A048277 -id:A048277 - cp's OEIS frontend

A048278 Positive numbers n such that the numbers binomial(n,k) are squarefree for all k = 0..n.

1, 2, 3, 5, 7, 11, 23

Offset: 1

It has been shown by Granville and Ramaré that the sequence is complete.
These are all the positive integers m that, when m is represented in binary, contain no composites represented in binary as substrings. - Leroy Quet, Oct 30 2008
This is a number-theoretic sequence, so it automatically assumes that n is positive. To quote Granville and Ramaré, "From Theorem 2 it is evident that there are only finitely many rows of Pascal's Triangle in which all of the entries are squarefree. In section 2 we show that this only occurs in rows 1, 2, 3, 5, 7, 11 and 23 (a result proved by Erdős long ago)." - N. J. A. Sloane, Mar 06 2014
See also comment in A249441. - Vladimir Shevelev, Oct 29 2014
This sequence is equivalent to: Positive integers n such that Fibonacci(n+1) divides n!. This comment depends on the finiteness of A019532. - Altug Alkan, Mar 31 2016

			n=11: C[11,k] = 1, 11, 55, 165, 330, 462, ... are all squarefree (or 1).

A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].
H. N. Ramaswamy and R. Siddaramu, On the Stufe, unit Stufe and Pythagoras number of the ring of integers modulo n, Adv. Stud. Contemp. Math. (Kyungshang) 20:3 (2010), pp. 373-388.

Cf. A005117, A046098, A048276, A048277, A048279, A249441.
Cf. A000225, A055010, A249452.
Cf. A019532.

select(n -> andmap(t -> numtheory:-issqrfree(binomial(n,t)),[$1..floor(n/2)]),[$1..100]); # Robert Israel, Oct 29 2014

Do[m = Prime[n]; k = 2; While[k < m/2 + .5 && Union[ Transpose[ FactorInteger[ Binomial[m, k]]] [[2]]] [[ -1]] < 2, k++ ]; If[k >= m/2 + .5, Print[ Prime[n]]], {n, 1, PrimePi[10^6]} ]
Select[Range[10^3], Function[n, AllTrue[Binomial[n, Range@ n], SquareFreeQ]]] (* Michael De Vlieger, Apr 01 2016, Version 10 *)

is(n)=for(k=0,n\2,if(!issquarefree(binomial(n,k)),return(0))); 1 \\ Charles R Greathouse IV, Mar 06 2014

Integers n>0 in set difference between union (A000225, A055010) and A249452. - Vladimir Shevelev, Oct 30 2014

a(n) = A018253(n+1) - 1. - Altug Alkan, Apr 26 2016

Edited by Ralf Stephan, Aug 03 2004

A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 3, 0, 4, 2, 0, 2, 1, 0, 12, 6, 0, 8, 4, 0, 4, 2, 0, 24, 21, 18, 19, 14, 9, 14, 7, 0, 28, 20, 12, 20, 13, 6, 12, 6, 0, 32, 19, 6, 21, 12, 3, 10, 5, 0, 48, 42, 36, 38, 28, 18, 28, 14, 0, 50, 37, 24, 36, 24, 12, 22, 11, 0, 52, 32, 12, 34, 20, 6, 16, 8, 0

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9).

This should have a formula. See for example A062296, A006047 and A048967.

			Row 9 of Pascal's triangle is {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}. The terms 9, 36, and 126 are the only multiples of nine, and each of them occurs two times on that row, thus a(9) = 2*3 = 6.
Row 10 of Pascal's triangle is {1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1}. The terms 45 (= 9*5) and 252 (= 9*28) are the only multiples of nine, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

Antti Karttunen, Table of n, a(n) for n = 0..6561
James G. Huard, Blair K. Spearman and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica (1997), Volume: 78, Issue: 4, page 331-349.

Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047.

