Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #98 Jan 19 2024 07:09:46
%S 1,3,4,1,7,3,1,6,4,3,1,1,12,7,4,3,3,1,1,8,6,7,4,4,3,3,1,1,1,1,15,12,6,
%T 7,7,4,4,3,3,3,3,1,1,1,1,13,8,12,6,6,7,7,4,4,4,4,3,3,3,3,1,1,1,1,1,1,
%U 1,18,15,8,12,12,6,6,7,7,7,7,4,4,4,4,3,3,3,3,3,3,3,1,1,1,1,1,1,1,1
%N Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which the partition number A000041(n-1) is the length of row n and every column k is A000203, the sum of divisors function.
%C The sum of row n equals A138879(n), the sum of all parts in the last section of the set of partitions of n.
%C T(n,k) is also the number of cubic cells (or cubes) added at the n-th stage in the k-th level starting from the base in the tower described in A221529, assuming that the tower is an object under construction (see the example). - _Omar E. Pol_, Jan 20 2022
%H Paolo Xausa, <a href="/A339278/b339278.txt">Table of n, a(n) for n = 1..11732 (rows 1..27 of triangle, flattened)</a>
%F a(m) = A000203(A336811(m)).
%F T(n,k) = A000203(A336811(n,k)).
%e Triangle begins:
%e 1;
%e 3;
%e 4, 1;
%e 7, 3, 1;
%e 6, 4, 3, 1, 1;
%e 12, 7, 4, 3, 3, 1, 1;
%e 8, 6, 7, 4, 4, 3, 3, 1, 1, 1, 1;
%e 15, 12, 6, 7, 7, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1;
%e 13, 8, 12, 6, 6, 7, 7, 4, 4, 4, 4, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
%e ...
%e From _Omar E. Pol_, Jan 13 2022: (Start)
%e Illustration of the first six rows of triangle showing the growth of the symmetric tower described in A221529:
%e Level k: 1 2 3 4 5 6 7
%e Stage
%e n _ _ _ _ _ _ _ _
%e | _ |
%e 1 | |_| |
%e |_ _ _ _ _ _ _ _|
%e | _ |
%e | | |_ |
%e 2 | |_ _| |
%e |_ _ _ _ _ _ _ _|_ _ _ _ _ _
%e | _ | _ |
%e | | | | |_| |
%e 3 | |_|_ _ | |
%e | |_ _| | |
%e |_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _
%e | _ | _ | _ |
%e | | | | | |_ | |_| |
%e 4 | | |_ | |_ _| | |
%e | |_ |_ _ | | |
%e | |_ _ _| | | |
%e |_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _ _ _ _ _
%e | _ | _ | _ | _ | _ |
%e | | | | | | | | |_ | |_| | |_| |
%e | | | | |_|_ _ | |_ _| | | |
%e 5 | |_|_ | |_ _| | | | |
%e | |_ _ _ | | | | |
%e | |_ _ _| | | | | |
%e |_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _|_ _ _ _|_ _ _ _ _ _
%e | _ | _ | _ | _ | _ | _ | _ |
%e | | | | | | | | | | | |_ | | |_ | |_| | |_| |
%e | | | | | |_ | |_|_ _ | |_ _| | |_ _| | | |
%e | | |_ _ | |_ |_ _ | |_ _| | | | | |
%e 6 | |_ | | |_ _ _| | | | | | |
%e | |_ |_ _ _ | | | | | | |
%e | |_ _ _ _| | | | | | | |
%e |_ _ _ _ _ _ _ _|_ _ _ _ _ _|_ _ _ _ _|_ _ _ _|_ _ _ _|_ _ _|_ _ _|
%e .
%e Every cell in the diagram of the symmetric representation of sigma represents a cubic cell or cube.
%e For n = 6 and k = 3 we add four cubes at 6th stage in the third level of the structure of the tower starting from the base so T(6,3) = 4.
%e For n = 9 another connection with the tower is as follows:
%e First we take the columns from the above triangle and build a new triangle in which all columns start at row 1 as shown below:
%e .
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3;
%e 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4;
%e 7, 7, 7, 7, 7, 7, 7;
%e 6, 6, 6, 6, 6;
%e 12, 12, 12;
%e 8, 8;
%e 15;
%e 13;
%e .
%e Then we rotate the triangle by 90 degrees as shown below:
%e _
%e 1; | |
%e 1; | |
%e 1; | |
%e 1; | |
%e 1; | |
%e 1; | |
%e 1; |_|_
%e 1, 3; | |
%e 1, 3; | |
%e 1, 3; | |
%e 1, 3; |_ _|_
%e 1, 3, 4; | | |
%e 1, 3, 4; | | |
%e 1, 3, 4; | | |
%e 1, 3, 4; |_ _|_|_
%e 1, 3, 4, 7; | | |
%e 1, 3, 4, 7; |_ _ _| |_
%e 1, 3, 4, 7, 6; | | |
%e 1, 3, 4, 7, 6; |_ _ _|_ _|_
%e 1, 3, 4, 7, 6, 12; |_ _ _ _| | |_
%e 1, 3, 4, 7, 6, 12, 8; |_ _ _ _|_|_ _|_ _
%e 1, 3, 4, 7, 6, 12, 8, 15; 13; |_ _ _ _ _|_ _|_ _|
%e .
%e Lateral view
%e of the tower
%e . _ _ _ _ _ _ _ _ _
%e |_| | | | | | | |
%e |_ _|_| | | | | |
%e |_ _| _|_| | | |
%e |_ _ _| _|_| |
%e |_ _ _| _| _ _|
%e |_ _ _ _| |
%e |_ _ _ _| _ _|
%e | |
%e |_ _ _ _ _|
%e .
%e Top view
%e of the tower
%e .
%e The sum of the m-th row of the new triangle equals A024916(j) where j is the length of the m-th row, equaling the number of cubic cells in the m-th level of the tower. For example: the last row of triangle has 9 terms and the sum of the last row is 1 + 3 + 4 + 7 + 6 + 12 + 8 + 15 + 13 = A024916(9) = 69, equaling the number of cubes in the base of the tower. (End)
%t A339278[rowmax_]:=Table[Flatten[Table[ConstantArray[DivisorSigma[1,n-m],PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]],{n,rowmax}];
%t A339278[15] (* Generates 15 rows *) (* _Paolo Xausa_, Feb 17 2023 *)
%o (PARI) f(n) = numbpart(n-1);
%o T(n, k) = {if (k > f(n), error("invalid k")); if (k==1, return (sigma(n))); my(s=0); while (k <= f(n-1), s++; n--;); sigma(1+s);}
%o tabf(nn) = {for (n=1, nn, for (k=1, f(n), print1(T(n,k), ", ");); print;);} \\ _Michel Marcus_, Jan 13 2021
%o (PARI) A339278(rowmax)=vector(rowmax,n,concat(vector(n,m,vector(numbpart(m-1)-numbpart(m-2),i,sigma(n-m+1)))));
%o A339278(15) \\ Generates 15 rows \\ _Paolo Xausa_, Feb 17 2023
%Y Sum of divisors of A336811.
%Y Row n has length A000041(n-1).
%Y Every column gives A000203.
%Y The length of the m-th block in row n is A187219(m), m >= 1.
%Y Row sums give A138879.
%Y Cf. A337209 (another version).
%Y Cf. A272172 (analog for the stepped pyramid described in A245092).
%Y Cf. A024916, A135010, A138121, A138137, A138785, A221529, A235791, A236104, A237270, A237591, A237593, A238442, A338156, A340423, A345023.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Nov 29 2020