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.

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A377455 Numbers k such that k and k+1 are both terms in A377385.

Original entry on oeis.org

1, 1224, 126191, 428519, 649727, 1015416, 1988064, 3425856, 4542740, 4574240, 4743900, 4813668, 5131008, 6899840, 7001315, 7172424, 7356096, 8020583, 10206000, 11146421, 11566800, 11597999, 11693807, 12556700, 13742624, 13745759, 13831487, 14365120, 16939799, 20561400
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			1224 is a term since both 1224 and 1225 are in A377385: 1224/A034968(1224) = 204 and 204/A034968(204) = 34 are integers, and 1225/A034968(1225) = 175 and 175/A034968(175) = 35 are integers.

Links

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205 and A377385.
Subsequences: A377456, A377457.
Analogous sequences: A376793 (binary), A377271 (Zeckendorf).

Programs

  • Mathematica
    fdigsum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s += r; m++]; s]; q[k_] := q[k] = Module[{f = fdigsum[k]}, Divisible[k, f] && Divisible[k/f, fdigsum[k/f]]]; Select[Range[2*10^6], q[#] && q[#+1] &]
  • 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)); !(k % f) && !((k/f) % fdigsum(k/f));}
lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

A226171 Smallest base in which n is not Niven (or zero if n is Niven in every base).

Original entry on oeis.org

0, 0, 2, 0, 2, 0, 2, 6, 2, 4, 2, 8, 2, 2, 2, 6, 2, 8, 2, 7, 5, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 6, 2, 3, 2, 8, 2, 2, 2, 12, 2, 3, 2, 2, 2, 2, 2, 14, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 8, 2, 2, 2, 6, 2, 3, 2, 3, 3, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 8, 5, 2, 2, 5, 2, 2
Offset: 1

Views

Author

Sergio Pimentel, May 29 2013

Keywords

Comments

Niven numbers (in base b) are divisible by the sum of their digits (in base b).
Questions: are 1, 2, 4 and 6 the only zeros in this sequence? Where are the records or high water marks?
From Bert Dobbelaere, Oct 08 2018: (Start)
1,2,4,6 are the only numbers that are Niven in every base.
Proof: Suppose n is Niven in every base, then consider the base-b representations of n for (n/2) < b <= n. These are all 2-digit numbers with 1 as 1st digit and (n-b) as last digit. Then 1+n-b is a divisor of n for all b, meaning that all numbers between 1 up to n/2 are divisors of n. Clearly there are no such numbers larger than 6.
a(n) < 60 for n < 10^13.
(End)

Examples

			The sum of digits of 24 in bases 1 through 14 are:  24, 2, 4, 3, 8, 4, 6, 3, 8, 6, 4, 2, 12, 11.  24 is divisible by all these numbers except the last one; therefore a(24) = 14.

Links

Crossrefs

Cf. A049445, A118363, A005349, A144261, A144262, A226169.
Cf. A225427 (least Niven number for all bases from 1 to n).

Programs

  • Mathematica
    Table[b = 2; While[s = Total[IntegerDigits[n, b]]; s < n && Mod[n, s] == 0, b++]; If[s == n, b = 0]; b, {n, 100}] (* T. D. Noe, May 30 2013 *)
  • PARI
    a(n) = {for (b=2, n-1, if (frac(n/sumdigits(n,b)), return(b));); 0;} \\ Michel Marcus, Oct 23 2018

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.

Links

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);}

A076428 Numbers k such that the sum of digits of k in superfactorial base divides k.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 12, 14, 15, 20, 24, 28, 30, 36, 40, 42, 48, 50, 56, 60, 63, 70, 72, 80, 84, 90, 96, 100, 108, 110, 120, 121, 132, 144, 150, 153, 156, 168, 180, 192, 200, 204, 216, 220, 228, 231, 240, 250, 252, 264, 276, 288, 290, 291, 295, 300, 304, 305, 312, 315
Offset: 1

Views

Author

Floor van Lamoen, Oct 10 2002

Keywords

Comments

We define the superfactorial base as a variant of the factorial base where place values are superfactorials (A000178) instead of factorials (A000142). - Rémy Sigrist, Mar 20 2018

Examples

			28 written in superfactorial base is 220, the sum of digits is thus 4 and 4 divides 28, so 28 is included in the sequence.

