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Number of partitions of n with at most one part in the interval [i*(i+1)/2, i+(i*(i+1)/2)] for all nonnegative integers i.
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%I #55 Apr 19 2019 10:23:07

%S 1,1,1,1,2,3,3,3,3,4,6,9,11,12,12,12,13,15,18,22,27,34,42,50,56,60,63,

%T 66,70,76,84,94,106,120,136,154,177,206,241,279,317,352,381,404,423,

%U 442,464,492,528,574,630,694,764,839,920,1008,1104,1213,1341,1494,1674,1878

%N Number of partitions of n with at most one part in the interval [i*(i+1)/2, i+(i*(i+1)/2)] for all nonnegative integers i.

%C The intervals are: [1,2], [3,5], [6,9], [10,14], [15,20], [21,27], [28,35], [36,44], [45,54], [55,65], ... .

%F G.f.: Product_{n>=0} (1 + Sum_{k=(n*(n+1)/2)..(n*(n+3)/2)} x^k).

%e a(0)=1 by definition of the empty partition.

%e a(10)=6 because 10=9+1=8+2=7+3=6+4=6+3+1 (for example, you cannot take 5+5 or 7+2+1 because of the definition of a(n)).

%p f:= n-> 1+add(x^j, j=n*(n+1)/2..n*(n+3)/2):

%p a:= n-> coeff(mul(f(k), k=1..ceil((sqrt(9+8*n)-3)/2)), x, n):

%p seq(a(n), n=0..61);

%o (PARI) f(n, x) = (1+sum(j=n*(n+1)/2, n*(n+3)/2, x^j));

%o a(n) = polcoef(prod(k=1, ceil((sqrt(9+8*n)-3)/2), f(k, x)), n, x); \\ version 2.11.0 or newer; _Michel Marcus_, Apr 08 2019

%o (PARI) first(n) = v = Vecrev(Vec(a(n))); vector(n, i, v[i]) \\ using a(n) from above \\ _David A. Corneth_, Apr 08 2019

%Y Cf. A000217, A057944.

%K nonn

%O 0,5

%A _Igor Haladjian_, Apr 06 2019