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 A049777 Triangular array read by rows: T(m,n) = n + n+1 + ... + m = (m+n)(m-n+1)/2. 15
 1, 3, 2, 6, 5, 3, 10, 9, 7, 4, 15, 14, 12, 9, 5, 21, 20, 18, 15, 11, 6, 28, 27, 25, 22, 18, 13, 7, 36, 35, 33, 30, 26, 21, 15, 8, 45, 44, 42, 39, 35, 30, 24, 17, 9, 55, 54, 52, 49, 45, 40, 34, 27, 19, 10, 66, 65, 63, 60, 56, 51, 45, 38, 30, 21, 11, 78, 77, 75, 72, 68, 63, 57, 50 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Triangle read by rows, T(n,k) = A000217(n) - A000217(k), 0 <= k < n. - Philippe Deléham, Mar 07 2013 Subtriangle of triangle in A049780. - Philippe Deléham, Mar 07 2013 No primes and all composite numbers (except 2^x) are generated after the first two columns of the square array for this sequence. In other words, no primes and all composites except 2^x are generated when m-n >= 2. - Bob Selcoe, Jun 18 2013 Diagonal sums in the square array equal partial sums of squares (A000330). - Bob Selcoe, Feb 14 2014 From Bob Selcoe, Oct 27 2014: (Start) The following apply to the triangle as a square array read by rows unless otherwise specified (see Table link); Conjecture: There is at least one prime in interval [T(n,k), T(n,k+1)]. Since T(n,k+1)/T(n,k) decreases to (k+1)/k as n increases, this is true for k=1 ("Bertrand's Postulate", first proved by P. Chebyshev), k=2 (proved by El Bachraoui) and k=3 (proved by Loo). Starting with T(1,1), The falling diagonal of the first 2 numbers in each column (read by column) are the generalized pentagonal numbers (A001318). That is, the coefficients of T(1,1), T(2,1), T(2,2), T(3,2), T(3,3), T(4,3), T(4,4) etc. are the generalized pentagonal numbers. These are A000326 and A005449 (Pentagonal and Second pentagonal numbers: n*(3*n+1)/2, respectively), interweaved. Let D(n,k) denote falling diagonals starting with T(n,k): Treating n as constant: pentagonal numbers of the form n*k + 3*k*(k-1)/2 are D(n,1); sequences A000326, 005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542 are formed by n = 1 through 12, respectively. Treating k as constant: D(1,k) are (3*n^2 + (4k-5)*n + (k-1)*(k-2))/2. When k = 2(mod3), D(1,k), is same as D(k+1,1) omitting the first (k-2)/3 numbers in the sequences. So D(1,2) is same as D(3,1); D(1,5) is same as D(6,1) omitting the 6; D(1,8) is same as D(9,1) omitting the 9 and 21; etc. D(1,3) and D(1,4) are sequences A095794 and A140229, respectively. (End) LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..5000 M. El Bachraoui, Primes in the interval [2n,3n], Int. J. Contemp. Math. Sciences 1:13 (2006), pp. 617-621. A. Loo, On the primes in the interval [3n,4n], Int. J. Contemp. Math. Sciences 6 (2011), no. 38, 1871-1882. S. Ramanujan, A proof of Bertrand's postulate, J. Indian Math. Soc., 11 (1919), 181-182. Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv, arXiv:1212.2785 [math.NT], 2012. FORMULA Partial sums of A002260 row terms, starting from the right; e.g., row 3 of A002260 = (1, 2, 3), giving (6, 5, 3). - Gary W. Adamson, Oct 23 2007 Sum_{k=0..n-1} (-1)^k*(2*k+1)*A000203(T(n,k)) = (-1)^(n-1)*A000330(n). - Philippe Deléham, Mar 07 2013 Read as a square array: T(n,k) = k*(k+2n-1)/2. - Bob Selcoe, Oct 27 2014 EXAMPLE Rows: {1}; {3,2}; {6,5,3}; ... Triangle begins:    1;    3,  2;    6,  5,  3;   10,  9,  7,  4;   15, 14, 12,  9,  5;   21, 20, 18, 15, 11,  6;   28, 27, 25, 22, 18, 13,  7;   36, 35, 33, 30, 26, 21, 15,  8;   45, 44, 42, 39, 35, 30, 24, 17,  9;   55, 54, 52, 49, 45, 40, 34, 27, 19, 10; ... MATHEMATICA Flatten[Table[(n+k) (n-k+1)/2, {n, 15}, {k, n}]] (* Harvey P. Dale, Feb 27 2012 *) PROG (PARI) {T(n, k) = if( k<1 || n

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