Matthew Conroy - cp's OEIS frontend

A381708 a(n) is the smallest nonnegative integer k such that sigma_k(n) > sigma_k(j) for all 1 <= j < n.

0, 0, 1, 0, 2, 0, 3, 1, 2, 1, 3, 0, 3, 2, 2, 1, 3, 1, 3, 1, 3, 2, 3, 0, 4, 3, 3, 2, 4, 1, 4, 2, 4, 2, 4, 0, 4, 3, 4, 2, 4, 1, 4, 2, 4, 2, 4, 0, 4, 3, 4, 2, 4, 2, 4, 2, 4, 3, 4, 0, 5, 3, 4, 2, 5, 2, 5, 3, 4, 2, 5, 1, 5, 3, 4, 3, 5, 2, 5, 2, 5, 3, 5, 1, 5, 3, 5, 3, 5, 1, 5, 3, 5, 3, 5, 1, 5, 3, 5, 2, 5, 2, 5, 3, 5, 3, 5, 1, 5

Offset: 1

Matthew Conroy, Mar 04 2025

nonn

sigma_k(n) is the sum of the k-th powers of the divisors of n.

a(n) exists since one can prove that for k > n*(log 2 + 1/2 log(n-1)), sigma_k sets a record at n.

			For n = 1, k = 0 is enough so a(1) = 0.
For n = 2, k = 0 works since sigma_0(2) = 2 > 1 = sigma_0(1) so a(2) = 0.
For n = 3, sigma_0(3) = 2 = sigma_0(2), but sigma_1(3) = 1^1+3^1 = 4 > 3 = sigma_1(2) > 1 = sigma_1(1) so a(3) = 1.
For n = 4, sigma_0(4) = 1^0+2^0+4^0 = 3 > 2 = sigma_0(3) = sigma_0(2) > 1 = sigma_0(1) so a(4) = 0.
For n = 5, sigma_0(5) = 2 = sigma_0(2) and sigma_1(5) = 6 < sigma_1(4) = 7 but sigma_2(5) = 26 > sigma_2(4) > sigma_2(3) > sigma_2(2) > sigma_2(1) so a(5) = 2.

Cf. A000203, A001157, A002093, A002182, A098475, A109974, A193988.

check(n,k) =  my(m=0);for(i=1,n-1, my(s=sigma(i,k)); if(s>m,m=s)); if(sigma(n,k)>m,return(1),return(0));
a(n) = my(ii=0); while(!check(n, ii), ii++);  ii;

a(n) = 0 precisely when n is highly composite number A002182.
a(n) = 1 precisely when n is highly abundant A002093 and not highly composite.
a(n) = 2 precisely when n is in A193988 and is not highly composite and is not highly abundant.
a(n) <= m if n < A098475(m). Empirically, it appears that a(A098475(m)) = m+1. - Pontus von Brömssen, Mar 16 2025

A288677 Every element of Z/nZ can be expressed as a sum of no more than a(n) squares.

Original entry on oeis.org

1, 2, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 4, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 4, 3, 2, 2, 3, 2, 3, 2, 4, 2, 2, 2, 3, 2, 2, 3, 4, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 4, 2, 3, 3, 3

Offset: 2

Matthew Conroy, Jun 13 2017

nonn

			0^2 = 0 and 1^2 = 1 mod 2, so each element of Z/2Z is a square, so a(2)=1;
0^2 = 0, 1^2 = 2^2 = 1 mod 3, so 2 = 1^2 + 1^2 requires two squares to sum to 2, so a(3)=2.

Matthew Conroy, Table of n, a(n) for n = 2..10000
Charles Small, Waring's problem mod n, Amer. Math. Monthly 84 (1977), no. 1, 12--25.

Cf. A287286.

a[n_] := Which[n == 2, 1, Mod[n, 4] != 0 && AllTrue[Select[Divisors[n] // Sqrt, IntegerQ], Mod[#, 4] == 1&], 2, Mod[n, 8] != 0, 3, True, 4];
Table[a[n], {n, 2, 140}] (* Jean-François Alcover, Jun 13 2017 *)

c(n) = A=factor(n);ok=1;for(i=1,matsize(A)[1],if(A[i,1]%4==3&&A[i,2]>1,ok=0));return(ok); wn(n) = if(n==2,1,if(n%4>0&&c(n)==1,2,if(n%8>0,3,4))); for(ii=2,140,print1(wn(ii),","))

From Small's paper, theorem 3.1: a(n)=1 if n=2; else a(n)=2 if n != 0 mod 4 and p^2|n implies p=1 mod 4; else a(n)=3 if n!=0 mod 8; else a(n)=4.

