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 1-10 of 10 results.

A081456 Product of entries in n-th row of triangle in A081454.

Original entry on oeis.org

1, 4, 16, 36, 144, 1296, 14400, 57600, 518400, 6350400, 101606400, 914457600, 12294374400, 196709990400, 3292047360000, 44259747840000, 708155965440000, 17703899136000000, 269276305858560000, 4308420893736960000, 107710522343424000000, 3377801980689776640000
Offset: 1

Views

Author

Amarnath Murthy, Mar 21 2003

Keywords

Crossrefs

Formula

a(n) equals smallest integer square m^2 such that A086435(m^2)=n-1. - Max Alekseyev, Jul 16 2009

Extensions

More terms from David Garber, Jun 17 2003
More terms from Ray G. Opao, Aug 01 2005
Corrected and extended by R. J. Mathar, Nov 12 2006
More terms from Max Alekseyev, Jun 05, Jul 16 2009
Corrected and extended by Max Alekseyev, Jun 03 2023

A081457 a(n) = A081456(n)^(1/2).

Original entry on oeis.org

1, 2, 4, 6, 12, 36, 120, 240, 720, 2520, 10080, 30240, 110880, 443520, 1814400, 6652800, 26611200, 133056000, 518918400, 2075673600, 10378368000, 58118860800, 261534873600, 1270312243200, 5928123801600, 29640619008000, 168951528345600, 844757641728000, 4505374089216000, 25342729251840000
Offset: 1

Views

Author

Amarnath Murthy, Mar 21 2003

Keywords

Crossrefs

Formula

a(n) equals smallest integer m>0 such that A086435(m^2)=n-1. - Max Alekseyev, Jul 16 2009

Extensions

More terms from David Garber, Jun 17 2003
More terms from Ray G. Opao, Aug 01 2005
Corrected and extended by R. J. Mathar, Nov 12 2006
More terms from Max Alekseyev, Jun 05, Jul 16 2009
Corrected and extended by Max Alekseyev, Jun 03 2023

A338160 Number of ways to represent n as a product of the greatest number of distinct factors.

Original entry on oeis.org

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

Views

Author

Vladimir Letsko, Oct 14 2020

Keywords

Comments

a(n) = A058060(n) for 1 < n < 60; a(60) = 3, A058060(60) = 1.
a(n) is the number of factorizations of n into A086435(n) distinct factors > 1.
a(n) depends only on the prime signature of n.

Examples

			a(72) = 3 because 72 = 2*3*12 = 2*4*9 = 3*4*6 and 72 cannot be represented as a product of 4 distinct factors each greater than 1 (adding the factor 1 to each product doesn't change anything).
		

Crossrefs

Programs

  • PARI
    a(n)={my(d=divisors(n)); my(F(r,k)=if(r==1, [0,1], my(b=-1,c=0); for(k=2, k, if(r%d[k]==0, my([tb,tc]=self()(r/d[k], k-1)); if(tb>b, b=tb; c=0); if(tb==b, c+=tc))); [b+1, c])); F(n, #d)[2]} \\ Andrew Howroyd, Oct 14 2020

Extensions

More terms from Andrew Howroyd, Oct 14 2020

A375272 The number of factors of n of the form p^Fibonacci(k), where p is a prime and k >= 2, when the factorization is uniquely done using the dual Zeckendorf representation of the exponents in the prime factorization of n.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Aug 09 2024

Keywords

Comments

First differs from A086435 at n = 36. Differs from A266226 at n = 1, 36, ... .
The number of dual-Zeckendorf-infinitary divisors of n (defined in A331109) that are prime powers (A246655).
a(n) depends only on the prime signature of n.
Analogous to A064547 (binary representation) and A318464 (Zeckendorf representation).

Examples

			For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.
For n = 256 = 2^8, the dual Zeckendorf representation of 8 is 1011, i.e., 8 = Fibonacci(2) + Fibonacci(3) + Fibonacci(5) = 1 + 2 + 5. Therefore 256 = 2^(1+2+5) = 2^1 * 2^2 * 2^5, and a(256) = 3.
		

Crossrefs

Programs

  • Mathematica
    toDualZeck[n_] := Module[{s = 0, v = 0, i = 0, f}, While[s < n, s += Fibonacci[i + 2]; v += 2^i; i++]; i--; While[i >= 0, f = Fibonacci[i + 2]; If[s - f >= n, s -= f; v -= 2^i]; i--]; v]; (* A003754, after Rémy Sigrist's PARI code in A112309 *)
    f[p_, e_] := DigitCount[toDualZeck[e], 2, 1]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    todualzeck(n) = {my (s=0, v=0); for (i=0, oo, if (s>=n, forstep (j=i-1, 0, -1, if (s-fibonacci(2+j)>=n, s-=fibonacci(2+j); v-=2^j;);); return (v);); s+=fibonacci(2+i); v+=2^i;);} \\ A003754, Rémy Sigrist's code in A112309
    a(n) = vecsum(apply(x -> hammingweight(todualzeck(x)), factor(n)[, 2]));

Formula

Additive with a(p^e) = A112310(e).
a(n) = log_2(A331109(n)).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} (A112310(k)-A112310(k-1)) * P(k) = 0.18790467121403662496..., and P(s) is the prime zeta function.

