cp's OEIS Frontend

A285898 Triangle read by row: T(n,k) = number of partitions of n into exactly k consecutive parts (1 <= k <= n).

%I A285898 #61 Jan 10 2025 04:29:43
%S A285898 1,1,0,1,1,0,1,0,0,0,1,1,0,0,0,1,0,1,0,0,0,1,1,0,0,0,0,0,1,0,0,0,0,0,
%T A285898 0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,
%U A285898 1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N A285898 Triangle read by row: T(n,k) = number of partitions of n into exactly k consecutive parts (1 <= k <= n).
%C A285898 To partition n into k parts, we see if m exists such that m + (m + 1) + ... + (m + k - 1) = k*m + binomial(k, 2) = n exists. a(n) = 1 if and only if (n - binomial(k, 2)) / k is an integer and larger than 0. - _David A. Corneth_, Apr 28 2017
%C A285898 It appears that this a full version of the irregular triangle A237048. - _Omar E. Pol_, Apr 28 2017
%C A285898 The value of a(n) can never exceed 1, since that would imply the existence of distinct equal-length ranges of consecutive integers that add up to the same number, which is impossible. - _Sidney Cadot_, Jan 22 2023
%F A285898 A000203(n) = Sum_{k=1..n} (-1)^(k-1) * ((Sum_{j=k..n} T(j,k))^2 - (Sum_{j=k..n} T(j-1,k))^2), assuming that T(k-1,k) = 0. - _Omar E. Pol_, Oct 10 2018
%e A285898 Triangle begins:
%e A285898 1;
%e A285898 1, 0;
%e A285898 1, 1, 0;
%e A285898 1, 0, 0, 0;
%e A285898 1, 1, 0, 0, 0;
%e A285898 1, 0, 1, 0, 0, 0;
%e A285898 1, 1, 0, 0, 0, 0, 0;
%e A285898 1, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 1, 1, 0, 0, 0, 0, 0, 0;
%e A285898 1, 0, 0, 1, 0, 0, 0, 0, 0, 0;
%e A285898 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e A285898 ...
%e A285898 For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. These partitions are formed by 1, 2, 3 and 5 consecutive parts respectively, so the 15th row of the triangle is [1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0].
%p A285898 A285898 := proc(n)
%p A285898     corn := (n-binomial(k,2))/k ;
%p A285898     if type(corn,'integer') then
%p A285898         if corn > 0 then
%p A285898             1 ;
%p A285898         else
%p A285898             0;
%p A285898         end if;
%p A285898     else
%p A285898         0 ;
%p A285898     end if;
%p A285898 end proc: # _R. J. Mathar_, Apr 30 2017
%t A285898 Table[Function[t, Function[s, ReplacePart[s, Map[# -> 1 &, t]]]@ ConstantArray[0, n]]@ Map[Length, Select[IntegerPartitions@ n, Length@ # == 1 || Union@ Differences@ # == {-1} &]], {n, 15}] // Flatten (* _Michael De Vlieger_, Apr 28 2017 *)
%o A285898 (PARI) T(n, k) = n-=binomial(k, 2); if(n>0,n%k==0) \\ _David A. Corneth_, Apr 28 2017
%o A285898 (Python)
%o A285898 from sympy import binomial
%o A285898 def T(n, k):
%o A285898     n=n - binomial(k, 2)
%o A285898     if n>0:
%o A285898         return 1 if n%k==0 else 0
%o A285898     return 0
%o A285898 for n in range(1, 21): print([T(n, k) for k in range(1, n + 1)]) # _Indranil Ghosh_, Apr 28 2017
%Y A285898 Row sums give A001227.
%Y A285898 Cf. A000203, A038547, A067742, A082647, A131576, A204217, A237048, A237593, A245579, A261699, A279387, A281009, A204217.
%K A285898 nonn,tabl,easy
%O A285898 1,1
%A A285898 _Omar E. Pol_ and _N. J. A. Sloane_, Apr 28 2017