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A128139
Triangle read by rows: matrix product A004736 * A128132.
2
1, 1, 2, 1, 3, 3, 1, 4, 5, 4, 1, 5, 7, 7, 5, 1, 6, 9, 10, 9, 6, 1, 7, 11, 13, 13, 11, 7, 1, 8, 13, 16, 17, 16, 13, 8, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10
OFFSET
0,3
COMMENTS
A077028 with the final term in each row omitted.
Interchanging the factors in the matrix product leads to A128140 = A128132 * A004736.
From Gary W. Adamson, Jul 01 2012: (Start)
Alternatively, antidiagonals of an array A(n,k) of sequences with arithmetic progressions as follows:
1, 2, 3, 4, 5, 6, ...
1, 3, 5, 7, 9, 11, ...
1, 4, 7, 10, 13, 16, ...
1, 5, 9, 13, 17, 21, ...
... (End)
From Gary W. Adamson, Jul 02 2012: (Start)
A summation generalization for Sum_{k>=1} 1/(A(n,k)*A(n,k+1)) (formulas copied from A002378, A000466, A085001, A003185):
1 = 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + ...
1 = 2/(1)*(3) + 2/(3)*(5) + 2/(5)*(7) + ...
1 = 3/(1)*(4) + 3/(4)*(7) + 3/(7)*(10) + ...
1 = 4/(1)*(5) + 4/(5)*(9) + 4/(9)*(13) + ...
...
As a summation of terms equating to a definite integral:
Integral_{0..1} dx/(1+x) = ... 1 - 1/2 + 1/3 - 1/4 + ... = log(2).
Integral_{0..1} dx/(1+x^2) = 1 - 1/3 + 1/5 - 1/7 + ... = Pi/4 (see A157142)
Integral_{0..1} dx/(1+x^3) = 1 - 1/4 + 1/7 - 1/10 + ... (see A016777)
Integral_{0..1} dx/(1+x^4) = 1 - 1/5 + 1/9 - 1/13 + ... (see A016813). (End)
FORMULA
A004736 * A128132 as infinite lower triangular matrices.
T(n,k) = k*(1+n-k)+1 = 1 + A094053(n+1,1+n-k). - R. J. Mathar, Jul 09 2012
EXAMPLE
First few rows of the triangle:
1;
1, 2;
1, 3, 3;
1, 4, 5, 4;
1, 5, 7, 7, 5;
1, 6, 9, 10, 9, 6;
1, 7, 11, 13, 13, 11, 7;
1, 8, 13, 16, 17, 16, 13, 8;
1, 9, 15, 19, 21, 21, 19, 15, 9;
1, 10, 17, 22, 25, 26, 25, 22, 17, 10;
...
CROSSREFS
Cf. A004736, A128132, A128140, A004006 (row sums).
Sequence in context: A186974 A286312 A278492 * A208721 A208777 A104732
KEYWORD
nonn,easy,tabl
AUTHOR
Gary W. Adamson, Feb 16 2007
STATUS
approved