Links

Crossrefs

Cf. A000142, A000178, A118363.

Programs

  • Maple
    A076428 := proc(n) local i,j,m,t,t1; t := NULL; for j from 1 to n do m := j; i := 2; t1 := 0; while m>0 do t1 := t1 + (m mod i!); m := floor(m/i!); i := i+1; od; if j mod t1 = 0 then t := t,j fi; od; t; end;
  • Mathematica
    max = 4; bases = Range[max, 1, -1]!; nmax = Times @@ bases - 1; sumdig[n_] := Plus @@ IntegerDigits[n, MixedRadix[bases]]; Select[Range[nmax], Divisible[#, sumdig[#]] &] (* Amiram Eldar, Sep 07 2020 *)

Extensions

Definition corrected by Rémy Sigrist, Mar 20 2018

A377456 Starts of runs of 3 consecutive integers that are all terms of A377385.

Original entry on oeis.org

39998374960, 326660221888, 520935101440, 723006782783, 923072388208, 977932351240, 1134397887874, 1351753892944, 1864828904536, 2171452161023
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			39998374960 is a term since 39998374960, 39998374961 and 39998374962 are all in A377385: 39998374960/A034968(39998374960) = 999959374, and 999959374/A034968(999959374) = 32256754 are integers, 39998374961/A034968(39998374961) = 975570121, and 975570121/A034968(975570121) = 33640349 are integers, and 39998374962/A034968(39998374962) = 1025599358, and 1025599358/A034968(1025599358) = 30164687 are integers.

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205, A377385 and A377455.
Analogous sequences: A376794 (binary), A377273 (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)); !(k % f) && !((k/f) % fdigsum(k/f));}
lista(kmax) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, kmax, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}

A377387 a(n) is the least number k such that A377384(k) = n, or -1 if no such number exists.

Original entry on oeis.org

1, 3, 8, 27, 135, 1215, 15795, 328050, 4920750, 127764000, 5826168000, 126097171200
Offset: 0

Views

Author

Amiram Eldar, Oct 27 2024

Keywords

Comments

a(12) > 2.2*10^12, if it exists.
a(12) <= 5160284236800, a(13) <= 227052506419200. - David A. Corneth, Oct 27 2024
All the terms except for 3 are factorial-base Niven numbers (A118363).

Examples

			  n | The n iterations
  --+------------------------------------------------------
  1 | 3 -> 3/2
  2 | 8 -> 4 -> 2 = 2!
  3 | 27 -> 9 -> 3 -> 3/2
  4 | 135 -> 27 -> 9 -> 3 -> 3/2
  5 | 1215 -> 135 -> 27 -> 9 -> 3 -> 3/2
  6 | 15795 -> 1215 -> 135 -> 27 -> 9 -> 3 -> 3/2
  7 | 328050 -> 18225 -> 1215 -> 135 -> 27 -> 9 -> 3 -> 3/2

Crossrefs

Cf. A118363, A377384.
Analogous sequences: A376619 (binary), A377211 (Zeckendorf).

Programs

  • Mathematica
    fdigsum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s += r; m++]; s]; f[n_] := f[n] = Module[{s = fdigsum[n]}, If[s == 1, 0, If[!Divisible[n, s], 1, 1 + f[n/s]]]]; seq[len_] := Module[{v = Table[0, {len}], c = 0, k = 1, i}, While[c < len, i = f[k] + 1; If[v[[i]] == 0, c++; v[[i]] = k]; k++]; v]; seq[8]
  • 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;}
f(n) = {my(s = fdigsum(n)); if(s == 1, 0, if(n % s, 1, 1 + f(n/s)));}
lista(len) = {my(v = vector(len), c = 0, k = 1, i); while(c < len, i = f(k) + 1; if(v[i] == 0, c++; v[i] = k); k++); v; }
Previous Showing 31-36 of 36 results.

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