Total/@Table[If[Mod[Binomial[n,k],9]==0,1,0],{n,0,80},{k,0,n}] (* Harvey P. Dale, Feb 12 2020 *)

A249733(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%9),c += (if(k<(n/2),2,1)))); return(c); } \\ Unoptimized.
for(n=0, 6561, write("b249733.txt", n, " ", A249733(n)));

import re
from gmpy2 import digits
def A249733(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    return n+1-(((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<Chai Wah Wu, Jul 24 2025

For all n >= 0, the following holds:
a(n) <= A048277(n).
a(n) <= A062296(n).
a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0.
a(n) is odd if and only if n is one of the terms of A249720.

A249442 a(n) is the smallest m such that binomial(n,m) is not squarefree, or a(n)=0, if there is no such m.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 3, 0, 1, 1, 2, 0, 1, 5, 3, 7, 1, 2, 1, 2, 1, 4, 3, 0, 1, 1, 2, 1, 1, 3, 3, 5, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 8, 1, 1, 2, 21, 1, 1, 1, 2, 1, 4, 1, 2, 1, 2, 3, 6, 1, 6, 3, 1, 1, 2, 3, 4, 1, 6, 3, 8, 1, 2, 3, 1, 1, 3, 3, 8, 1, 1, 2, 3, 1, 5, 3

Offset: 0

Antti Karttunen and Vladimir Shevelev, Oct 28 2014

nonn

The sequence gives the position of the first zero on row n (both starting from zero) in the triangular table A103447, and zero if there is no zero on that row. After a(0) = 0, A048278 gives the positions of seven other zeros in the sequence.

Records are 0,1,3,5,7,8,21,... (A249439) in positions 0,4,6,13,15,43,47,... (A249440).

Antti Karttunen, Table of n, a(n) for n = 0..10000

A249439 gives the record values, A249440 the positions where they occur for the first time.
Cf. A005117, A007913, A048277, A048278, A103447, A249716, A249717, A249718.
Differs from A249695 for the first time at n=9.

Table[If[#>n,0,#]&[NestWhile[#+1&,1,SquareFreeQ[Binomial[n,#]]&]],{n,0,100}] (* Peter J. C. Moses, Nov 04 2014 *)

A249442(n) = { for(k=0,n\2,if(0==moebius(binomial(n,k)),return(k))); return(0); }
for(n=0, 10000, write("b249442.txt", n, " ", A249442(n)));
\\ Antti Karttunen, Nov 04 2014

Other identities:

A249716(n) = binomial(n, a(n)). [A249716(n) gives the corresponding minimal nonsquarefree binomial coefficient, or 1 when n is one of the terms of A048278].

More terms from Peter J. C. Moses, Oct 28 2014

A048276 a(n) = number of squarefree numbers among C(n,k), k=0..n.

Original entry on oeis.org

1, 2, 3, 4, 3, 6, 6, 8, 3, 2, 6, 12, 4, 10, 12, 14, 2, 6, 2, 8, 8, 10, 12, 24, 4, 4, 8, 2, 4, 12, 6, 12, 2, 4, 8, 8, 2, 8, 14, 12, 4, 12, 14, 26, 16, 8, 20, 42, 2, 2, 2, 4, 6, 18, 4, 6, 2, 6, 10, 22, 8, 26, 40, 8, 2, 4, 6, 8, 8, 16, 12, 18, 2, 8, 18, 4, 6, 14, 18, 34, 2, 2, 4, 6, 4, 10, 12, 16, 4

Offset: 0

Labos Elemer

nonn

The only odd numbers are at n = 0, 2, 4, and 8. So this sequence is essentially twice A238337. - T. D. Noe, Mar 07 2014

			If n=20, then C(20, k) is squarefree for k = 0,2,4,8,12,16,18,20, that is, for 8 cases of k, so a(20)=8.

T. D. Noe, Table of n, a(n) for n = 0..5000

Cf. A005117, A046098, A048277, A238337.

