cp's OEIS Frontend

A168019 Square array A(n,k) read by antidiagonals, in which row n lists the number of partitions of n into parts divisible by k+1.

%I A168019 #9 Jan 28 2023 12:34:55
%S A168019 1,1,1,2,0,1,3,1,0,1,5,0,0,0,1,7,2,1,0,0,1,11,0,0,0,0,0,1,15,3,0,1,0,
%T A168019 0,0,1,22,0,2,0,0,0,0,0,1,30,5,0,0,1,0,0,0,0,1,42,0,0,0,0,0,0,0,0,0,1,
%U A168019 56,7,3,2,0,1,0,0,0,0,0,1,77,0,0,0,0,0,0,0,0,0,0,0,1
%N A168019 Square array A(n,k) read by antidiagonals, in which row n lists the number of partitions of n into parts divisible by k+1.
%C A168019 Note that column k lists each partition number A000041 followed by k zeros. See also A168020 and A168021.
%C A168019 Let A(n,k) denote the number of partitions of n into parts divisible by k+1. Let p(n) denote the number of partitions of n. If k+1 is a divisor of n then A(n,k) = p(n/(k+1)) otherwise A(n,k) = 0. [Conjectured by _Omar E. Pol_, Nov 25 2009] - this is trivial, just divide each part size by k - _Franklin T. Adams-Watters_, May 14 2010.
%H A168019 G. C. Greubel, <a href="/A168019/b168019.txt">Antidiagonals n = 0..50, flattened</a>
%F A168019 From _G. C. Greubel_, Jan 13 2023: (Start)
%F A168019 A(n, k) = A000041(n/(k+1)) if (k+1)|n, otherwise 0 (array).
%F A168019 T(n, k) = A000041((n-k)/(k+1)) if (k+1)|(n-k), otherwise 0 (antidiagonals).
%F A168019 A(n, 0) = T(n, 0) = A000041(n).
%F A168019 T(2*n, n) = A(n, n) = A000007(n).
%F A168019 Sum_{k=0..n} T(n, k) = A083710(n+1). (End)
%e A168019 The array, A(n, k), begins:
%e A168019 ==================================================
%e A168019 ... Column k: 0.. 1. 2. 3. 4. 5. 6. 7. 8. 9 10 11
%e A168019 . Row ...........................................
%e A168019 ...n ............................................
%e A168019 ==================================================
%e A168019 .. 0 ........ 1,  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
%e A168019 .. 1 ........ 1,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
%e A168019 .. 2 ........ 2,  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
%e A168019 .. 3 ........ 3,  0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
%e A168019 .. 4 ........ 5,  2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
%e A168019 .. 5 ........ 7,  0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
%e A168019 .. 6 ....... 11,  3, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0,
%e A168019 .. 7 ....... 15,  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
%e A168019 .. 8 ....... 22,  5, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0,
%e A168019 .. 9 ....... 30,  0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 0,
%e A168019 . 10 ....... 42,  7, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0,
%e A168019 . 11 ....... 56,  0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0,
%e A168019 . 12 ....... 77, 11, 5, 3, 0, 2, 0, 0, 0, 0, 0, 1,
%e A168019 ...
%e A168019 Antidiagonal triangle, T(n, k), begins as:
%e A168019    1;
%e A168019    1, 1;
%e A168019    2, 0, 1;
%e A168019    3, 1, 0, 1;
%e A168019    5, 0, 0, 0, 1;
%e A168019    7, 2, 1, 0, 0, 1;
%e A168019   11, 0, 0, 0, 0, 0, 1;
%e A168019   15, 3, 0, 1, 0, 0, 0, 1;
%e A168019   22, 0, 2, 0, 0, 0, 0, 0, 1;
%e A168019   30, 5, 0, 0, 1, 0, 0, 0, 0, 1;
%e A168019   42, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
%t A168019 T[n_, k_]:= If[IntegerQ[(n-k)/(k+1)], PartitionsP[(n-k)/(k+1)], 0];
%t A168019 Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 13 2023 *)
%o A168019 (SageMath)
%o A168019 def A168019(n,k): return number_of_partitions((n-k)/(k+1)) if ((n-k)%(k+1))==0 else 0
%o A168019 flatten([[A168019(n,k) for k in range(n+1)] for n in range(16)]) # _G. C. Greubel_, Jan 13 2023
%Y A168019 Cf. A000041, A035363, A083710, A135010.
%Y A168019 Cf. A168016, A168120, A168021, A168121.
%K A168019 easy,nonn,tabl
%O A168019 0,4
%A A168019 _Omar E. Pol_, Nov 21 2009
%E A168019 Edited by _Charles R Greathouse IV_, Mar 23 2010
%E A168019 Edited by _Franklin T. Adams-Watters_, May 14 2010