OFFSET
1,2
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)
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)
EXAMPLE
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; ...
MATHEMATICA
a[1]=1; a[n_] := a[n] = #/2^IntegerExponent[#, 2] &@ (n + a[n-1]); Array[a, 70] (* Giovanni Resta, Jun 25 2020 *)
PROG
(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), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 02 2012
STATUS
approved