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 A091602 Triangle: T(n,k) = number of partitions of n such that some part is repeated k times and no part is repeated more than k times. 14
 1, 1, 1, 2, 0, 1, 2, 2, 0, 1, 3, 2, 1, 0, 1, 4, 3, 2, 1, 0, 1, 5, 4, 3, 1, 1, 0, 1, 6, 7, 3, 3, 1, 1, 0, 1, 8, 8, 6, 3, 2, 1, 1, 0, 1, 10, 12, 7, 5, 3, 2, 1, 1, 0, 1, 12, 15, 11, 6, 5, 2, 2, 1, 1, 0, 1, 15, 21, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1, 18, 26, 20, 12, 9, 5, 4, 2, 2, 1, 1, 0, 1, 22, 35, 25, 18, 11, 8, 5, 4, 2, 2, 1, 1, 0, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS From Gary W. Adamson, Mar 13 2010: (Start) The triangle by rows = finite differences starting from the top, of an array in which row 1 = p(x)/p(x^2), row 2 = p(x)/p(x^3), ... row k = p(x)/p(x^k); such that p(x) = polcoeff A000041: (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...) Note that p(x)/p(x^2) = polcoeff A000009: (1 + x + x^2 + 2x^3 + 2x^4 + ...). Refer to the example. (End) LINKS Alois P. Heinz, Rows n = 1..141, flattened FORMULA G.f.: G = G(t,x) = sum(k>=1, t^k*(prod(j>=1, (1-x^((k+1)*j))/(1-x^j) ) -prod(j>=1, (1-x^(k*j))/(1-x^j) ) ) ). - Emeric Deutsch, Mar 30 2006 Sum_{k=1..n} k * T(n,k) = A264397(n). - Alois P. Heinz, Nov 20 2015 EXAMPLE Triangle starts: 01:  1, 02:  1, 1, 03:  2, 0, 1, 04:  2, 2, 0, 1, 05:  3, 2, 1, 0, 1, 06:  4, 3, 2, 1, 0, 1, 07:  5, 4, 3, 1, 1, 0, 1, 08:  6, 7, 3, 3, 1, 1, 0, 1, 09:  8, 8, 6, 3, 2, 1, 1, 0, 1, 10: 10, 12, 7, 5, 3, 2, 1, 1, 0, 1, 11: 12, 15, 11, 6, 5, 2, 2, 1, 1, 0, 1, 12: 15, 21, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1, 13: 18, 26, 20, 12, 9, 5, 4, 2, 2, 1, 1, 0, 1, 14: 22, 35, 25, 18, 11, 8, 5, 4, 2, 2, 1, 1, 0, 1, ... In the partition 5+2+2+2+1+1, 2 is repeated 3 times, no part is repeated more than 3 times. From Gary W. Adamson, Mar 13 2010: (Start) First few rows of the array = ... 1,..1,..1,..2,..2,..3,...4,...5,...6,...8,...10,...; = p(x)/p(x^2) = A000009 1,..1,..2,..2,..4,..5,...7,...9,..13,..16,...22,...; = p(x)/p(x^3) 1,..1,..2,..3,..4,..6,...9,..12,..16,..22,...29,...; = p(x)/p(x^4) 1,..1,..2,..3,..5,..6,..10,..13,..19,..25,...34,...; = p(x)/p(x^5) 1,..1,..2,..3,..5,..7,..10,..14,..20,..27,...37,...; = p(x)/p(x^6) ... Finally, taking finite differences from the top and deleting the first "1", we obtain triangle A091602 with row sums = A000041 starting with offset 1: 1; 1, 1; 2, 0, 1; 2, 2, 0, 1; 3, 2, 1, 0, 1; 4, 3, 2, 1, 0, 1; ... (End) MAPLE g:=sum(t^k*(product((1-x^((k+1)*j))/(1-x^j), j=1..50)-product((1-x^(k*j))/(1-x^j), j=1..50)), k=1..50): gser:=simplify(series(g, x=0, 20)): for n from 1 to 13 do P[n]:=coeff(gser, x^n) od: for n from 1 to 13 do seq(coeff(P[n], t^j), j=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Mar 30 2006 b:= proc(n, i, k) option remember; `if`(n=0, 1,       `if`(i>n, 0, add(b(n-i*j, i+1, min(k,        iquo(n-i*j, i+1))), j=0..min(n/i, k))))     end: T:= (n, k)-> b(n, 1, k) -`if`(k=0, 0, b(n, 1, k-1)): seq(seq(T(n, k), k=1..n), n=1..20); # Alois P. Heinz, Nov 27 2013 MATHEMATICA b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i>n, 0, Sum[b[n-i*j, i+1, Min[k, Quotient[n-i*j, i+1]]], {j, 0, Min[n/i, k]}]]]; t[n_, k_] := b[n, 1, k] - If[k == 0, 0, b[n, 1, k-1]]; Table[t[n, k], {n, 1, 20}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 17 2014, after Alois P. Heinz's second Maple program *) CROSSREFS Row sums: A000041. Inverse: A091603. Square: A091604. Columns 1-6: A000009, A091605-A091609. Convergent of columns: A002865. Cf. A000009. - Gary W. Adamson, Mar 13 2010 T(2n,n) gives: A232697. Cf. A213177, A264397. Sequence in context: A128187 A266477 A133121 * A035465 A096144 A118401 Adjacent sequences:  A091599 A091600 A091601 * A091603 A091604 A091605 KEYWORD nonn,tabl AUTHOR Christian G. Bower, Jan 23 2004 STATUS approved

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Last modified October 15 15:14 EDT 2019. Contains 328030 sequences. (Running on oeis4.)