A322919 Numbers k such that k and k-1 both first appear in the same power of 2 (in base 10).

59, 74, 111, 785, 793, 914, 957, 985, 1070, 1467, 2019, 2099, 2332, 2610, 2934, 3028, 3083, 3311, 3334, 3973, 4198, 4208, 4334, 4590, 4689, 4785, 5247, 5350, 5535, 6166, 6335, 6669, 6761, 7167, 7340, 7707, 7980, 8668, 8990, 9180, 9840, 11110, 13096, 16285, 17418, 18091, 18361, 19219, 20522, 21494, 21827

Offset

1,1

Links

David A. Corneth, Table of n, a(n) for n = 1..3106

Example

For instance 2019 is in the sequence since 2018 and 2019 both appear in 2^212 and neither appear in any smaller power of two.

Prog

(C)
#include <stdio.h>
int main() {
  int n = 1000001;  /* Highest term */
  int p = 2;        /* Powers of two.  Test throughly if you change it. */
  int r = 10;       /* Base ten.  Test throughly if you change it. */
  char a[n];
  int b, c, i, j, k, k2, l, lk, m, ok, ok2, u, d[7], f[n], g[n], v[n];
  u = n;
  for (j=0; j<n; j++) {
    a[j] = v[j] = 0; /* corrected v[n] to v[j]. - M. F. Hasler, Jul 05 2021*/
    f[j] = g[j] = -1;
  }
  a[0] = 1;
  for (m=0; (m<n) && u; m++) {
    for (j=0; j<n; j++) if (a[j]) b = j;
    for (k=0; k<n; k++) {
      k2 = k;
      for (j=0; k2; j++) {
        d[j] = k2 % r;
        k2 /= r;
      }
      lk = j;
      if (!j) {
        d[0] = 0;
        lk = 1;
      }
      ok2 = 0;
      if ((f[k] == -1) && (lk<=b+1)) {
        ok2 = 0;
        for (l=b-lk+1; l>-1; l--) {
          ok = 1;
          for (j=lk-1; j>-1; j--) if (a[l+j] != d[j]) ok = 0;
          if (ok) ok2 = 1;
        }
        if (ok2) {
          f[k] = m;
          u--;
        }
      }
      if ((g[k]==-1) && (lk<=b+1)) {
        ok = 1;
        for (j=lk-1; j>-1; j--) if (a[b-lk+j+1] != d[j]) ok = 0;
        if (ok) g[k] = m;
      }
    }
    c = 0;
    for (j=0; j<b+2; j++) {
      a[j] = a[j]*p + c;
      c = 0;
      if (a[j] > r-1) {
        c = a[j] / r;
        a[j] %= r;
      }
    }
  }
  for (i=1; i<n; i++) if (f[i] == f[i-1]) printf("%d\n", i);
  return(0);
}
(PARI) uptoQdigits(n) = {v = vector(10^n); p = 1/2; todo = 10^n; my(res = List());
for(i = 1, oo, p<<=1; process(p, n); if(todo <= 0, break)); for(i = 1, #v - 1,
if(v[i] == v[i+1], listput(res, i))); res}
process(p, n) = {my(dp = digits(p), vd, lp = logint(p, 2)); qdp = #dp; my(t = min(n, qdp)); for(qd = 1, t, for(j = 1, qdp - qd + 1, vd = fromdigits(vector(qd, i, dp[j+i-1])); if(v[vd + 1] == 0, v[vd + 1] = lp; todo--)))} \\ David A. Corneth, Dec 31 2018

Crossrefs

Indices of consecutive repeats in A030000.

Sequence in context: A033235 A106913 A212266 * A026050 A347804 A304356

Adjacent sequences: A322916 A322917 A322918 * A322920 A322921 A322922

Keyword

nonn,base

Author

Keith F. Lynch, Dec 30 2018

Status

approved