%I A154921
%S A154921 1,1,1,3,2,1,13,9,3,1,75,52,18,4,1,541,375,130,30,5,1,4683,3246,1125,
%T A154921 260,45,6,1,47293,32781,11361,2625,455,63,7,1,545835,378344,131124,
%U A154921 30296,5250,728,84,8,1
%N A154921 Triangle read by rows. Matrix inverse of A154926.
%C A154921 A000670 appears in the first column. A052882 appears in the second column. 
               A000027 and A045943 appear as diagonals. An alternative to calculating 
               the matrix inverse of A154926 is to move the term in the lower right 
               corner to a position in the same column and calculate the determinant 
               instead, which yields the same answer.
%C A154921 Contribution from Peter Bala (pbala(AT)talktalk.net), Jul 01 2009: (Start)
%C A154921 Matrix inverse of (2*I - P), where P is Pascal's triangle and I the
%C A154921 identity matrix. See A162312 for the matrix inverse of (2*P - I) and
%C A154921 some general remarks about arrays of the form M(a) := (I - a*P)^-1
%C A154921 and their connection with weighted sums of powers of integers. The
%C A154921 present array equals 1/2*M(1/2).
%C A154921 (End)
%C A154921 The values in this triangle can be seen as permanents of the Pascal triangle 
               analogous to the method in the Redheffer matrix. The elements satisfy 
               T(n,k)/T(n,k-1)*k=T(n-1,k)/T(n,k)*n which converges to log(2) as 
               n-->infinity and k-->0. More generally to calculate log(x) multiply 
               the negative values in A154926 with 1/(x-1) and calculate the matrix 
               inverse. Then T(n,k)/T(n,k-1)*k and T(n-1,k)/T(n,k)*n in the resulting 
               triangle converge to log(x). [From Mats Granvik (mats.granvik(AT)abo.fi), 
               Aug 11 2009]
%C A154921 This method for calculating log(x) converges faster than the Taylor series 
               when x is greater than 5 or so. See chapter on Taylor series in Spiegel 
               for comparison. [From Mats Granvik (mats.granvik(AT)abo.fi), Aug 
               11 2009]
%D A154921 R. B. Nelsen, Problem E3062: Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 
               1987), 376-377. [From Peter Bala (pbala(AT)talktalk.net), Jul 01 
               2009]
%D A154921 R. B. Nelsen and H. Schmidt, Jr., Chains in Power Sets, Mathematics Magazine, 
               Vol. 64, No. 1 (Feb., 1991), 23-31. [From Peter Bala (pbala(AT)talktalk.net), 
               Jul 01 2009]
%D A154921 Murray R. Spiegel, Mathematical handbook, Schaum's Outlines, p. 111. 
               [From Mats Granvik (mats.granvik(AT)abo.fi), Aug 11 2009]
%F A154921 Contribution from Peter Bala (pbala(AT)talktalk.net), Jul 01 2009: (Start)
%F A154921 TABLE ENTRIES
%F A154921 (1)... T(n,k) = binomial(n,k)*A000670(n-k).
%F A154921 GENERATING FUNCTION
%F A154921 (2)... exp(x*t)/(2-exp(t)) = 1 + (1+x)*t + (3+2*x+x^2)*t^2/2! + ....
%F A154921 PROPERTIES OF THE ROW POLYNOMIALS
%F A154921 The row generating polynomials R_n(x) form an Appell sequence. They
%F A154921 appear in the study of the poset of power sets [Nelsen and Schmidt].
%F A154921 The first few values are R_0(x) = 1, R_1(x) = 1+x, R_2(x) = 3+2*x+x^2
%F A154921 and R_3(x) = 13+9*x+3*x^2+x^3.
%F A154921 The row polynomials may be recursively computed by means of
%F A154921 (3)... R_n(x) = x^n + sum {k = 0..n-1} binomial(n,k)*R_k(x).
%F A154921 Explicit formulas include
%F A154921 (4)... R_n(x) = 1/2*sum {k = 0..inf}(1/2)^k*(x+k)^n,
%F A154921 (5)... R_n(x) = sum {j = 0..n} sum {k = 0..j} (-1)^(j-k)*binomial(j,k)
%F A154921 *(x+k)^n,
%F A154921 and
%F A154921 (6)... R_n(x) = sum {j = 0..n} sum{k = j..n} k!*Stirling2(n,k)
%F A154921 *binomial(x,k-j).
