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.

A294306 Irregular triangle read by rows: T(n, m) = total number of times the different values appear in row n of A280269, where 0 <= m <= A280274(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1, 1, 1, 5, 2, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 4, 1, 1, 1, 5, 2, 1, 1, 1, 1, 1, 5, 2, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 7, 3, 1, 2, 1, 3, 1, 1, 1, 1, 3, 1, 5, 2, 1, 1, 1, 7, 6, 3, 1, 1, 1, 1, 5
Offset: 1

Views

Author

Michael De Vlieger, Oct 30 2017

Keywords

Comments

The numbers i in A162306(n) divide n^k with k >= 0; these k are listed in row n of A280269.
Row 1 = 1 and T(n, 0) = 1 for all n, since 1 is the empty product and divides n^0.
Row p = 1, 1, (row length = 2) since the only divisors of p are 1 and p; 1 | p^0, and p | p^1.
Row p^e = 1, e, since the only numbers in A162306(p^e) are 1 and p^k for 1 <= k <= e.
Row length of a(n) > 2 for n with omega(n) > 1.
Total of row n = A010846(n).
Sum of terms of T(n, m) with m <= 1 in row n = A000005(n).
Sum of terms of T(n, m) with m > 1 = A243822(n).
Terms in row n of A294306 start at 1, generally quickly rise to a maximum, then gradually decline at m = A280274(n).

Examples

			Row n of A280269(10) = 0, 1, 2, 1, 3, 1, corresponding to A162306(10) = 1, 2, 4, 5, 8, 10, since 1 | 10^0, 2 | 10^1, 4 | 10^2, 5 | 10^1, 8 | 10^3, and 10 | 10^1. There is 1 zero, 3 ones, 1 two, and 1 three, thus a(10) = 1, 3, 1, 1. sum(a(10)) = A010846(10) = 6. Length of a(10) = A280274(10) + 1 = 4.
Triangle begins:
   1:  1
   2:  1  1
   3:  1  1
   4:  1  2
   5:  1  1
   6:  1  3  1
   7:  1  1
   8:  1  3
   9:  1  2
  10:  1  3  1  1
  11:  1  1
  12:  1  5  2
  13:  1  1
  14:  1  3  1  1
  15:  1  3  1
  16:  1  4
  17:  1  1
  18:  1  5  2  1  1
  19:  1  1
  20:  1  5  2
  ...

Links

Crossrefs

Cf. A000005, A010846, A162306, A243822, A280269, A280274.

Programs

  • Mathematica
    Table[Tally[#][[All, -1]] &@ Map[SelectFirst[Range[0, Floor@ Log2@ n], Function[k, Divisible[n^k, #]]] &, Select[Range@ n, PowerMod[n, Floor@ Log2@ n, #] == 0 &]], {n, 32}] // Flatten (* Michael De Vlieger, Oct 30 2017 *)

A280269 Irregular triangle T(n,m) read by rows: smallest power e of n that is divisible by m = term k in row n of A162306.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 2, 1, 3, 1, 0, 1, 0, 1, 1, 1, 1, 2, 2, 1, 0, 1, 0, 1, 2, 1, 3, 1, 0, 1, 1, 2, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 3, 1, 2, 4, 1, 0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1, 2, 3, 1, 4, 1, 0, 1, 0, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Michael De Vlieger, Dec 30 2016

Keywords

Comments

This table eliminates the negative values in row n of A279907.
Let k = A162306(n,m), i.e., the value in column m of row n.
T(n,1) = 0 since 1 | n^0.
T(n,p) = 1 for prime divisors p of n since p | n^1.
T(n,d) = 1 for divisors d > 1 of n since d | n^1.
Row n for prime p have two terms, {0,1}, the maximum value 1, since all k < p are coprime to p, and k | p^1 only when k = p.
Row n for prime power p^i have (i+1) terms, one zero and i ones, since all k that appear in corresponding row n of A162306 are divisors d of p^i.
Values greater than 1 pertain only to composite k of composite n > 4, but not in all cases. T(n,k) = 1 for squarefree kernels k of composite n.
Numbers k > 1 coprime to n and numbers that are products of at least one prime q coprime to n and one prime p | n do not appear in A162306; these do not divide n^e evenly.
T(n,k) is nonnegative for all numbers k for which n^k (mod k) = 0, i.e., all the prime divisors p of k also divide n.
The largest possible value s in row n of T = floor(log_2(n)), since the largest possible multiplicity of any number m <= n pertains to perfect powers of 2, as 2 is the smallest prime. This number s first appears at T(2^s + 2, 2^s) for s > 1.
1/k terminates T(n,k) digits after the radix point in base n for values of k that appear in row n of A162306.
Originally from Robert Israel at A279907: (Start)
T(a*b,c*d) = max(T(a,c),T(b,d)) if GCD(a,b)=1, GCD(b,d)=1,T(a,c)>=0 and T(b,d)>=0.
T(n,a*b) = max(T(n,a),T(n,b)) if GCD(a,b)=1 and T(n,a)>=0 and T(n,b)>=0.
(End)

