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A305992
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The sequence whose indicator function is I in conjectured formula A300997(n) = 2*n - Sum_{k=1..n} I(k), as long as the conjecture holds.
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1
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1, 2, 4, 8, 15, 24, 32, 48, 62, 80, 101, 122, 147, 171, 202, 230, 267, 299, 339, 377, 418, 464, 509, 559, 611, 664, 719, 776, 836, 896, 960, 1024, 1098, 1167, 1240, 1315, 1392, 1471, 1553, 1642, 1724, 1816, 1906, 1999, 2094, 2190, 2290, 2392, 2499, 2599, 2713, 2818, 2937, 3048, 3166, 3288
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OFFSET
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1,2
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COMMENTS
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A300997(n) is believed to be equal to 2*n - Sum_{k=1..n} I(k), where I is the indicator function of some other sequence -- let it be this sequence. This sequence is finite if the conjecture is false.
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LINKS
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PROG
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(C)
#include <stdio.h>
#include <string.h>
#define N 10000
void e(int *t, int *s) {
int T[N], i = 0; memset(T, 0, sizeof(T));
while (i < *s) {
int f = t[i] / 2;
T[i] += f + (t[i] % 2);
T[++ i] += f;
}
if (T[*s] != 0) { *s += 1; }
for (i = 0; i < *s; i ++) { t[i] = T[i]; }
}
int f(int n) {
int t[N], s = 1, i = 0; t[0] = n;
while (s != n) { i ++; e(t, &s); }
return 2 * n - i;
}
int main() {
int n, last = 1, current;
for (n = 1; n <= N; n ++) {
current = f(n);
switch (current - last) {
case 0: break;
case 1: printf("%d, ", n); fflush(stdout); break;
default: fprintf(stderr, "CONJECTURE IS FALSE"); return;
}
last = current;
}
printf("\n");
}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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