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A123753
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Partial sums of A070941.
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11
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1, 3, 6, 9, 13, 17, 21, 25, 30, 35, 40, 45, 50, 55, 60, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343
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refs;
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history;
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internal format)
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OFFSET
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0,2
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COMMENTS
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a(n) = A003314(n+1) + 1.
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LINKS
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Peter Luschny, Table of n, a(n) for n = 0..10000
Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47
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FORMULA
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a(n) = A003314(n+1)+1. - Reinhard Zumkeller, Oct 12 2006
Let bil(n) = floor(log_2(n)) + 1 for n>0, bil(0) = 0 and b(n) = n + n*bil(n) - 2^bil(n) + 1 then a(n) = b(n+1). (This suggests to prepend '0' to this sequence.) - Peter Luschny, Dec 02 2017
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MAPLE
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A123753 := proc(n) local i, J, z; i := n+1: J := i; i := i-1; z := 1;
while 0 <= i do J := J+i; i := i-z; z := z+z od; J end:
seq(A123753(n), n=0..57); # Peter Luschny, Nov 30 2017
# Alternatively:
a := n -> (n+1)*(1 + ilog2(2*n+3)) - 2^ilog2(2*n+3) + 1:
seq(a(n), n=0..57); # Peter Luschny, Dec 02 2017
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MATHEMATICA
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a[n_] := (n + 1)(1 + IntegerLength[n + 1, 2]) - 2^IntegerLength[n + 1, 2] + 1;
Table[a[n], {n, 0, 57}] (* Peter Luschny, Dec 02 2017 *)
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PROG
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(Python)
def A123753(n):
s, i, z = n+1, n, 1
while 0 <= i: s += i; i -= z; z += z
return s
print([A123753(n) for n in range(0, 58)]) # Peter Luschny, Nov 30 2017
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CROSSREFS
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Cf. A001855, A003314, A033156, A054248, A061168, A083652, A097383, A295508.
Sequence in context: A278449 A006590 A061781 * A124288 A256966 A280944
Adjacent sequences: A123750 A123751 A123752 * A123754 A123755 A123756
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KEYWORD
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nonn
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AUTHOR
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Reinhard Zumkeller, Oct 12 2006
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STATUS
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approved
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