Examples

			Triangle T(n,m) begins:  Triangle A162306(n,k):
1:  0                    1
2:  0  1                 1  2
3:  0  1                 1  3
4:  0  1  1              1  2  4
5:  0  1                 1  5
6:  0  1  1  2  1        1  2  3  4  6
7:  0  1                 1  7
8:  0  1  1  1           1  2  4  8
9:  0  1  1              1  3  9
10: 0  1  2  1  3  1     1  2  4  5  8  10
...

Links

Crossrefs

Cf. A162306, A279907 (T(n,k) with values for all 1 <= k <= n), A280274 (maximum values in row n), A010846 (number of nonnegative k in row n), A051731 (k with e <= 1), A000005 (number of k in row n with e <= 1), A272618 (k with e > 1), A243822 (number of k in row n with e > 1), A007947.

Programs

  • Mathematica
    Table[SelectFirst[Range[0, #], PowerMod[n, #, k] == 0 &] /. m_ /; MissingQ@ m -> Nothing &@ Floor@ Log2@ n, {n, 24}, {k, n}] // Flatten (* Version 10.2, or *)
DeleteCases[#, -1] & /@ Table[If[# == {}, -1, First@ #] &@ Select[Range[0, #], PowerMod[n, #, k] == 0 &] &@ Floor@ Log2@ n, {n, 24}, {k, n}] // Flatten (* or *)
DeleteCases[#, -1] & /@ Table[Boole[k == 1] + (Boole[#[[-1, 1]] == 1] (-1 + Length@ #) /. 0 -> -1) &@ NestWhileList[Function[s, {#1/s, s}]@ GCD[#1, #2] & @@ # &, {k, n}, And[First@# != 1, ! CoprimeQ @@ #] &], {n, 24}, {k, n}] // Flatten

A280363 a(n) = floor(log_p(n)) where p = A020639(n), i.e., the least prime factor of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 1, 3, 2, 4, 1, 4, 1, 4, 2, 4, 1, 4, 2, 4, 3, 4, 1, 4, 1, 5, 3, 5, 2, 5, 1, 5, 3, 5, 1, 5, 1, 5, 3, 5, 1, 5, 2, 5, 3, 5, 1, 5, 2, 5, 3, 5, 1, 5, 1, 5, 3, 6, 2, 6, 1, 6, 3, 6, 1, 6, 1, 6, 3, 6, 2, 6, 1, 6, 4, 6, 1, 6, 2, 6, 4, 6, 1, 6, 2, 6, 4, 6, 2, 6, 1, 6, 4, 6, 1, 6, 1, 6, 4, 6, 1, 6, 1, 6, 4, 6, 1, 6, 2, 6, 4, 6, 2, 6
Offset: 1

Views

Author

Michael De Vlieger, Jan 01 2017

Keywords

Comments

a(1) = 0 since 1 is the empty product.
a(p) = 1 since the exponent e of the largest power p^e of the prime divisor p is p^1 (i.e., p itself).
a(p^m) = m since the largest power p^e of the prime divisor p is p^m, (p^m itself), i.e., e = m.
a(n) is the greatest value of the power e of p^e across the prime divisors p of n such that p^e <= n.
Consider integers 1<=r<=n with all prime divisors p of r also dividing n. Let m be the smallest power n^m | r, and let e be the largest value of m across 1<=r<=n. This is A280274(n). This sequence underlies A280274: A280274(1) = 0, A280274(n) = 1 with n having omega(n) = 1. A280274(n) = a(n) for squarefree n. A280274(n) for all other n is ceiling(a(n)/k), with k being the multiplicity of p = A020639(n) in the prime decomposition of n.

Examples

			a(10) = 3, because 2^3 = 8 and 5^1 = 5 are less than 10 = 2*5, and of the multiplicities of these numbers, 3 is the greatest.
a(12) = 3, because 2^3 = 8 and 3^2 = 9 are less than 12 = 2*2*3, and of the multiplicities of these numbers, 3 is the greatest.
a(16) = 4, because 2^4 = 16 = n, and is the largest power of the distinct prime divisor 2 of 16.

Links

Crossrefs

Cf. A020639, A007947, A280274.

Programs

  • Mathematica
    Table[If[n == 1, 0, Floor[Log[FactorInteger[n][[1, 1]], n]]], {n, 120}]
  • PARI
    a(n) = if (n==1, 0, logint(n, vecmin(factor(n)[,1]))); \\ Michel Marcus, Jan 01 2017
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