A239447 Partial sums of A030101.
0, 1, 2, 5, 6, 11, 14, 21, 22, 31, 36, 49, 52, 63, 70, 85, 86, 103, 112, 137, 142, 163, 176, 205, 208, 227, 238, 265, 272, 295, 310, 341, 342, 375, 392, 441, 450, 491, 516, 573, 578, 615, 636, 689, 702, 747, 776, 837, 840, 875, 894, 945, 956, 999, 1026, 1085
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000
- Project Euler, Problem 463: A weird recurrence relation
Crossrefs
Cf. A030101.
Programs
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Mathematica
Accumulate[Table[FromDigits[Reverse[IntegerDigits[n,2]],2],{n,0,80}]] (* Harvey P. Dale, Jan 10 2023 *) -
Python
A = {0: 0, 1: 1, 2: 2, 3: 5} def a(n): a_n = A.get(n) if a_n is not None: return a_n q, r = divmod(n, 4) if r == 0: a_n = a(q*2)*6 - a(q)*5 - a(q - 1)*3 - 1 elif r == 1: a_n = a(q*2 + 1)*2 + a(q*2)*4 - a(q)*6 - a(q - 1)*2 - 1 elif r == 2: a_n = a(q*2 + 1)*3 + a(q*2)*3 - a(q)*6 - a(q - 1)*2 - 1 else: a_n = a(q*2 + 1)*6 - a(q)*8 - 1 A[n] = a_n # memoization return a_n
Formula
a(4n) = 6 a(2n) - 5 a(n) - 3 a(n-1) - 1.
a(4n + 1) = 2 a(2n+1) + 4 a(2n) - 6 a(n) - 2 a(n-1) - 1.
a(4n + 2) = 3 a(2n+1) + 3 a(2n) - 6 a(n) - 2 a(n-1) - 1.
a(4n + 3) = 6 a(2n+1) - 8 a(n) - 1.
Extensions
More terms from Alois P. Heinz, May 19 2014