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A195310 Triangle read by rows with T(n,k) = n - A001318(k), n >= 1, k >= 1, if (n - A001318(k)) >= 0.

%I A195310 #64 Feb 16 2025 08:33:15
%S A195310 0,1,0,2,1,3,2,4,3,0,5,4,1,6,5,2,0,7,6,3,1,8,7,4,2,9,8,5,3,10,9,6,4,
%T A195310 11,10,7,5,0,12,11,8,6,1,13,12,9,7,2,14,13,10,8,3,0,15,14,11,9,4,1,16,
%U A195310 15,12,10,5,2,17,16,13,11,6,3,18,17,14,12,7,4
%N A195310 Triangle read by rows with T(n,k) = n - A001318(k), n >= 1, k >= 1, if (n - A001318(k)) >= 0.
%C A195310 Also triangle read by rows in which column k lists the nonnegative integers A001477 starting at the row A001318(k). This sequence is related to Euler's Pentagonal Number Theorem. A000041(a(n)) gives the absolute value of A175003(n). To get the number of partitions of n see the example.
%H A195310 L. Euler, <a href="http://www.math.dartmouth.edu/~euler/docs/originals/E542.pdf">De mirabilibus proprietatibus numerorum pentagonalium</a>
%H A195310 L. Euler, <a href="https://arxiv.org/abs/math/0505373">On the remarkable properties of the pentagonal numbers</a>, arXiv:math/0505373 [math.HO], 2005.
%H A195310 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PentagonalNumberTheorem.html">Pentagonal Number Theorem</a>
%H A195310 Wikipedia, <a href="http://en.wikipedia.org/wiki/Pentagonal_number_theorem">Pentagonal number theorem</a>
%F A195310 A175003(n,k) = A057077(k-1)*A000041(T(n,k)), n >= 1, k >= 1.
%e A195310 Written as a triangle:
%e A195310    0;
%e A195310    1,  0;
%e A195310    2,  1;
%e A195310    3,  2;
%e A195310    4,  3,  0;
%e A195310    5,  4,  1;
%e A195310    6,  5,  2,  0;
%e A195310    7,  6,  3,  1;
%e A195310    8,  7,  4,  2;
%e A195310    9,  8,  5,  3;
%e A195310   10,  9,  6,  4;
%e A195310   11, 10,  7,  5,  0;
%e A195310   12, 11,  8,  6,  1;
%e A195310   13, 12,  9,  7,  2;
%e A195310   14, 13, 10,  8,  3,  0;
%e A195310 .
%e A195310 For n = 15, consider row 15 which lists the numbers 14, 13, 10, 8, 3, 0. From Euler's Pentagonal Number Theorem we have that the number of partitions of 15 is p(15) = p(14) + p(13) - p(10) - p(8) + p(3) + p(0) = 135 + 101 - 42 - 22 + 3 + 1 = 176.
%t A195310 rows = 20;
%t A195310 a1318[n_] := If[EvenQ[n], n(3n/2+1)/4, (n+1)(3n+1)/8];
%t A195310 T[n_, k_] := n - a1318[k];
%t A195310 Table[DeleteCases[Table[T[n, k], {k, 1, n}], _?Negative], {n, 1, rows}] // Flatten (* _Jean-François Alcover_, Sep 22 2018 *)
%Y A195310 Row sums give A195311.
%Y A195310 Cf. A000041, A001318, A010815, A026741, A057077, A175003.
%K A195310 nonn,tabf
%O A195310 1,4
%A A195310 _Omar E. Pol_, Sep 21 2011
%E A195310 Name essentially suggested by _Franklin T. Adams-Watters_ (see history), Sep 21 2011