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A285898 Triangle read by row: T(n,k) = number of partitions of n into exactly k consecutive parts (1 <= k <= n). 11

%I #57 Jan 22 2023 08:44:21

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

%C 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 integer and larger than 0. - _David A. Corneth_, Apr 28 2017

%C It appears that this a full version of the irregular triangle A237048. - _Omar E. Pol_, Apr 28 2017

%C 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 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 Triangle begins:

%e 1;

%e 1, 0;

%e 1, 1, 0;

%e 1, 0, 0, 0;

%e 1, 1, 0, 0, 0;

%e 1, 0, 1, 0, 0, 0;

%e 1, 1, 0, 0, 0, 0, 0;

%e 1, 0, 0, 0, 0, 0, 0, 0;

%e 1, 1, 1, 0, 0, 0, 0, 0, 0;

%e 1, 0, 0, 1, 0, 0, 0, 0, 0, 0;

%e 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

%e ...

%e 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 := proc(n)

%p corn := (n-binomial(k,2))/k ;

%p if type(corn,'integer') then

%p if corn > 0 then

%p 1 ;

%p else

%p 0;

%p end if;

%p else

%p 0 ;

%p end if;

%p end proc: # _R. J. Mathar_, Apr 30 2017

%t 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 (PARI) T(n, k) = n-=binomial(k, 2); if(n>0,n%k==0) \\ _David A. Corneth_, Apr 28 2017

%o (Python)

%o from sympy import binomial

%o def T(n, k):

%o n=n - binomial(k, 2)

%o if n>0:

%o return 1 if n%k==0 else 0

%o return 0

%o for n in range(1, 21): print([T(n, k) for k in range(1, n + 1)]) # _Indranil Ghosh_, Apr 28 2017

%Y Row sums give A001227.

%Y Cf. A000203, A038547, A067742, A082647, A131576, A204217, A237048, A237593, A245579, A261699, A279387, A281009, A204217.

%K nonn,tabl,easy

%O 1,1

%A _Omar E. Pol_ and _N. J. A. Sloane_, Apr 28 2017

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)