cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A376793 Numbers k such that k and k+1 are both in A376616.

Original entry on oeis.org

1, 1424, 2484, 4304, 4655, 6223, 10624, 13824, 20624, 22784, 29448, 33424, 33984, 36944, 41535, 43263, 45184, 45324, 65744, 66199, 68624, 69632, 70784, 74304, 74627, 99584, 103103, 103424, 105720, 114704, 120680, 139904, 147636, 164224, 166144, 192576, 199968
Offset: 1

Views

Author

Amiram Eldar, Oct 04 2024

Keywords

Examples

			1424 is a term since both 1424 and 1425 are in A376616: 1424/A000120(1424) = 356 and 356/A000120(356) = 89 are integers, and 1425/A000120(1425) = 285 and 285/A000120(285) = 57 are integers.
		

Crossrefs

Subsequence of A330931 and A376616.
Subsequences: A376794, A376795.
Cf. A000120.

Programs

  • Mathematica
    q[k_] := q[k] = Module[{w = DigitCount[k, 2, 1]}, Divisible[k, w] && Divisible[k/w, DigitCount[k/w, 2, 1]]]; Select[Range[20000], q[#] && q[#+1] &]
  • PARI
    is1(k) = {my(w = hammingweight(k)); !(k % w) && !((k/w) % hammingweight(k/w));}
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

A377272 Numbers k such that k and k+1 are both terms in A377210.

Original entry on oeis.org

1, 2, 3, 4, 5, 12, 47375, 2310399, 3525200, 6506367, 9388224, 17613504, 29373839, 41534800, 48191759, 48344120, 66927384, 68094999, 71982999, 92547279, 95497919, 110146959, 110395439, 126123920, 148865535, 152546030, 154451583, 171570069, 193628799, 232058519
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2024

Keywords

Examples

			47375 is a term since both 47375 and 47376 are in A377210: 47375/A007895(47375) = 9475, 9475/A007895(9475) = 1895 and 1895/A007895(1895) = 379 are integers, and 47376/A007895(47376) = 15792, 15792/A007895(15792) = 3948 and 3948/A007895(3948) = 1316 are integers.
		

Crossrefs

Cf. A007895, A376795 (binary analog).
Subsequence of A328208, A328209, A377210 and A377271.

Programs

  • Mathematica
    zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
    q[k_] := q[k] = Module[{z = zeck[k], z2, m, n}, IntegerQ[m = k/z] && Divisible[m, z2 = zeck[m]] && Divisible[n = m/z2, zeck[n]]]; Select[Range[50000], q[#] && q[#+1] &]
  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
    is1(k) = {my(z = zeck(k), z2, m); if(k % z, return(0)); m = k/z; z2 = zeck(m); !(m % z2) && !((m/z2) % zeck(m/z2)); }
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2); }

A377457 Numbers k such that k and k+1 are both terms in A377386.

Original entry on oeis.org

1, 12563307224, 15897851550, 30412355999, 37706988600, 52576459775, 67673545631, 118533901904, 244316235000, 297265003100, 332110595000, 340800265728, 349358409503, 375624917760, 378624889440, 416375389115, 450026519903, 561162864248, 596004199840, 728643460544
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			12563307224 is a term since both 12563307224 and 12563307225 are in A377386: 12563307224/A034968(12563307224) = 369509036, 369509036/A034968(369509036) = 9723922 and 9723922/A034968(9723922) = 373997 are integers, and 12563307225/A034968(12563307225) = 358951635, 358951635/A034968(358951635) = 7976703 and 7976703/A034968(7976703) = 257313 are integers.
		

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205, A377385, A377386 and A377455.
Analogous sequences: A376795 (binary), A377272 (Zeckendorf).

Programs

  • PARI
    fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
    is1(k) = {my(f = fdigsum(k), f2, m); if(k % f, return(0)); m = k/f; f2 = fdigsum(m); !(m % f2) && !((m/f2) % fdigsum(m/f2));}
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}
Showing 1-3 of 3 results.