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 A101447 Triangle read by rows: T(n,k) = (2*k+1)*(n+1-k), 0<=k
 1, 2, 3, 3, 6, 5, 4, 9, 10, 7, 5, 12, 15, 14, 9, 6, 15, 20, 21, 18, 11, 7, 18, 25, 28, 27, 22, 13, 8, 21, 30, 35, 36, 33, 26, 15, 9, 24, 35, 42, 45, 44, 39, 30, 17, 10, 27, 40, 49, 54, 55, 52, 45, 34, 19, 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21, 12, 33, 50, 63, 72, 77, 78, 75, 68 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The triangle is generated from the product of matrix A and matrix B, I.e. A * B where A = the infinite lower triangular matrix: 1 0 0 0 0 ... 1 1 0 0 0 ... 1 1 1 0 0 ... 1 1 1 1 0 ... 1 1 1 1 1 ... ... and B = the infinite lower triangular matrix: 1 0 0 0 0... 1 3 0 0 0... 1 3 5 0 0... 1 3 5 7 0... 1 3 5 7 9... ... Row sums give the square pyramidal numbers A000330. T(n+0,0)=1*n=A000027(n+1); T(n+1,1)=3*n=A008585(n); T(n+2,2)=5*n=A008587(n); T(n+3,3)=7*n=A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=A028896(n) (6 times triangular numbers.) T(n,1)*T(n,2)/10=3*n*(n+1)/2=A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=A024966(n) for n>1 (7 times triangular numbers.) etc. Contribution from Gary W. Adamson, Apr 25 2010: (Start) Consider the following array, signed as shown: ... 1,...3,...5,...7,...9,...11,... 2,..-6,..10,.-14,..18,..-22,... 3,...9,..15,..21,..27,...33,... 4,.-12,..20,.-28,..36,..-44,... 5,..15,..25,..35,..45,...55,... 6,.-18,..30,.-42,..54,..-66,... 7,..21,..35,..49,..63,...77,... ... Let each term (+, -)k = (+, -) phi^(-k). Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7,...). By way of example, let q = phi = 1.6180339,...; then ... 1/1. = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^-9) + ... 1/3. = q^(^2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ... 1/4. = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +... 1/7. = q^(-4) - q^-12) + q^(-20) - q^(-28) + q^(-36) + ... 1/11 = q^(-5) + q^-15) + q^(-25) + q^(-35) + q^(-45) + ... ... Relating to the Pell series, the corresponding "Lucas"-like series is: (2, 6, 14, 34, 82, 198,...) such that herein, q = 2.414213,...= (1 + sqrt(2)) Then analogous to the previous set, ... 1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ... 1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ... ... (End) LINKS EXAMPLE Triangle begins: 1, 2,  3, 3,  6,  5, 4,  9,  10, 7, 5,  12, 15, 14, 9, 6,  15, 20, 21, 18, 11, 7,  18, 25, 28, 27, 22, 13, 8,  21, 30, 35, 36, 33, 26, 15, 9,  24, 35, 42, 45, 44, 39, 30, 17, 10, 27, 40, 49, 54, 55, 52, 45, 34, 19, 11, 30, 45, 56, 63, 66, 65, 60, 51, 38, 21, etc. [Bruno Berselli, Feb 10 2014] MATHEMATICA t[n_, k_] := If[n < k, 0, (2*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 20 2005 *) PROG (PARI) T(n, k)=if(n

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Last modified January 28 10:02 EST 2020. Contains 331319 sequences. (Running on oeis4.)