.
The total number of parts in the 4th partitions (k = 4) is 6, so T(6,4) = 6.
So the 6th row of the triangle is [6, 14, 7, 6].
MATHEMATICA
nn = 20; s[_] = 0; m[0] = 0; Do[Set[m[n], Max[m[n - 1], DivisorSigma[0, n]]], {n, nn}]; Do[MapIndexed[(s[First[#2]] += #1; Set[t[n, First[#2]], s[First[#2]]]) &, PadRight[Divisors[n], m[n]]], {n, nn}]; Table[t[n, k], {n, nn}, {k, m[n]}] // Flatten (* Michael De Vlieger, Mar 04 2023 *)
A360461[rowmax_]:=DeleteCases[Accumulate[PadRight[Divisors[Range[rowmax]]]], 0, {2}];
A360461[20] (* Generates 20 rows *) (* Paolo Xausa, Mar 05 2023 *)
PROG
(PARI) rowlen(n) = vecmax(vector(n, k, numdiv(k))); \\ A070319
row(n) = my(vd=vector(n, i, divisors(i)), nb=rowlen(n)); vector(nb, k, sum(i=1, #vd, if (#(vd[i]) >= k, vd[i][k]))); \\ Michel Marcus, Mar 06 2023
CROSSREFS
Row sums give A024916.
Row lengths give A070319.
Column 1 gives A000027.
Column 2 gives A088821.
The sum of the first n rows gives A175254.
Main sequences: A027750 and A244051.
Sequence in context: A117267 A086363 A174094 * A284114 A323480 A330404
KEYWORD
nonn,tabf,look
AUTHOR
Omar E. Pol, Feb 07 2023
STATUS
approved