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 A015739 Number of 4's in all the partitions of n into distinct parts. 3
 0, 0, 0, 1, 1, 1, 2, 1, 2, 3, 3, 5, 6, 7, 9, 10, 12, 15, 18, 22, 26, 31, 36, 42, 50, 58, 68, 80, 92, 107, 124, 142, 164, 189, 216, 248, 284, 323, 369, 420, 476, 541, 613, 693, 784, 885, 997, 1123, 1264, 1419, 1593, 1787, 2000, 2239, 2504, 2795, 3120, 3479 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,7 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..1000 FORMULA G.f.: x^4*prod(j>=1, 1+x^j)/(1+x^4). - Emeric Deutsch, Apr 17 2006 Corresponding g.f. for "number of k's" is x^k/(1+x^k)*prod(n>=1, 1+x^n ). [Joerg Arndt, Feb 20 2014] EXAMPLE a(9) = 2 because in the 8 (=A000009(9)) partitions of 9 into distinct parts, namely [9], [8,1], [7,2], [6,3], [6,2,1], [5,4], [5,3,1] and [4,3,2] we have altogether two parts equal to 4. MAPLE g:=x^4*product(1+x^j, j=1..60)/(1+x^4): gser:=series(g, x=0, 57): seq(coeff(gser, x, n), n=1..54); # Emeric Deutsch, Apr 17 2006 b:= proc(n, i) option remember; local g; if n=0 then [1, 0] elif i<1 then [0, 0] else g:= `if`(i>n, [0\$2], b(n-i, i-1)); b(n, i-1) +g +[0, `if`(i=4, g[1], 0)] fi end: a:= n-> b(n, n)[2]: seq (a(n), n=1..100); # Alois P. Heinz, Oct 27 2012 MATHEMATICA \$RecursionLimit = 1000; b[n_, i_] := b[n, i] = Module[{g}, If[n==0, {1, 0}, If[i<1 , {0, 0}, g = If[i>n, {0, 0}, b[n-i, i-1]]; b[n, i-1] + g + {0, If[i == 4, g[[1]], 0]}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Apr 01 2015, after Alois P. Heinz *) Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 4], {n, 58}] (* Robert Price, May 16 2020 *) CROSSREFS Sequence in context: A180235 A239493 A331849 * A015746 A109266 A022876 Adjacent sequences: A015736 A015737 A015738 * A015740 A015741 A015742 KEYWORD nonn AUTHOR Clark Kimberling STATUS approved

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