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%I #96 Oct 02 2019 21:30:41
%S 1,2,3,3,5,6,4,7,9,10,5,9,12,14,15,6,11,15,18,20,21,7,13,18,22,25,27,
%T 28,8,15,21,26,30,33,35,36,9,17,24,30,35,39,42,44,45,10,19,27,34,40,
%U 45,49,52,54,55
%N Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.
%C As a rectangle, the accumulation array of A051340.
%C From _Clark Kimberling_, Feb 05 2011: (Start)
%C Here all the weights are divided by two where they aren't in Cahn.
%C As a rectangle, A141419 is in the accumulation chain
%C ... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
%C (See A144112 for the definition of accumulation array.)
%C row 1: A000027
%C col 1: A000217
%C diag (1,5,...): A000326 (pentagonal numbers)
%C diag (2,7,...): A005449 (second pentagonal numbers)
%C diag (3,9,...): A045943 (triangular matchstick numbers)
%C diag (4,11,...): A115067
%C diag (5,13,...): A140090
%C diag (6,15,...): A140091
%C diag (7,17,...): A059845
%C diag (8,19,...): A140672
%C (End)
%C Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n]_(1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery_, Nov 20 2011
%C Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x) (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N]_(i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N]_(k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - _L. Edson Jeffery_, Jan 20 2012
%C Row sums: n*(n+1)*(2*n+1)/6. - _L. Edson Jeffery_, Jan 25 2013
%C n-th row = partial sums of n-th row of A004736. - _Reinhard Zumkeller_, Aug 04 2014
%C T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - _Derek Orr_, Nov 26 2014
%C A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - _Hartmut F. W. Hoft_, Apr 14 2016
%C Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - _L. Edson Jeffery_, Jan 30 2018
%C From _Peter Munn_, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
%C A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
%C The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
%C Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
%C (End)
%D R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.
%H Reinhard Zumkeller, <a href="/A141419/b141419.txt">Rows n = 1..125 of triangle, flattened</a>
%H Johann Cigler, <a href="https://arxiv.org/abs/1611.05252">Some elementary observations on Narayana polynomials and related topics</a>, arXiv:1611.05252 [math.CO], 2016. See p. 24.
%H Carlton Gamer, David W. Roeder, and John J. Watkins, <a href="https://www.jstor.org/stable/2689901">Trapezoidal Numbers</a>, Mathematics Magazine 58:2 (1985), pp. 108-110.
%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>
%H M. A. Nyblom, <a href="https://www.fq.math.ca/Scanned/39-3/nyblom.pdf">On the representation of the integers as a difference of nonconsecutive triangular numbers</a>, Fibonacci Quarterly 39:3 (2001), pp. 256-263.
%F t(n,m) = m*(2*n - m + 1)/2.
%F t(n,m) = A000217(n) - A000217(n-m). - _L. Edson Jeffery_, Jan 16 2013
%F Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - _Hartmut F. W. Hoft_, Apr 14 2016
%F G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - _Stefano Spezia_, Oct 14 2018
%F T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10. T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - _Stefano Spezia_, Oct 14 2018
%F T(2*n-k, k) = A070543(n, k). - _Peter Munn_, Aug 21 2019
%e As a triangle:
%e 1,
%e 2, 3,
%e 3, 5, 6,
%e 4, 7, 9, 10,
%e 5, 9, 12, 14, 15,
%e 6, 11, 15, 18, 20, 21,
%e 7, 13, 18, 22, 25, 27, 28,
%e 8, 15, 21, 26, 30, 33, 35, 36,
%e 9, 17, 24, 30, 35, 39, 42, 44, 45,
%e 10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
%e As a rectangle:
%e 1 2 3 4 5 6 7 8 9 10
%e 3 5 7 9 11 13 15 17 19 21
%e 6 9 12 15 18 21 24 27 30 33
%e 10 14 18 22 26 30 34 38 42 46
%e 15 20 25 30 35 40 45 50 55 60
%e 21 27 33 39 45 51 57 63 69 75
%e 28 35 42 49 56 63 70 77 84 91
%e 36 44 52 60 68 76 84 92 100 108
%e 45 54 63 72 81 90 99 108 117 126
%e 55 65 75 85 95 105 115 125 135 145
%e Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016
%p a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # _Muniru A Asiru_, Oct 14 2018
%t T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]
%o (Haskell)
%o a141419 n k = k * (2 * n - k + 1) `div` 2
%o a141419_row n = a141419_tabl !! (n-1)
%o a141419_tabl = map (scanl1 (+)) a004736_tabl
%o -- _Reinhard Zumkeller_, Aug 04 2014
%Y Cf. A000330 (row sums), A004736, A057059, A070543.
%Y A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
%Y A141418 is a variant.
%Y Cf. A001227, A209260. - _Hartmut F. W. Hoft_, Apr 14 2016
%Y A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
%Y A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.
%K nonn,tabl,easy
%O 1,2
%A _Roger L. Bagula_, Aug 05 2008
%E Simpler name by _Stefano Spezia_, Oct 14 2018
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