A261752 Minimum number of knights on an n X n chessboard such that every square is attacked.

Original entry on oeis.org

6, 7, 8, 10, 14, 18, 22, 25, 28, 32, 36, 43, 48, 54, 58, 66, 70, 78, 84, 91, 98, 107, 112, 123, 128, 139, 146, 156, 164

Offset: 4

Matthew Conroy, Aug 31 2015

Total domination number of n X n knight graph.
Distinct from A006075 since here all squares must be attacked, whereas, in A006075, all squares are either attacked or occupied.
a(34) = 182, a(36) = 202, a(38) = 224. - Andy Huchala, Jun 04 2021

			An example for the 4 X 4 case:
  ....
  .NNN
  .N..
  NN..
and for the 5 x 5 case:
  .....
  ..N..
  .NN..
  NNN..
  N....

Matthew Conroy, Examples of minimum knight arrangements, n = 4 through n = 14
Andy Huchala, Python program
Giovanni Resta, Examples of minimum knight arrangements, from n = 15 to n = 18
Andy Huchala, Examples of minimum knight arrangements, from n = 25 to n = 34
Eric Weisstein's World of Mathematics, Knight Graph
Eric Weisstein's World of Mathematics, Total Domination Number

Cf. A006075.

a(15)-a(18) from Giovanni Resta, Aug 31 2015
a(19)-a(26) from Andy Huchala, Oct 16 2017
a(27)-a(30) from Andy Huchala, Oct 18 2017
a(31)-a(32) from Andy Huchala, Jun 04 2021

A178830 Number of distinct partitions of {1,...,n} into two nonempty sets P and S with the product of the elements of P equal to the sum of elements in S.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 1, 1, 3, 1, 2, 1, 3, 3, 3, 3, 1, 4, 4, 3, 2, 2, 4, 3, 5, 3, 2, 4, 5, 4, 5, 6, 1, 4, 2, 5, 4, 7, 4, 4, 3, 3, 6, 14, 3, 4, 10, 6, 3, 6, 4, 4, 4, 8, 7, 6, 8, 7, 10, 5, 11, 8, 5, 11, 4, 7, 7, 5, 8, 12, 5, 6, 9, 8, 11, 8, 5, 8, 9, 8, 10, 8, 12, 7, 11, 8, 7, 7, 8, 6, 7, 8, 11, 8, 16, 9, 12, 13, 8

Offset: 1

Matthew Conroy, Dec 31 2010

That the terms are nonzero for n>4 is shown by the fact that for n odd, P={1,(n-1)/2,n-1} works, and for n even, P={1,(n-2)/2,n} works.

a(A207852(n)) = n and a(m) <> n for m < A207852(n). - Reinhard Zumkeller, Feb 21 2012

			{1,2,3} can be partitioned into P={3} and S={1,2} with 3=1+2.  This is the only such partition, so a(3)=1.
{1,2,3,4,5} can be partitioned into P={1,2,4} and S={3,5} with 1*2*4=3+5.  This is the only such partition, so a(5)=1.
{1,...,10} can be partitioned into subsets P={1,2,3,7} and S={4,5,6,8,9,10} since 1*2*3*7=4+5+6+8+9+10. P can also be taken as P={6,7} or P={1,4,10}, and so there are 3 distinct ways to partition {1,...,10}, and so a(10)=3.

Problem 2826, Crux Mathematicorum, May 2004, 178-179

Alois P. Heinz, Table of n, a(n) for n = 1..1800, (first 300 terms from Reinhard Zumkeller)

a178830 1 = 0
a178830 n = z [1..n] (n * (n + 1) `div` 2) 1 where
   z []     s p             = fromEnum (s == p)
   z (x:xs) s p | s > p     = z xs (s - x) (p * x) + z xs s p
                | otherwise = fromEnum (s == p)
-- Reinhard Zumkeller, Feb 21 2012

b:= proc(n, s, p)
      `if`(s=p, 1, `if`(n<1, 0, b(n-1, s, p)+
      `if`(s-n `if`(n=1, 0, b(n, n*(n+1)/2, 1)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 07 2012

b[, s, s_] = 1; b[n_ /; n < 1, , ] = 0; b[n_, s_, p_] := b[n, s, p] = b[n-1, s, p] + If[s-n < p*n, 0, b[n-1, s-n, p*n]]; a[1] = 0; a[n_] := a[n] = b[n, n*(n+1)/2, 1]; Table[Print[a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Aug 16 2013, after Alois P. Heinz *)

maxprodset(n)=ii=1;while(ii!+ii*(ii+1)/2

A114061 Numbers n such that n = (product of digits of n) * (sum of digits of n) in some base.

