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 A196020 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. 185
 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 step 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 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 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. 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. Cf. A000203, A000217, A001227, A003056, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112, A237048, A237270, A237591, A237593, A238442, A239660, A244050, A245092, A261699, A262626, A286000. Sequence in context: A302204 A065077 A118788 * A028510 A122053 A124084 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 19 16:17 EDT 2019. Contains 328223 sequences. (Running on oeis4.)