A385378 The maximum possible number of distinct factors in the factorization of n into prime powers (A246655).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jun 27 2025

Keywords

Comments

Differs from A376885 and A384422 at n = 32, 64, 96, 128, 160, 192, ... .
Differs from A086435 at n = 36, 100, 144, 180, 196, 225, ... .
Differs from A375272 at n = 128, 384, 640, 896, 1024, 1152, ... .
a(n) depends only on the prime signature of n (A118914).
The indices of records in this sequence are the partial products of the sequence of powers of primes (A000961), i.e., the terms in A024923.
The least index n such that a(n) = k, for k = 0, 1, 2, ..., is A024923(k+1).

Examples

			      n | a(n) | factorization
  ------+------+--------------------------------
      2 |  1   | 2
      6 |  2   | 2 * 3
     24 |  3   | 2 * 3 * 2^2
    120 |  4   | 2 * 3 * 2^2 * 5
    840 |  5   | 2 * 3 * 2^2 * 5 * 7
   6720 |  6   | 2 * 3 * 2^2 * 5 * 7 * 2^3
  60480 |  7   | 2 * 3 * 2^2 * 5 * 7 * 2^3 * 3^2
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Floor[(Sqrt[8*e + 1] - 1)/2]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> (sqrtint(8*x+1)-1)\2 , factor(n)[, 2]));

Formula

Additive with a(p^e) = A003056(e).
a(n) >= A001221(n), with equality if and only if n is cubefree (A004709).
a(n) >= 1 for n >= 2, with equality if and only if n is a prime or a square of a prime (A000430).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761), C = Sum_{k>=2} P(k*(k+1)/2) = 0.19285739770001405035..., and P is the prime zeta function.

A378126 Array read by antidiagonals: T(n, m) is the maximal size of partitions of (n, m) into sums of distinct pairs of nonnegative integers, excluding (0, 0).

Original entry on oeis.org

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

Views

Author

Jimin Park, Nov 17 2024

Keywords

Comments

A378379(n) is the least number x such that T(x, x) >= n, and as the growth rate of A378379(n) is Theta(n^(3/2)), the growth rate of T(n, m) is O((n+m)^(2/3)).

Examples

			Table begins:
  0 1 1 2 2 2 3 3 3 3 ...
  1 2 2 3 3 3 4 4 4 4 ...
  1 2 3 3 4 4 4 5 5 5 ...
  2 3 3 4 4 4 5 5 5 6 ...
  2 3 4 4 5 5 5 6 6 6 ...
  2 3 4 4 5 5 6 6 6 7 ...
  3 4 4 5 5 6 6 6 7 7 ...
  3 4 5 5 6 6 6 7 7 7 ...
  3 4 5 5 6 6 7 7 7 8 ...
  3 4 5 6 6 7 7 7 8 8 ...
  ...
T(9, 5) = 7, as (9, 5) can be expressed as the sum (0, 1) + (0, 2) + (1, 0) + (1, 1) + (2, 0) + (2, 1) + (3, 0), which is the longest for (9, 5).
		

Crossrefs

Maximal size among partitions considered by A054242 and A201377.
The first row is T(n, 0) = A003056(n).

Programs

  • Python
    import functools
    @functools.cache
    def A378126(n: int, m: int, t: tuple[int, int] = (0, 0)) -> int:
      if (n, m) <= t: return 0
      v = 1
      for nn in range(t[0], n//2+1):
        for nm in range(m+1):
          if (nn, nm) <= t: continue
          rn, rm = n-nn, m-nm
          if (rn, rm) <= (nn, nm): continue
          nv = 1 + A378126(rn, rm, (nn, nm))
          if nv > v: v = nv
      return v

Formula

T(n, m) = T(m, n).
T(n, m) >= T(n, 0) + T(0, m).
T(n, m) = A086435(p^n * q^m) for any distinct primes p and q.
T(n, 0) = A003056(n).
T(n, 1) = T(n, 0) + 1.
T(n, 2) = T(n-1, 0) + 2, for n >= 1.
T(n, m) = O((n+m)^(2/3)).

A378379 Minimal x such that there is a partition of (x, x) into sums of distinct pairs of nonnegative integers with size at least n, excluding (0, 0).

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 7, 9, 10, 12, 14, 16, 18, 20, 23, 25, 28, 30, 33, 35, 38, 41, 44, 47, 50, 53, 56, 60, 63, 67, 70, 74, 77, 81, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 125, 129, 134, 138, 143, 147, 152, 156, 161, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220
Offset: 1

Views

Author

Jimin Park, Nov 24 2024

Keywords

Comments

For (n, n), there is at least one maximal partition P that's symmetric: (x, y) in P <=> (y, x) in P. This can be proven by manipulating integer sequences c(i) (i >= 1) such that 0 <= c(i) <= i+1 for all i and Sum_{i > 0} i*c(i) = 2n, which correspond to partitions P of (n, n) with size |P| = Sum_{i > 0} c(i), where c(i) is equal to number of (x, y) in P such that x+y = i.

