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A321440
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Number of partitions of n into consecutive parts, all singletons except the largest.
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7
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1, 1, 2, 3, 3, 4, 5, 4, 5, 7, 5, 6, 8, 5, 8, 10, 5, 8, 10, 7, 10, 11, 7, 8, 13, 9, 9, 14, 7, 12, 15, 6, 12, 13, 11, 15, 14, 8, 10, 19, 10, 12, 18, 8, 16, 19, 9, 12, 17, 14, 16, 16, 10, 15, 21, 15, 14, 20, 7, 16, 25, 7, 20, 21, 14, 18, 18, 14, 12, 26, 16, 17
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OFFSET
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0,3
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COMMENTS
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Number of representations of n as the difference of two distinct triangular numbers, plus any multiple of the order of the larger triangular number.
For n > 0, a(n) is also equal to the Hurwitz class number H(8n-1).
a(n) is also equal to the number of partitions y of n having no repeated even parts and smallest part odd, counted according to the weight w(y) = (-1)^(the number of even parts)*(the number of occurrences of the smallest part). For example, the partitions of 6 having no repeated even parts and smallest part odd are [5,1], [4,1,1], [3,3], [3,2,1], [3,1,1,1], [2,1,1,1,1], and [1,1,1,1,1,1], which are counted with weights 1,-2,2,-1,3,-4, and 6, giving a(6) = 1-2+2-1+3-4+6 = 5. (End)
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LINKS
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N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)
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FORMULA
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G.f.: 1 + Sum_{n>=0} x^(n+1)*Product_{k=1..n} (1-x^(2*k))/Product_{k=1..n+1} (1-x^(2*k-1)).
G.f.: 1 + Sum_{n>=1} (-1)^(n+1)*x^(n^2)/((1-x^(2*n-1))*Product_{k=1..n} (1-x^(2*k-1))). (End)
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EXAMPLE
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Here are the derivations of the terms given. Partitions are listed as strings of digits.
n = 0: (empty partition)
n = 1: 1
n = 2: 11, 2
n = 3: 111, 12, 3
n = 4: 1111, 22, 4
n = 5: 11111, 122, 23, 5
n = 6: 111111, 123, 222, 33, 6
n = 7: 1111111, 1222, 34, 7
n = 8: 11111111, 2222, 233, 44, 8
n = 9: 111111111, 12222, 1233, 234, 333, 45, 9
n = 10: 1111111111, 1234, 22222, 55, (10)
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PROG
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(Python)
from sympy.utilities.iterables import partitions
return 1 if n == 0 else sum(1 for s, p in partitions(n, size=True) if len(p)-1 == max(p)-min(p) == s-p[max(p)]) # Chai Wah Wu, Nov 09 2018
from __future__ import division
def A321440(n): # a faster program based on the characterization in the comments
if n == 0:
return 1
c = 0
for i in range(n):
mi = i*(i+1)//2 + n
for j in range(i+1, n+1):
k = mi - j*(j+1)//2
if k < 0:
break
if not k % j:
c += 1
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CROSSREFS
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See comment by Emeric Deutsch at A001227 (partitions into consecutive parts, all singletons); the partitions considered in the present sequence are a superset of those described by Deutsch.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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