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A390315
Numbers m such that A390254(m) > c*(4/3)^m, where c = lim_{m->oo} A390254(m)/(4/3)^m = A390321.
4
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 26, 31, 32, 33, 34, 35, 36, 46, 47, 48, 73, 74, 75, 76, 77, 78, 79, 80, 81, 97, 101, 115, 116, 117, 118, 119, 120, 126, 127, 128, 129, 130, 131, 132, 138, 139, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 153, 154, 155, 156, 157, 161, 162, 165, 166, 167, 168, 169, 170, 171, 172, 174, 175, 176
OFFSET
1,2
COMMENTS
In general, let {f_n}_{n>=0} be the sequence defined by f_{n+1} = alpha*f_n + e_n, where alpha > 1, a <= e_n <= b, then f_n = (f_0 + e_0/alpha + ... + e_{n-1}/alpha^n)*alpha^n = c*alpha^n - (e_n/alpha + e_{n+1}/alpha^2 + ...), where c = f_0 + Sum_{n>=0} e_n/alpha^{n+1}. Fix a natural number n. For each k >= 0,
(*) -(e_n/alpha + e_{n+1}/alpha^2 + ... + e_{n+k-1}/alpha^k) - (b/(alpha - 1))/alpha^k <= f_n - c*alpha^n <= -(e_n/alpha + e_{n+1}/alpha^2 + ... + e_{n+k-1}/alpha^k) - (a/(alpha - 1))/alpha^k.
As a result:
- If there exists k such that e_n + e_{n+1}/alpha + ... + e_{n+k-1}/alpha^(k-1) > -(a/(alpha - 1))/alpha^(k-1), then f_n < c*alpha^n;
- If there exists k such that e_n + e_{n+1}/alpha + ... + e_{n+k-1}/alpha^(k-1) < -(b/(alpha - 1))/alpha^(k-1), then f_n > c*alpha^n;
- Otherwise we have e_n/alpha + e_{n+1}/alpha^2 + ... = 0, so f_n = c*alpha^n.
Here we have alpha = 4/3, a = -1/3, b = 1/3, and e_n = 0, -1/3, 1/3 for A390254(n) == 0, 1, 2 (mod 3) respectively. Since we have neither e_n = -1/3 for all sufficiently large n nor e_n = 1/3 for all sufficiently large n, the inequalities in (*) are strict. So n is a term in this sequence iff there exists k such that
e_n + (3/4)*e_{n+1} + ... + (3/4)^(k-1)*e_{n+k-1} <= -(3/4)^(k-1).
LINKS
Jianing Song, Table of n, a(n) for n = 1..10120 (terms up to 20000)
EXAMPLE
1 through 10 and 20 are terms, as witnessed by
m = 1: e_1 + (3/4)*e_2 + (3/4)^2*e_3 + (3/4)^3*e_4 + (3/4)^4*e_5 + (3/4)^5*e_6 + (3/4)^6*e_7 <= -(3/4)^6;
m = 2: e_2 + (3/4)*e_3 + (3/4)^2*e_4 + (3/4)^3*e_5 + (3/4)^4*e_6 + (3/4)^5*e_7 <= -(3/4)^5;
m = 3: e_3 + (3/4)*e_4 + (3/4)^2*e_5 + (3/4)^3*e_6 + (3/4)^4*e_7 + (3/4)^5*e_8 + (3/4)^6*e_9 + (3/4)^7*e_10 + (3/4)^8*e_11 <= -(3/4)^8;
m = 4: e_4 + (3/4)*e_5 + (3/4)^2*e_6 + (3/4)^3*e_7 <= -(3/4)^3;
m = 5: e_5 + (3/4)*e_6 + (3/4)^2*e_7 + (3/4)^3*e_8 + (3/4)^4*e_9 + (3/4)^5*e_10 <= -(3/4)^5;
m = 6: e_6 + (3/4)*e_7 + (3/4)^2*e_8 + (3/4)^3*e_9 + (3/4)^4*e_10 <= -(3/4)^4;
m = 7: e_7 + (3/4)*e_8 + (3/4)^2*e_9 + (3/4)^3*e_10 <= -(3/4)^3;
m = 8: e_8 + (3/4)*e_9 + (3/4)^2*e_10 + (3/4)^3*e_11 <= -(3/4)^3;
m = 9: e_9 + (3/4)*e_10 + (3/4)^2*e_11 <= -(3/4)^2;
m = 10: e_10 + (3/4)*e_11 + (3/4)^2*e_12 + (3/4)^3*e_13 + (3/4)^4*e_14 + (3/4)^5*e_15 + (3/4)^6*e_16 <= -(3/4)^6;
m = 20: e_20 + (3/4)*e_21 + (3/4)^2*e_22 + (3/4)^3*e_23 <= -(3/4)^3.
PROG
(PARI) A390315_up_to_N(N) = { \\ gives terms <= N
my(A390254 = vector(N+1), e = vector(1+N), res = vector(1+N), len = N, sumk); \\ res[1+n] = -1 (resp. 1) means A390254(n) > (resp. -1) c*(4/3)^n
A390254[1] = 2; e[1] = 1/3;
for(n=1, N, A390254[1+n] = round(4*A390254[n]/3); e[1+n] = -kronecker(A390254[1+n], 3)/3);
for(n=0, N, sumk = 0; for(k=0, oo, if(n+k > len, A390254 = concat(A390254, round(4*A390254[1+len]/3)); e = concat(e, -kronecker(A390254[1+len+1], 3)/3); len++); sumk += (3/4)^k*e[1+n+k]; if(abs(sumk) >= (3/4)^k, res[1+n] = sign(sumk); break())));
select(n->res[n+1]==-1, vector(N+1, i, i-1));
}
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Nov 01 2025
STATUS
approved