Examples

			For n = 8, a(n) = 9, as (9, 9) can be expressed as the sum (0, 1) + (0, 2) + (0, 3) + (1, 0) + (2, 0) + (3, 0) + (1, 2) + (2, 1), but the longest sum for (8, 8) has 7 pairs.
		

Crossrefs

Maximal size among partitions considered by A054242 and A201377.
Minimal x such that A378126(x, x) >= n.
Cf. A086435.

Programs

  • Python
    import math
    def A378379(n: int) -> int:
      l = (math.isqrt(1+8*n)-1)//2 # l = A003056(n), min. possible largest pair norm
      r = n - (l-1)*(l+2)//2 # r = n - A000096(l-1), number of pairs with norm l
      return ((l-1)*l*(l+1)//3 + l*r + 1)//2 # ceil((A007290(l+1) + l*r) / 2)

Formula

a(n*(n+3)/2) = n*(n+1)*(n+2)/6.

A083489 Triangle read by rows where the n-th row gives the reverse-lexicographically earliest sequence of n distinct integers > 1 whose product equals the smallest possible square (=A081456(n)).

Original entry on oeis.org

4, 2, 8, 2, 3, 6, 2, 3, 4, 6, 2, 3, 4, 6, 9, 2, 3, 5, 6, 8, 10, 2, 3, 4, 5, 6, 8, 10, 2, 3, 4, 5, 6, 8, 9, 10, 2, 3, 4, 5, 6, 7, 9, 10, 14, 2, 3, 4, 5, 7, 8, 9, 10, 12, 14, 2, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15, 2, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15, 16, 2, 3, 4, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 2
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 03 2003

Keywords

Examples

			4
2 8
2 3 6
2 3 4 6
2 3 4 6 9
2 3 5 6 8 10
...
		

Crossrefs

Formula

Results from the triangle in A081454 by removing first column (equivalently, by removing all 1's from the sequence).

Extensions

Corrected by R. J. Mathar, Jul 18 2007
Corrected and extended by Max Alekseyev, Jul 16 2009

A119249 Indices of Fibonacci numbers in A081952.

Original entry on oeis.org

4, 7, 10, 17, 22, 31, 41, 47, 49, 71, 71, 98, 119, 161, 169, 191, 191, 239, 239, 239, 338, 338, 338, 431, 431, 449, 529, 599, 649, 649, 769, 769, 769, 769, 769, 769, 1009, 1009, 1009, 1409, 1441, 1441, 1441, 1441, 1729, 1889, 1889, 1889, 1889, 2159, 2209, 2209, 2209, 2591, 2642, 2642
Offset: 1

Views

Author

Ryan Propper, Jul 22 2006

Keywords

Examples

			A081952(4) = 1597 and 1597 is the 17th Fibonacci number, so a(4) = 17.
		

Crossrefs

Programs

  • Mathematica
    f[n_, m_, 0] := n == 1; f[n_, m_, d_] := (b = False; l = Select[Divisors[n], #>=m&]; For[i = 1, i <= Length[l], i++, k = l[[i]]; If[f[n / k, k + 1, d - 1], b = True; Break[]]]; b); Do[x = 3; While[ !f[Fibonacci[x] - 1, 2, y], x++ ]; Print[x], {y, 1, 50}]

Formula

A081952(n) = F(a(n)), where F(k) is the k-th Fibonacci number.
a(n) = the smallest k>3 such that A086435(A000045(k)-1)>=n. - Max Alekseyev, Apr 23 2010

Extensions

a(24)-a(26) from Max Alekseyev, Apr 23 2010
a(27)-a(56) from Amiram Eldar, Jun 24 2023

A287549 Total number of unordered factorizations of all positive integers <= n into distinct factors greater than 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 17, 18, 20, 22, 24, 25, 28, 29, 32, 34, 36, 37, 42, 43, 45, 47, 50, 51, 56, 57, 60, 62, 64, 66, 71, 72, 74, 76, 81, 82, 87, 88, 91, 94, 96, 97, 104, 105, 108, 110, 113, 114, 119, 121, 126, 128, 130, 131, 140, 141, 143, 146, 150, 152, 157, 158, 161, 163, 168, 169, 178, 179, 181, 184
Offset: 1

Views

Author

Ilya Gutkovskiy, May 26 2017

Keywords

Comments

Partial sums of A045778.

Examples

			a(6) = 7 because we have [1], [2], [3], [4], [5], [2*3] and [6] (the factorization [2*2] is not permitted because the factor 2 is present twice).
		

Crossrefs

Programs

  • Mathematica
    Accumulate[gd[m_, 1] := 1; gd[1, n_] := 0; gd[1, 1] := 1; gd[0, n_] := 0; gd[m_, n_] := gd[m, n] = Total[gd[# - 1, n/#] & /@ Select[Divisors[n], # <= m &]]; Array[ gd[#, #] &, 75]]

Formula

a(p^k) = a(p^k-1) + A000009(k), where p is a prime.
Showing 1-10 of 10 results.