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 A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2. 227
 1, 3, 5, 1, 7, 0, 9, 3, 11, 0, 1, 13, 5, 0, 15, 0, 0, 17, 7, 3, 19, 0, 0, 1, 21, 9, 0, 0, 23, 0, 5, 0, 25, 11, 0, 0, 27, 0, 0, 3, 29, 13, 7, 0, 1, 31, 0, 0, 0, 0, 33, 15, 0, 0, 0, 35, 0, 9, 5, 0, 37, 17, 0, 0, 0, 39, 0, 0, 0, 3, 41, 19, 11, 0, 0, 1, 43, 0, 0, 7, 0, 0, 45, 21, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link. Row n has length A003056(n) hence column k starts in row A000217(k). The number of positive terms in row n is A001227(n), the number of odd divisors of n. If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1. If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2. If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1. The partial sums of column k give the column k of A236104. The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270. Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015 Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018 From Omar E. Pol, Nov 24 2020: (Start) T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section). For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End) The number of zeros in the n-th row equals A238005(n). - Omar E. Pol, Sep 11 2021 Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021 Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022 LINKS G. C. Greubel, Table of n, a(n) for the first 200 rows, flattened Max Alekseyev, Proof of the alternating sum property of A196020, SeqFan Mailing List, Nov 17 2013. Paul D. Hanna, About an identity for sigma, SeqFan Mailing List, Nov 18 2013. FORMULA A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k). T(n,k) = 2*A211343(n,k) - 1, if A211343(n,k) >= 1 otherwise T(n,k) = 0. If n==k/2 (mod k) and n>=k(k+1)/2, then T(n,k) = 2*n/k - k; otherwise T(n,k) = 0. - Max Alekseyev, Nov 18 2013 T(n,k) = A236104(n,k) - A236104(n-1,k), assuming that A236104(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 14 2018 T(n,k) = A237048(n,k)*A338721(n,k). - Omar E. Pol, Feb 22 2022 EXAMPLE Triangle begins:    1;    3;    5,  1;    7,  0;    9,  3;   11,  0,  1;   13,  5,  0;   15,  0,  0;   17,  7,  3;   19,  0,  0,  1;   21,  9,  0,  0;   23,  0,  5,  0;   25, 11,  0,  0;   27,  0,  0,  3;   29, 13,  7,  0,  1;   31,  0,  0,  0,  0;   33, 15,  0,  0,  0;   35,  0,  9,  5,  0;   37, 17,  0,  0,  0;   39,  0,  0,  0,  3;   41, 19, 11,  0,  0,  1;   43,  0,  0,  7,  0,  0;   45, 21,  0,  0,  0,  0;   47,  0, 13,  0,  0,  0;   49, 23,  0,  0,  5,  0;   51,  0,  0,  9,  0,  0;   53, 25, 15,  0,  0,  3;   55,  0,  0,  0,  0,  0,  1;   ... For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 29, 13, 7, 0, 1, so the alternating row sum is 29 - 13 + 7 - 0 + 1 = 24, equaling the sum of divisors of 15. If n is even then the alternating sum of the n-th row is simpler to evaluate than the sum of divisors of n. For example the sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60, and the alternating sum of the 24th row of triangle is 47 - 0 + 13 - 0 + 0 - 0 = 60. From Omar E. Pol, Nov 24 2020: (Start) For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616 (see also the theorem there). For n = 15 the diagram with first 15 levels looks like this: . Level                         "Double-staircases" diagram .                                          _ 1                                        _|1|_ 2                                      _|1 _ 1|_ 3                                    _|1  |1|  1|_ 4                                  _|1   _| |_   1|_ 5                                _|1    |1 _ 1|    1|_ 6                              _|1     _| |1| |_     1|_ 7                            _|1      |1  | |  1|      1|_ 8                          _|1       _|  _| |_  |_       1|_ 9                        _|1        |1  |1 _ 1|  1|        1|_ 10                     _|1         _|   | |1| |   |_         1|_ 11                   _|1          |1   _| | | |_   1|          1|_ 12                 _|1           _|   |1  | |  1|   |_           1|_ 13               _|1            |1    |  _| |_  |    1|            1|_ 14             _|1             _|    _| |1 _ 1| |_    |_             1|_ 15            |1              |1    |1  | |1| |  1|    1|              1| . The first largest double-staircase has 29 horizontal steps, the second double-staircase has 13 steps, the third double-staircase has 7 steps, and the fifth double-staircases has only one step. Note that the fourth double-staircase does not count because it does not have horizontal steps in the 15th level, so the 15th row of triangle is [29, 13, 7, 0, 1]. For a connection with the "Ziggurat" diagram and the parts and subparts of the symmetric representation of sigma(15) see also A237270. (End) MAPLE T_row := proc(n) local T; T := (n, k) -> if modp(n-k/2, k) = 0 and n >= k*(k+1)/2 then 2*n/k-k else 0 fi; seq(T(n, k), k=1..floor((sqrt(8*n+1)-1)/2)) end: seq(print(T_row(n)), n=1..24); # Peter Luschny, Oct 27 2015 MATHEMATICA T[n_, k_] := If[Mod[n - k*(k+1)/2, k] == 0 , 2*n/k - k, 0] row[n_] := Floor[(Sqrt[8n+1]-1)/2] line[n_] := Map[T[n, #]&, Range[row[n]]] a196020[m_, n_] := Map[line, Range[m, n]] Flatten[a196020[1, 22]] (* data *) (* Hartmut F. W. Hoft, Oct 26 2015 *) A196020row = Function[n, Table[If[Divisible[Numerator[n-k/2], k] && CoprimeQ[ Denominator[n- k/2], k], 2*n/k-k, 0], {k, 1, Floor[(Sqrt[8 n+1]-1)/2]}]] Flatten[Table[A196020row[n], {n, 1, 24}]] (* Peter Luschny, Oct 28 2015 *) PROG (Sage) def T(n, k):     q = (2*n-k)/2     b = k.divides(q.numerator()) and gcd(k, q.denominator()) == 1     return 2*n/k - k if b else 0 for n in (1..24): [T(n, k) for k in (1..floor((sqrt(8*n+1)-1)/2))] # Peter Luschny, Oct 28 2015 CROSSREFS Columns 1-2: A005408, A193356. Compare with A027750, A238442, A237270, A237273, A252117, A280851, A296508. Cf. A000203, A000217, A001227, A001318, A003056, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112, A237048, A237271, A237591, A237593, A238005, A239660, A244050, A245092, A261699, A262626, A286000, A286001, A280850, A335616, A338721. Sequence in context: A302204 A065077 A118788 * A346775 A028510 A338721 Adjacent sequences:  A196017 A196018 A196019 * A196021 A196022 A196023 KEYWORD nonn,tabf AUTHOR Omar E. Pol, Feb 02 2013 STATUS approved

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Last modified October 2 20:53 EDT 2022. Contains 357230 sequences. (Running on oeis4.)