A208884 a(n) = (a(n-1) + n)/2^k where 2^k is the largest power of 2 dividing a(n-1) + n, for n>1 with a(1)=1.

1, 3, 3, 7, 3, 9, 1, 9, 9, 19, 15, 27, 5, 19, 17, 33, 25, 43, 31, 51, 9, 31, 27, 51, 19, 45, 9, 37, 33, 63, 47, 79, 7, 41, 19, 55, 23, 61, 25, 65, 53, 95, 69, 113, 79, 125, 43, 91, 35, 85, 17, 69, 61, 115, 85, 141, 99, 157, 27, 87, 37, 99, 81, 145, 105, 171

Offset: 1

Author

Paul D. Hanna, Mar 02 2012

Keywords

• nonn

Comments

In other words, to get a(n), add n to a(n-1) and compute the odd part (A000265) of the sum. - Ralf Stephan, Oct 27 2013

POSITIONS of odd numbers in the initial 7000000 terms begin:

1: [1, 7, 69, 285, 3601, 5167, 92989, 112651, 6933175, ...];

3: [2, 3, 5, 613, 8461, 46749, 81237, 102171, 126661, 3309589, ...];

5: [13, 97, 2431, 92095, ...];

7: [4, 33, 3167, 78095, 2723179, ...];

9: [6, 8, 9, 21, 27, 303, 2017, 3239, 3765, 6753, 28387, 251451, ...];

11: [75, 15823, 28221, 4091959, 5820487, ...];

13: [22975, 42391, 3729249, ...];

15: [11, 22587, 2527579, 6954893, ...];

17: [15, 51, 3121, 13433, 74763, 376853, 576439, 896899, ...];

19: [10, 14, 25, 35, 291, 77747, 757319, 1227595, 2307099, ...];

21: [1417, 1557, 712229, 2563807, ...];

23: [37, 127, 609, 2211, 5563, 199901, ...];

25: [17, 39, 221, 1145, 3425, 17593, 4318897, ...];

27: [12, 23, 59, 73, 289, 1149, 3393, 20439, 37107, ...];

29: [573, 33315, 61505, 467047, 491359, 1170709, 1492309, 2498593, 3017011, ...];

31: [19, 22, 229, 409, 6199, 60529, 3602675, 4108215, 4604929, ...]; ...

From Ya-Ping Lu, Jun 25 2020: (Start)

Conjecture: For any given odd number m, there exists a number n_max such that all odd numbers <= m can be found in the sequence a(n) with n <= n_max. For example:

m = 1, n_max = 1;

m = 3, n_max = 2;

m = 5, n_max = 13;

m = 11, n_max = 75

m = 13, n_max = 22975;

m = 305, n_max = 1025715;

m = 749, n_max = 14695985;

m = 795, n_max = 150788015;

m = 7525, n_max = 31129547917;

...

If the conjecture above is true, this sequence contains all odd numbers. (End)

Examples

			a(2) = 1 + 2 = 3;
a(3) = (3 + 3)/2 = 3;
a(4) = 3 + 4 = 7;
a(5) = (7 + 5)/4 = 3;
a(6) = 3 + 6 = 9;
a(7) = (9 + 7)/16 = 1; ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Colored scatterplot of the first 100000 terms (where the color is function of the parity of n)

Crossrefs

Cf. A069834, A090895, A114216, A335817.

Programs

Mathematica

a[1]=1; a[n_] := a[n] = #/2^IntegerExponent[#, 2] &@ (n + a[n-1]); Array[a, 70] (* Giovanni Resta, Jun 25 2020 *)

PARI

{a(n)=if(n==1, 1, (a(n-1)+n)/2^valuation(a(n-1)+n,2))}

PARI

{A=vector(1024); a(n)=A[n]=if(n==1, 1, (A[n-1]+n)/2^valuation(A[n-1]+n,2))}
for(n=1,#A,print1(a(n),", "))