%I
%S 1,1,1,1,3,2,1,5,11,6,1,7,26,50,24,1,9,47,154,274,120,1,11,74,342,
%T 1044,1764,720,1,13,107,638,2754,8028,13068,5040,1,15,146,1066,5944,
%U 24552,69264,109584,40320,1,17,191,1650,11274,60216,241128,663696,1026576
%N Array read by antidiagonals: a(m,n) = m!*H(n,m), where H(n,m) is a higherorder harmonic number (H(0,m) = 1/m; H(n,m) = Sum_{k=1..m} H(n1,k)).
%C In the array, the m index runs from 1 on, n index runs from 0 on.
%C Antidiagonal sums are A093345 (n! * (1 + Sum_{i=1..n}((1/i)*Sum_{j=0..i1} 1/j!))).  _Gerald McGarvey_, Aug 27 2005
%C A recasting of A093905 and A067176.  _R. J. Mathar_, Mar 01 2009
%C The triangular array of this sequence is the reversal of A165675 which is related to the asymptotic expansion of the higher order exponential integral E(x,m=2,n); see also A165674.  _Johannes W. Meijer_, Oct 16 2009
%H G. C. Greubel, <a href="/A105954/b105954.txt">Table of n, a(n) for the first 27 rows, flattened</a>
%H Arthur T. Benjamin, David Gaebler and Robert Gaebler, <a href="http://www.emis.de/journals/INTEGERS/papers/d15/d15.Abstract.html">A Combinatorial Approach to Hyperharmonic Numbers</a>, (#A15)
%F a(m, n) = (H_{m+n1}  H_{n1})(m+n1)!/(n1)!, where H_k = H(1, k), a standard harmonic number. Array is read off by diagonals.
%F E.g.f. for column n: log(1x)/(1x)^n.  _Gerald McGarvey_, Aug 27 2005
%F 4th row is 4n^3 + 6n^2  2n  2. 5th row is 5n^4 + 20n^3 + 15n^2  10n  6. 6th row is 6n^5 + 45n^4 + 100n^3 + 45n^2  52n  24. 7th row is 7n^6 + 84n^5 + 350n^4 + 560n^3 + 147n^2  308n  120. 8th row is 8n^7 + 140n^6 + 924n^5 + 2800n^4 + 3556n^3 + 420n^2  2088n  720. The sum of the polynomial coefficients for the mth row is (m1)!. A005564 begins as 6, 20, 45, 84, 140, ...  _Gerald McGarvey_, Aug 27 2005
%F A(m, n) = Sum_{k=1..m} n*A094645(m, n)*(n+1)^(k1). (A094645 is Generalized Stirling number triangle of first kind, e.g.f.: (1y)^(1x).)  _Gerald McGarvey_, Aug 27 2005
%F If we replace n with (n+1) in Gerard McGarvey's formulas for the row coefficients we find Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; see A165674.  _Johannes W. Meijer_, Oct 16 2009
%e a(2,3) = (1 + (1 + 1/2) + (1 + 1/2 + 1/3))*6 = 26.
%e Array begins:
%e 1 1 1 1 1 1 1 1 1 ...
%e 1 3 5 7 9 11 13 15 17 ...
%e 2 11 26 47 74 107 146 191 242 ...
%e 6 50 154 342 638 1066 1650 2414 3382 ...
%e 24 274 1044 2754 5944 11274 19524 31594 48504 ...
%t H[0, m_] := 1/m; H[n_, m_] := Sum[H[n  1, k], {k, m}]; a[n_, m_] := m!H[n, m]; Flatten[ Table[ a[i, n  i], {n, 10}, {i, n  1, 0, 1}]]
%t Table[ a[n, m], {m, 8}, {n, 0, m + 1}] // TableForm (* to view the table *)
%t (* _Robert G. Wilson v_, Jun 27 2005 *)
%Y Cf. A000254.
%Y Column 0 = A000142 (factorial numbers).
%Y Column 1 = A000254 (Stirling numbers of first kind s(n, 2)) starting at n=1.
%Y Column 2 = A001705 (Generalized Stirling numbers: a(n) = n!*Sum_{k=0..n1}(k+1)/(nk)), starting at n=1.
%Y Column 3 = A001711 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(1)^(n+k)*(k+1)*3^k*stirling1(n+1, k+1)).
%Y Column 4 = A001716 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(1)^(n+k)*(k+1)*4^k*stirling1(n+1, k+1)).
%Y Column 5 = A001721 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(1)^(n+k)*binomial(k+1, 1)*5^k*stirling1(n+1, k+1)).
%Y Column 6 = A051524 (second unsigned column of triangle A051338) starting at n=1.
%Y Column 7 = A051545 (second unsigned column of triangle A051339) starting at n=1.
%Y Column 8 = A051560 (second unsigned column of triangle A051379) starting at n=1.
%Y Column 9 = A051562 (second unsigned column of triangle A051380) starting at n=1.
%Y Column 10= A051564 (second unsigned column of triangle A051523) starting at n=1.
%Y 2nd row is A005408 (2n  1, starting at n=1).
%Y 3rd row is A080663 (3n^2  1, starting at n=1).
%Y Cf. A165674 and A165675; A028421 and A126671.  _Johannes W. Meijer_, Oct 16 2009
%K nonn,tabl,easy
%O 0,5
%A _Leroy Quet_, Jun 26 2005
%E More terms from _Robert G. Wilson v_, Jun 27 2005