A048276 := proc(n)
    local a,k ;
    a := 0 ;
    for k from 0 to n do
        if issqrfree(binomial(n,k)) then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A048276(n),n=0..40) ; # R. J. Mathar, Jan 18 2018

Table[Length[Select[Binomial[n, Range[0, n]], SquareFreeQ[#] &]], {n, 0, 100}]

a(n) = sum(k=0, n, issquarefree(binomial(n, k))); \\ Michel Marcus, Dec 19 2013

a(n) = n+1-A048277(n). - R. J. Mathar, Jan 18 2018

A249732 Number of (not necessarily distinct) multiples of 4 on row n of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, 4, 3, 0, 7, 2, 3, 0, 14, 12, 11, 8, 13, 6, 7, 0, 19, 14, 11, 4, 17, 6, 7, 0, 30, 28, 27, 24, 29, 22, 23, 16, 33, 26, 23, 12, 29, 14, 15, 0, 43, 38, 35, 28, 37, 22, 23, 8, 45, 34, 27, 12, 37, 14, 15, 0, 62, 60, 59, 56, 61, 54, 55, 48, 65, 58, 55, 44, 61, 46, 47, 32

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

a(n) = Number of zeros on row n of A034931 (Pascal's triangle reduced modulo 4).

This should have a formula (see A048967).

			Row 9 of Pascal's triangle is: {1,9,36,84,126,126,84,36,9,1}. The terms 36 and 84 are only multiples of four, and both of them occur two times on that row, thus a(9) = 2*2 = 4.
Row 10 of Pascal's triangle is: {1,10,45,120,210,252,210,120,45,10,1}. The terms 120 (= 4*30) and 252 (= 4*63) are only multiples of four, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.

Cf. A007318, A034931, A048277, A048645, A072823, A048967, A062296, A187059, A249733.

A249732(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%4),c += (if(k<(n/2),2,1)))); return(c); } \\ Slow...
for(n=0, 8192, write("b249732.txt", n, " ", A249732(n)));

def A249732(n): return n+1-(2+((n>>1)&~n).bit_count()<>1) # Chai Wah Wu, Jul 24 2025

Other identities:
a(n) <= A048277(n) for all n.
a(n) <= A048967(n) for all n.

A249716 The least nonsquarefree number on row n of Pascal's triangle, or 1 if all the terms on that row are squarefree.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 20, 1, 8, 9, 45, 1, 12, 1287, 364, 6435, 16, 136, 18, 171, 20, 5985, 1540, 1, 24, 25, 325, 27, 28, 3654, 4060, 169911, 32, 528, 5984, 52360, 36, 666, 8436, 82251, 40, 820, 11480, 145008513, 44, 45, 1035, 12551759587422, 48, 49, 50, 1275, 52, 292825, 54, 1485, 56, 1596, 30856, 45057474, 60, 55525372, 37820, 63, 64, 2080

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

After a(0) = 1, A048278 gives the positions of seven other ones in the sequence.

			           Binomial coefficients     First squarefree     a(n)
                 A007318             occurs at index?      =
----------------------------------------------------------------------------
Row 0                1               no squarefrees        1 (by definition)
Row 1              1   1             no squarefrees        1
Row 2            1   2   1           no squarefrees        1
Row 3          1   3   3   1         no squarefrees        1
Row 4        1   4   6   4   1              1              4
Row 5      1   5  10  10   5   1     no squarefrees        1
Row 6    1   6  15  20  15   6   1          3             20

Antti Karttunen, Table of n, a(n) for n = 0..10000

A249717 and A249718 give the smallest and the largest prime whose square divides these numbers.

Cf. also A007318, A005117, A013929, A048277, A048278, A249442.

A249716(n) = { my(b); for(k=0,n\2,if(0==moebius(b=binomial(n,k)),return(b))); return(1); }
for(n=0, 10000, write("b249716.txt", n, " ", A249716(n)));

(define (A249716 n) (A007318tr n (A249442 n)))

a(n) = binomial(n, A249442(n)).