Original entry on oeis.org

0, 1, 6, 12, 16, 20, 30, 42, 48, 54, 56, 72, 90, 96, 110, 128, 132, 135, 144, 156, 160, 162, 176, 182, 210, 231, 240, 250, 272, 300, 306, 324, 336, 342, 380, 384, 420, 432, 448, 455, 462, 480, 495, 504, 506, 540, 552, 576, 600, 624, 650, 663, 686, 702, 720, 750

Offset: 1

Matthew Conroy, Feb 02 2006

This sequence is infinite since b^2+b is in the sequence for all b>1: in base b, b^2+b has digits {1,b} and (1*b)*(1+b)=b^2+b.

			12 is in the sequence since in base 9, 12 has digits {1,3} and (1*3)*(1+3)=12.

Cf. A038369.

A064693 Number of connected components remaining when decimal expansion of the number n is cut from a piece of paper.

Original entry on oeis.org

2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 2, 2, 2, 3, 2, 3, 2, 4, 3, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 2, 2, 2, 3, 2, 3, 2, 4, 3, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 4, 3, 3, 3, 4, 3, 4, 3, 5, 4, 3, 2, 2, 2, 3, 2, 3, 2, 4, 3, 3, 2, 2, 2, 3

Offset: 0

Matthew Conroy, Oct 11 2001

			We assume 1,2,3,5 have no hole; 0,4,6,9 have 1 hole; 8 has two holes. So cutting 8 from a piece of paper creates three connected components: one for each hole and one for the remainder of the paper. Hence a(8)=3.

Cf. A064531. Equals A064692 + 1.

A060324 a(n) is the minimal prime q such that n*(q+1)-1 is prime, that is, the smallest prime q so that n = (p+1)/(q+1) with p prime; or a(n) = -1 if no such q exists.

Original entry on oeis.org

2, 2, 3, 2, 3, 2, 5, 2, 5, 2, 3, 3, 7, 2, 3, 2, 3, 2, 5, 2, 3, 5, 5, 2, 5, 3, 3, 2, 5, 2, 13, 3, 3, 2, 3, 2, 11, 2, 5, 5, 3, 3, 5, 2, 3, 2, 5, 3, 5, 2, 19, 5, 3, 7, 7, 2, 3, 2, 5, 2, 7, 11, 3, 2, 5, 2, 5, 3, 11, 5, 3, 5, 13, 5, 5, 2, 3, 2, 7, 2, 7, 5, 3, 2, 5, 2, 3, 2, 17, 2, 7, 3, 5, 2, 3, 3, 11, 2, 5, 5

Offset: 1

Matthew Conroy, Mar 29 2001

A conjecture of Schinzel, if true, would imply that such a q always exists.

			1 = (2+1)/(2+1), so the first term is 2; 3(2+1) - 1 = 8 which is not prime, yet 3(3+1) - 1 = 11 is prime (3 = (11+1)/(3+1)) so the 3rd term is 3.

T. D. Noe, Table of n, a(n) for n = 1..10000
Matthew M. Conroy, A sequence related to a conjecture of Schinzel, J. Integ. Seqs. Vol. 4 (2001), #01.1.7.
Peter Luschny, Schinzel-Sierpinski conjecture and Calkin-Wilf tree.

Cf. A060424. Values of p are given in A062251.

Cf. A000040, A010051.

a060324 n = head [q | q <- a000040_list, a010051' (n * (q + 1) - 1) == 1]
-- Reinhard Zumkeller, Aug 28 2014

a:= proc(n) local q;
       q:= 2;
       while not isprime(n*(q+1)-1) do
          q:= nextprime(q);
       od; q
    end:
seq(a(n), n=1..300); # Alois P. Heinz, Feb 11 2011

a[n_] := (q = 2; While[!PrimeQ[n*(q + 1) - 1], q = NextPrime[q]]; q); a /@ Range[100] (* Jean-François Alcover, Jul 20 2011, after Maple prog. *)

a(n) = {my(q=2); while (!isprime(n*(q+1)-1), q = nextprime(q+1)); q;} \\ Michel Marcus, Nov 20 2017

a(n) = (A062251(n)+1) / n - 1. - Reinhard Zumkeller, Aug 28 2014

A060424 Record-setting n's for the function q(n), the minimum prime q such that n(q+1)-1 is prime p (i.e., q(n) > q(j) for all 0 < j < n).