%F A154921 SUMS OF POWERS OF INTEGERS
%F A154921 The row polynomials satisfy the difference equation
%F A154921 (7)... 2*R_m(x) - R_m(x+1) = x^m,
%F A154921 which easily leads to the evaluation of the weighted sums of powers
%F A154921 of integers
%F A154921 (8)... sum {k = 1..n-1} (1/2)^k*k^m = 2*R_m(0) - (1/2)^(n-1)*R_m(n).
%F A154921 For example, m = 2 gives
%F A154921 (9)... sum {k = 1..n-1} (1/2)^k*k^2 = 6 - (1/2)^(n-1)*(n^2+2*n+3).
%F A154921 More generally we have
%F A154921 (10)... sum {k=0..n-1} (1/2)^k*(x+k)^m = 2*R_m(x)-(1/2)^(n-1)*R_m(x+n).
%F A154921 RELATIONS WITH OTHER SEQUENCES
%F A154921 Sequences in the database given by particular values of the row
%F A154921 polynomials are
%F A154921 (11)... A000670(n) = R_n(0)
%F A154921 (12)... A052841(n) = R_n(-1)
%F A154921 (13)... A000629(n) = R_n(1)
%F A154921 (14)... A007047(n) = R_n(2)
%F A154921 (15)... A080253(n) = 2^n*R_n(1/2).
%F A154921 This last result is the particular case (x = 0) of the result that
%F A154921 the polynomials 2^n*R_n(1/2+x/2) are the row generating polynomials
%F A154921 for A162313.
%F A154921 The above formulas should be compared with those for A162312.
%F A154921 (End)
%e A154921 Contribution from Peter Bala (pbala(AT)talktalk.net), Jul 01 2009: (Start)
%e A154921 Triangle begins
%e A154921 ==============================================
%e A154921 n\k|.....0.....1.....2.....3.....4.....5.....6
%e A154921 ==============================================
%e A154921 0..|.....1
%e A154921 1..|.....1.....1
%e A154921 2..|.....3.....2.....1
%e A154921 3..|....13.....9.....3.....1
%e A154921 4..|....75....52....18.....4.....1
%e A154921 5..|...541...375...130....30.....5.....1
%e A154921 6..|..4683..3246..1125...260....45.....6.....1
%e A154921 ...
%e A154921 (End)
%e A154921 Contribution from Mats Granvik (mats.granvik(AT)abo.fi), Aug 11 2009: 
               (Start)
%e A154921 Row 4 equals 75,52,18,4,1 because permanents of:
%e A154921 1,0,0,0,1..1,0,0,0,0..1,0,0,0,0..1,0,0,0,0..1,0,0,0,0
%e A154921 1,1,0,0,0..1,1,0,0,1..1,1,0,0,0..1,1,0,0,0..1,1,0,0,0
%e A154921 1,2,1,0,0..1,2,1,0,0..1,2,1,0,1..1,2,1,0,0..1,2,1,0,0
%e A154921 1,3,3,1,0..1,3,3,1,0..1,3,3,1,0..1,3,3,1,1..1,3,3,1,0
%e A154921 1,4,6,4,0..1,4,6,4,0..1,4,6,4,0..1,4,6,4,0..1,4,6,4,1
%e A154921 are:
%e A154921 75.........52.........18.........4..........1
%e A154921 (End)
%Y A154921 A000629 (row sums), A000670, A007047, A052822 (column 1), A052841 (alt. 
               row sums), A080253, A162312, A162313. [From Peter Bala (pbala(AT)talktalk.net), 
               Jul 01 2009]
%Y A154921 Sequence in context: A134090 A132845 A129652 this_sequence A127126 A161133 
               A112911
%Y A154921 Adjacent sequences: A154918 A154919 A154920 this_sequence A154922 A154923 
               A154924
%K A154921 nonn,tabl
%O A154921 1,4
%A A154921 Mats Granvik (mats.granvik(AT)abo.fi), Jan 17 2009