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.

A161840 Number of noncentral divisors of n.

Original entry on oeis.org

0, 0, 0, 2, 0, 2, 0, 2, 2, 2, 0, 4, 0, 2, 2, 4, 0, 4, 0, 4, 2, 2, 0, 6, 2, 2, 2, 4, 0, 6, 0, 4, 2, 2, 2, 8, 0, 2, 2, 6, 0, 6, 0, 4, 4, 2, 0, 8, 2, 4, 2, 4, 0, 6, 2, 6, 2, 2, 0, 10, 0, 2, 4, 6, 2, 6, 0, 4, 2, 6, 0, 10, 0, 2, 4, 4, 2, 6, 0, 8, 4, 2, 0, 10, 2, 2, 2, 6, 0, 10, 2, 4, 2, 2, 2, 10, 0, 4, 4, 8
Offset: 1

Views

Author

Omar E. Pol, Jun 21 2009

Keywords

Comments

Noncentral divisors in the following sense: if we sort the divisors of n in natural order, there is one "central", median divisor if the number of divisors tau(n) = A000005(n) is odd, and there are two "central" divisors if tau(n) is even. a(n) is the number of divisors not counting the median or two central divisors.

Examples

			The divisors of 4 are 1, 2, 4 so the noncentral divisors of 4 are 1, 4 because its central divisor is 2.
The divisors of 12 are 1, 2, 3, 4, 6, 12 so the noncentral divisors of 12 are 1, 2, 6, 12 because its central divisors  are 3, 4.
		

Crossrefs

Programs

Formula

a(n) = tau(n)-2 + (tau(n) mod 2), tau = A000005.
a(n) = A000005(n) - A049240(n) - 1.
a(n) = A000005(n) + A010052(n) - 2.
a(n) = A000005(n) - A169695(n).
For n >= 2, a(n) = A200213(n) + 2*A010052(n). - Antti Karttunen, Jul 07 2017
a(n) = 2*A072670(n-1). - Omar E. Pol, Jul 08 2017
Sum_{k=1..n} a(k) ~ n * (log(n) + 2*gamma - 3), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 14 2024

Extensions

More terms from R. J. Mathar, Jul 04 2009

A183002 a(n) is the total number of noncentral divisors in all positive integers <= n.

Original entry on oeis.org

0, 0, 0, 2, 2, 4, 4, 6, 8, 10, 10, 14, 14, 16, 18, 22, 22, 26, 26, 30, 32, 34, 34, 40, 42, 44, 46, 50, 50, 56, 56, 60, 62, 64, 66, 74, 74, 76, 78, 84, 84, 90, 90, 94, 98, 100, 100, 108, 110, 114, 116, 120, 120, 126, 128, 134, 136, 138, 138, 148, 148, 150, 154
Offset: 1

Views

Author

Omar E. Pol, Jan 27 2011

Keywords

Comments

The original name was: Partial sums of A161840.

Crossrefs

Programs

  • Maple
    with(numtheory):
    b:= n-> (x-> x-2+(x mod 2))(tau(n)):
    a:= proc(n) option remember; b(n) +`if`(n=1, 0, a(n-1)) end:
    seq(a(n), n=1..100);
  • Mathematica
    Accumulate[Table[d = DivisorSigma[0, n]; If[OddQ[d], d - 1, d - 2], {n, 100}]]
  • PARI
    lista(nmax) = {my(s = 0, d); for(n = 1, nmax, d = numdiv(n); s += (d + d%2 - 2); print1(s, ", ")); } \\ Amiram Eldar, Jan 19 2024

Formula

a(n) = Sum_{k=1..n} (tau(k)-2 + (tau(k) mod 2)), tau = A000005.
a(n) ~ n * (log(n) + 2*gamma - 3), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 19 2024

Extensions

New name from Omar E. Pol, Jan 04 2022

A350497 Sum of the largest parts in all the partitions of n into 3 parts whose largest part is greater than or equal to the product of the other two.

Original entry on oeis.org

0, 0, 0, 1, 2, 5, 7, 12, 19, 27, 32, 48, 55, 68, 84, 109, 120, 149, 162, 196, 223, 249, 266, 323, 359, 392, 430, 484, 509, 586, 614, 678, 728, 775, 831, 952, 989, 1044, 1106, 1219, 1261, 1379, 1424, 1520, 1627, 1698, 1748, 1919, 2009, 2124, 2213, 2332, 2392, 2552, 2655, 2827
Offset: 0

Views

Author

Wesley Ivan Hurt, Jan 03 2022

Keywords

Examples

			a(7) = 12 since we have 7 = 1+1+5 = 1+2+4 = 1+3+3, and the sum of the largest parts in each partition is 5+4+3 = 12. The partition 2+2+3 is not included since 2*2 > 3.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(n - i - k) Sign[Floor[(n - i - k)/(i*k)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}]
    Table[Total[Select[IntegerPartitions[n,{3}],#[[1]]>=Times@@Rest[#]&][[All,1]]],{n,0,60}] (* Harvey P. Dale, Aug 22 2022 *)
  • PARI
    first(n) = my(res=vector(n, i, [0, 0])); for(i = 1, n\2, for(j = i, n\i, c = i + j + i * j; if(c <= n, res[c][1]++; res[c][2] += i*j))); forstep(i = n, 1, -1, for(j = i + 1, n, res[j][2] += ((j-i) * res[i][1] + res[i][2]))); concat(0, vector(#res, i, res[i][2])) \\ David A. Corneth, Jan 07 2022

Formula

a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} sign(floor((n-i-k)/(i*k))) * (n-i-k).

Extensions

a(0) = 0 prepended by David A. Corneth, Jan 09 2022

A350534 Sum of the largest parts of the partitions of n into 3 parts whose largest part is equal to the product of the other two.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 3, 4, 4, 0, 11, 0, 6, 8, 16, 0, 18, 0, 21, 12, 10, 0, 40, 16, 12, 16, 31, 0, 52, 0, 36, 20, 16, 24, 88, 0, 18, 24, 74, 0, 76, 0, 51, 60, 22, 0, 121, 36, 60, 32, 61, 0, 100, 40, 108, 36, 28, 0, 198, 0, 30, 88, 125, 48, 124, 0, 81, 44, 140, 0, 243, 0, 36, 104
Offset: 0

Views

Author

Wesley Ivan Hurt, Jan 04 2022

Keywords

Examples

			a(13) = 6 since we have 13 = 1+6+6, whose largest part is 6. Partitions not counted: 1+1+11, 1+2+10, 1+3+9, 1+4+8, 1+5+7, 2+2+9, 2+3+8, 2+4+7, 2+5+6, 3+3+7, 3+4+6, 3+5+5, 4+4+5.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(n - i - k) KroneckerDelta[(n - i - k), (i*k)], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 0, 100}]
  • PARI
    first(n) = {my(res = vector(n)); for(i = 1, n \ 2, for(j = i, n\i, c = i + j + i*j; if(c <= n, res[c] += i*j))); concat(0, res)} \\ David A. Corneth, Jan 07 2022

Formula

a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} [n-i-k = i*k] * (n-i-k), where [ ] is the Iverson bracket.
Showing 1-4 of 4 results.