Original entry on oeis.org

1, 3, 7, 13, 31, 51, 101, 146, 311, 1332, 2213, 6089, 10382, 11333, 32003, 83633, 143822, 176192, 246314, 386237, 450644, 1198748, 2302457, 5513867, 9108629, 11814707, 16881479, 18786623, 24911213, 28836722, 34257764, 196457309

Offset: 1

Matthew Conroy, Apr 10 2001

nonn

			a(3)=7, since q(7)=5 and q(j) < 5 for 0 < j < 7.

Amiram Eldar, Table of n, a(n) for n = 1..39
Matthew M. Conroy, A sequence related to a conjecture of Schinzel, J. Integ. Seqs. Vol. 4 (2001), #01.1.7.

Cf. A060324. See A062252 and A062256 for the corresponding values of q and p.

q[n_] := Module[{p = 2}, While[! PrimeQ[n*(p+1)-1], p = NextPrime[p]]; p]; record = 0; a[0] = 0; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[q[k] > record, record = q[k]; Print[k]; Return[k]]]; Table[a[n], {n, 1, 32}] (* Jean-François Alcover, Nov 18 2013 *)

A066419 Numbers k such that k! is not divisible by the sum of the decimal digits of k!.

Original entry on oeis.org

432, 444, 453, 458, 474, 476, 485, 489, 498, 507, 509, 532, 539, 541, 548, 550, 552, 554, 555, 556, 560, 565, 567, 576, 593, 597, 603, 608, 609, 610, 611, 612, 613, 624, 630, 632, 634, 640, 645, 657, 663, 665, 683, 685, 686, 692, 698, 703, 706, 708, 714

Offset: 1

Matthew Conroy, Dec 25 2001

			The sum of the decimal digits of 5! is 1+2+0=3 and 3 divides 120, so 5 is not in the sequence.
The sum of the decimal digits of 432! is 3897 = (9)(433) and 3897 does not divide 432!, so 432 is in the sequence.

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Matthew Conroy)
Dimitri Zucker, Factorial Fact Frenzy (!), Combo Class Youtube video (2022).

Cf. A004152 (sum of digits of n!).

Select[Range[1000], !Divisible[Factorial[#],Total[IntegerDigits[Factorial[#]]]] &], (* Tanya Khovanova, Jun 13 2021 *)

isA066419(n) = (Mod(n!, sumdigits(n!)) != 0) \\ Jianing Song, Aug 26 2024

from math import factorial
def sd(n): return sum(map(int, str(n)))
def ok(f): return f%sd(f) != 0
print([n for n in range(1, 715) if ok(factorial(n))]) # Michael S. Branicky, Jun 13 2021

A064692 Total number of holes in decimal expansion of the number n, assuming 4 has 1 hole.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 0, 0, 0, 1, 0, 1, 0, 2, 1, 3, 2, 2, 2, 3, 2, 3, 2, 4, 3, 2, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 1, 1, 2

Offset: 0

Matthew Conroy, Oct 11 2001

			We assume decimal digits 1,2,3,5,7 have no hole; 0,4,6,9 have one hole each; 8 has two holes. So a(148)=3.

Indranil Ghosh, Table of n, a(n) for n = 0..50000

Cf. A001742, A064532, A064693, A206439, A064692, A208568.

Table[DigitCount[x].{0, 0, 0, 1, 0, 1, 0, 2, 1, 1}, {x, 0, 104}] (* Zak Seidov, Jul 25 2015 *)

def A064692(n):
    x=str(n)
    return x.count("0")+x.count("4")+x.count("6")+x.count("8")*2+x.count("9") # Indranil Ghosh, Feb 02 2017

a(A001742(n)) = 0. - Michel Marcus, Jul 25 2015

cp's OEIS Frontend

User: Matthew Conroy

A381708 a(n) is the smallest nonnegative integer k such that sigma_k(n) > sigma_k(j) for all 1 <= j < n.

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A288677 Every element of Z/nZ can be expressed as a sum of no more than a(n) squares.

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A261752 Minimum number of knights on an n X n chessboard such that every square is attacked.

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A178830 Number of distinct partitions of {1,...,n} into two nonempty sets P and S with the product of the elements of P equal to the sum of elements in S.

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A114061 Numbers n such that n = (product of digits of n) * (sum of digits of n) in some base.

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A064693 Number of connected components remaining when decimal expansion of the number n is cut from a piece of paper.

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A060324 a(n) is the minimal prime q such that n*(q+1)-1 is prime, that is, the smallest prime q so that n = (p+1)/(q+1) with p prime; or a(n) = -1 if no such q exists.

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Maple

Mathematica

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A060424 Record-setting n's for the function q(n), the minimum prime q such that n(q+1)-1 is prime p (i.e., q(n) > q(j) for all 0 < j < n).

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