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 A116861 Triangle read by rows: T(n,k) is the number of partitions of n such that the sum of the parts, counted without multiplicities, is equal to k (n>=1, k>=1). 4
 1, 1, 1, 1, 0, 2, 1, 1, 1, 2, 1, 0, 2, 1, 3, 1, 1, 3, 1, 1, 4, 1, 0, 3, 2, 2, 2, 5, 1, 1, 3, 3, 2, 4, 2, 6, 1, 0, 5, 2, 3, 4, 4, 3, 8, 1, 1, 4, 3, 4, 7, 4, 5, 3, 10, 1, 0, 5, 3, 4, 7, 7, 6, 6, 5, 12, 1, 1, 6, 4, 3, 12, 6, 8, 7, 9, 5, 15, 1, 0, 6, 4, 5, 10, 10, 9, 10, 11, 10, 7, 18, 1, 1, 6, 4, 5, 15, 11, 13, 9, 16, 11, 13, 8, 22 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,6 COMMENTS Conjecture: Reverse the rows of the table to get an infinite lower-triangular matrix b with 1's on the main diagonal. The third diagonal of the inverse of b is minus A137719. - George Beck, Oct 26 2019 Proof: The reversed rows yield the matrix I+N where N is strictly lower triangular, N[i,j] = 0 for j >= i, having its 2nd diagonal equal to the 2nd column (1, 0, 1, 0, 1, ...): N[i+1,i] = A000035(i), i >= 1, and 3rd diagonal equal to the 3rd column of this triangle, (2, 1, 2, 3, 3, 3, ...): N[i+2,i] = A137719(i), i >= 1. It is known that (I+N)^-1 = 1 - N + N^2 - N^3 +- .... Here N^2 has not only the second but also the 3rd diagonal zero, because N²[i+2,i] = N[i+2,i+1]*N[i+1,i] = A000035(i+1)*A000035(i) = 0. Therefore the 3rd diagonal of (I+N)^-1 is equal to -A137719 without leading 0. - M. F. Hasler, Oct 27 2019 LINKS Alois P. Heinz, Rows n = 1..141, flattened P. J. Rossky, M. Karplus, The enumeration of Goldstone diagrams in many-body perturbation theory, J. Chem. Phys. 64 (1976) 1569, equation (16)(1). FORMULA G.f.: -1 + Product_{j>=1} (1 + t^j*x^j/(1-x^j)). Sum_{k=1..n} T(n,k) = A000041(n). T(n,n) = A000009(n). Sum_{k=1..n} k*T(n,k) = A014153(n-1). T(n,1) = 1. T(n,2) = A000035(n+1). T(n,3) = A137719(n-2). - R. J. Mathar, Oct 27 2019 T(n,4) = A002264(n-1)+A121262(n). - R. J. Mathar, Oct 28 2019 EXAMPLE T(10,7) = 4 because we have [6,1,1,1,1], [4,3,3], [4,2,2,1,1] and [4,2,1,1,1,1] (6+1=4+3=4+2+1=7). Triangle starts:   1;   1, 1;   1, 0, 2;   1, 1, 1, 2;   1, 0, 2, 1, 3;   1, 1, 3, 1, 1, 4; MAPLE g:= -1+product(1+t^j*x^j/(1-x^j), j=1..40): gser:= simplify(series(g, x=0, 18)): for n from 1 to 14 do P[n]:=sort(coeff(gser, x^n)) od: for n from 1 to 14 do seq(coeff(P[n], t^j), j=1..n) od; # yields sequence in triangular form # second Maple program: b:= proc(n, i) option remember; local f, g, j;       if n=0 then [1] elif i<1 then [ ] else f:= b(n, i-1);          for j to n/i do            f:= zip((x, y)->x+y, f, [0\$i, b(n-i*j, i-1)[]], 0)          od; f       fi     end: T:= n-> subsop(1=NULL, b(n, n))[]: seq(T(n), n=1..20);  # Alois P. Heinz, Feb 27 2013 MATHEMATICA max = 14; s = Series[-1+Product[1+t^j*x^j/(1-x^j), {j, 1, max}], {x, 0, max}, {t, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 17 2014 *) PROG (PARI) A116861(n, k, s=0)={forpart(X=n, vecsum(Set(X))==k&&s++, k); s} \\ M. F. Hasler, Oct 27 2019 CROSSREFS Cf. A000041 (row sums), A000009 (diagonal), A014153. Cf. A114638 (count partitions with #parts = sum(distinct parts)). Column 1: A000012, column 2: A000035(1..), column 3: A137719(1..). Sequence in context: A112400 A316523 A219185 * A327785 A105242 A336709 Adjacent sequences:  A116858 A116859 A116860 * A116862 A116863 A116864 KEYWORD nonn,tabl AUTHOR Emeric Deutsch, Feb 27 2006 STATUS approved

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Last modified September 25 16:13 EDT 2020. Contains 337344 sequences. (Running on oeis4.)