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A334541
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a(n) is the number of partitions of n into consecutive parts that differ by 5.
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8
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1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 3, 2, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 3, 3, 1, 3, 2, 2, 3, 2, 2, 3, 1, 3, 3, 2, 1, 3, 3, 2, 2, 2, 2, 4, 1, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 4, 1, 2, 4, 2, 1, 4, 2, 3, 2, 2, 2, 4, 2, 2, 3, 2, 1, 5
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OFFSET
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1,7
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COMMENTS
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Note that all sequences of this family as A000005, A001227, A038548, A117277, A334461, etc. could be prepended with a(0) = 1 when they are interpreted as sequences of number of partitions, since A000041(0) = 1. However here a(0) is omitted in accordance with the mentioned members of the same family.
For the relation to heptagonal numbers see also A334465.
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LINKS
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FORMULA
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The g.f. for "consecutive parts that differ by d" is Sum_{k>=1} x^(k*(d*k-d+2)/2) / (1-x^k); cf. A117277. - Joerg Arndt, Nov 30 2020
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EXAMPLE
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For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4], so a(27) = 3.
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MATHEMATICA
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first[n_] := Module[{res = Array[1&, n]}, For[i = 2, True, i++, start = i + 5 Binomial[i, 2]; If[start > n, Return[res]]; For[j = start, j <= n, j += i, res[[j]]++]]];
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PROG
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(PARI) seq(N, d)=my(x='x+O('x^N)); Vec(sum(k=1, N, x^(k*(d*k-d+2)/2)/(1-x^k)));
(PARI) first(n) = { my(res = vector(n, i, 1)); for(i = 2, oo, start = i + 5 * binomial(i, 2); if(start > n, return(res)); forstep(j = start, n, i, res[j]++ ) ); } \\ David A. Corneth, May 17 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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STATUS
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approved
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