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 1-4 of 4 results.

A381581 Numbers divisible by the sum of the digits in their Chung-Graham representation (A381579).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12, 16, 20, 21, 22, 24, 27, 28, 30, 40, 42, 44, 45, 48, 55, 56, 57, 58, 60, 66, 70, 72, 75, 76, 80, 84, 90, 92, 95, 96, 100, 102, 110, 111, 112, 115, 116, 120, 132, 135, 138, 140, 144, 150, 152, 153, 156, 168, 170, 175, 176, 180, 186, 190, 195, 198
Offset: 1

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Author

Amiram Eldar, Feb 28 2025

Keywords

Comments

Numbers k such that A291711(k) divides k.
Analogous to Niven numbers (A005349) with the Chung-Graham representation (A381579) instead of the decimal representation.
A001906(k) = Fibonacci(2*k) is a term for all k >= 1.
If k is not divisible by 3 (A001651), then Fibonacci(2*k) + 1 is a term.

Examples

			4 is a term since A291711(4) = 1 divides 4.
6 is a term since A291711(6) = 2 divides 6.

Links

Crossrefs

Cf. A000045, A001651, A001906, A291711.
Subsequences: A381582, A381583, A381584, A381585.
Similar sequences: A005349, A049445, A064150, A328208, A328212.

Programs

  • Mathematica
    f[n_] := f[n] = Fibonacci[2*n]; q[n_] := Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; Select[Range[200], q]
  • PARI
    mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
isok(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}

A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

Original entry on oeis.org

1, 2, 20, 55, 56, 110, 304, 364, 398, 846, 1024, 1084, 1744, 1854, 2044, 2104, 2105, 2527, 2824, 2862, 3870, 4374, 5222, 5223, 5243, 5718, 5928, 6488, 6784, 6844, 6894, 6978, 7142, 7924, 10590, 11240, 11889, 11975, 12248, 14284, 14915, 16638, 17710, 17714, 17824
Offset: 1

Views

Author

Amiram Eldar, Feb 28 2025

Keywords

Comments

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

Examples

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, and A291711(3) = 1 divides 3.
20 is a term since A291711(20) = 4 divides 20, A291711(21) = 1 divides 21, and A291711(22) = 2 divides 22.

Links

Crossrefs

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581 and A381582.
Subsequences: A381584, A381585.
Similar sequences: A154701, A328210, A330932, A351721.

Programs

  • Mathematica
    f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[45, 3]
  • PARI
    mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, lim, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}

A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

Original entry on oeis.org

1, 55, 2104, 5222, 24784, 63510, 64264, 69487, 95463, 121393, 184327, 327303, 374589, 463110, 468168, 561069, 572550, 596868, 671407, 740310, 759030, 819948, 902670, 956680, 1023009, 1036230, 1065030, 1259817, 1274910, 1359552, 1683154, 1714470, 1731750, 2182023
Offset: 1

Views

Author

Amiram Eldar, Feb 28 2025

Keywords

Comments

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

Examples

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, A291711(3) = 1 divides 3, and A291711(4) = 2 divides 4.
55 is a term since A291711(55) = 1 divides 55, A291711(56) = 2 divides 56, A291711(57) = 3 divides 57, and A291711(58) = 2 divides 58.

Links

Crossrefs

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581, A381582 and A381583.
A381585 is a subsequence.
Similar sequences: A141769, A328211, A328215, A330933.

Programs

  • Mathematica
    f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[12, 4]
  • PARI
    mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3 = is1(3), q4); for(k = 4, lim, q4 = is1(k); if(q1 && q2 && q3 && q4, print1(k-3, ", ")); q1 = q2; q2 = q3; q3 = q4);}

A381585 Starts of runs of 5 consecutive integers that are all terms in A381581.

Original entry on oeis.org

57744971, 159104411, 203738652, 212548572, 260463851, 361823291, 413644572, 431577521, 440353328, 520800012, 717222337, 726300972, 779825648, 843559091, 913313321, 945016812, 986681527, 1091786528, 1116032201, 1185786431, 1318751081, 1347208812, 1360423692, 1418212627
Offset: 1

Views

Author

Amiram Eldar, Feb 28 2025

Keywords

Comments

Are there 6 consecutive integers that are all terms in A381581?

Examples

			57744971 is a term since A291711(57744971) = 17 divides 57744971, A291711(57744972) = 18 divides 57744972, A291711(57744973) = 13 divides 57744973, A291711(57744974) = 14 divides 57744974, and A291711(57744975) = 15 divides 57744975.

Links

Crossrefs

Cf. A291711.
Subsequence of A381581, A381582, A381583 and A381584.
Similar sequences: A330928, A334373, A364220, A364383.

Programs

  • PARI
    mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3 = is1(3), q4 = is1(4), s5); for(k = 5, lim, q5 = is1(k); if(q1 && q2 && q3 && q4 && q5, print1(k-4, ", ")); q1 = q2; q2 = q3; q3 = q4; q4 = q5);}
Showing 1-4 of 4 results.

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