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A330889
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a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 3.
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12
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1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 5, 3, 4, 3, 5, 6, 1, 3, 8, 3, 1, 6, 5, 8, 4, 3, 5, 6, 6, 3, 8, 3, 1, 11, 5, 3, 4, 3, 10, 12, 1, 3, 8, 8, 1, 12, 5, 3, 9, 3, 5, 12, 1, 8, 8, 3, 1, 12, 17, 3, 4, 3, 5, 17, 1, 10, 8, 3, 6, 12, 5, 3, 11, 8, 5, 12, 1, 3, 13
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OFFSET
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1,5
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COMMENTS
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The one-part partition n = n is included in the count.
For the relation to pentagonal numbers see also A330888.
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LINKS
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FORMULA
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Conjecture: G.f.: Sum_{n>=1} n*x^(n*(3*n-1)/2)/(1-x^n).
The summation variable n runs over the number of parts in the partition.
For fixed n, the smallest such partition is:
1 + 4 + 7 + ... + (3n-2).
The above sum is equal to n * (3*n-1) / 2. That's where the x^(n*(3*n-1)/2) factor comes from.
Then we want to (add 1 to every part), (add 2 to every part), etc. to get 2 + 5 + 8 + ..., 3 + 6 + 9 + ..., which corresponds to adding n, 2*n, 3*n, etc. to the base partition. So we divide by (1 - x^n).
Multiply by n (to count the total number of parts) and we are done. QED
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EXAMPLE
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For n = 21 there are three partitions of 21 into consecutive parts that differ by 3, including 21 as a partition. They are [21], [12, 9] and [10, 7, 4]. The number of parts of these partitions are 1, 2 and 3 respectively. The total number of parts is 1 + 2 + 3 = 6, so a(27) = 6.
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MAPLE
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local a;
a := 0 ;
for k from 1 do
else
return a;
end if;
end do:
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MATHEMATICA
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nmax = 100;
CoefficientList[Sum[n x^(n(3n-1)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
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PROG
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(PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-1)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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