A064460 Number of distinct nonsquarefree entries in n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 0, 3, 4, 3, 0, 5, 2, 2, 1, 8, 6, 9, 6, 7, 6, 6, 0, 11, 11, 10, 13, 13, 9, 13, 10, 16, 15, 14, 14, 18, 15, 13, 14, 19, 15, 15, 9, 15, 19, 14, 3, 24, 24, 25, 24, 24, 18, 26, 25, 28, 26, 25, 19, 27, 18, 12, 28, 32, 31, 31, 30, 31, 27, 30, 27, 36

Offset: 0

Robert G. Wilson v, Oct 02 2001

			a(13) = 2 because C(13,5) = 3^2*11*13 and C(13,6) = 2^2*3*11*13.

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A048277, A064461, A064462.

f[ n_ ] := (c = 0; k = 1; While[ k < n/2 + .5, If[ Union[ Transpose[ FactorInteger[ Binomial[ n, k ] ] ] [ [ 2 ] ] ] [ [ -1 ] ] > 1, c++ ]; k++ ]; c); Table[ f[ n ], {n, 0, 100} ]

a(n) = sum(k=0, n\2, !issquarefree(binomial(n, k))); \\ Michel Marcus, Mar 05 2014

a(n) + A238337(n) = A008619(n). - R. J. Mathar, Jan 18 2018

A048279 Values of n for which no values of C(n,k) except k=0 and k=n are squarefree.

Original entry on oeis.org

0, 1, 9, 16, 18, 27, 32, 36, 48, 49, 50, 56, 64, 72, 80, 81, 96, 98, 99, 100, 108, 112, 121, 126, 128, 135, 136, 144, 147, 148, 153, 162, 169, 171, 175, 176, 180, 192, 196, 198, 200, 216, 225, 243, 244, 245, 248, 250, 252, 256, 264, 272, 276, 288, 289, 294, 300

Offset: 1

Labos Elemer

nonn

			n=9, C(n,k) = (1),9,36,84,126,126,84,36,9,(1); all values include squares.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A048276, A048277, A048278.

f[n_] := (c = 0; k = 1; While[k < n, If[ Union[ Transpose[ FactorInteger[ Binomial[n, k]]] [[2]]] [[ -1]] > 1, c++ ]; k++ ]; c); Select[Range[150], f[ # ] == # - 1 &]
a[ n_] := If[ n < 2, 0, Module[ {c = 1, k = 1}, While[ c < n, If[ {} == Select[ Table[ Binomial[k, i], {i, k - 1}], SquareFreeQ], c++]; k++]; k - 1]]; (* Michael Somos, Mar 07 2014 *)
Join[{0,1},Select[Range[3,300],NoneTrue[Binomial[#,Range[2,#-1]], SquareFreeQ] &]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 29 2019 *)

More terms from Robert G. Wilson v, Oct 02 2001

A064462 First row of Pascal's triangle that has n nonsquarefree entries, or -1 if no such row exists.

Original entry on oeis.org

0, 6, 4, 14, 13, 10, 8, -1, 9, 12, -1, 22, 17, 20, -1, 16, -1, 18, 29, 26, 31, 24, 25, 62, -1, 28, 27, 34, 35, 42, 33, 32, -1, -1, -1, 36, 53, 40, 45, -1, -1, -1, 95, -1, -1, -1, 79, 48, 49, 50, 55, 54, 57, 60, 69, 56, 63, 74, -1, 70, 67, 66, 65, 64, 77, -1

Offset: 0

Robert G. Wilson v, Oct 02 2001

sign

Numbers such that a(n) is -1: 7, 10, 14, 16, 24, 32, 33, 34, 39, 40, 41, 43, ... - Michel Marcus, Mar 05 2014

			a(4) = 13 because C(13,5) = C(13,8) = 3^2*11*13 and C(13,6) = C(13,7) = 2^2*3*11*13.

Cf. A048277, A064461.

f[ n_ ] := (c = 0; k = 1; While[ k < n, If[ Union[ Transpose[ FactorInteger[ Binomial[ n, k ] ] ] [ [ 2 ] ] ] [ [ -1 ] ] > 1, c++ ]; k++ ]; c); Do[ m = 2; While[ f[ m ] != n, m++ ]; Print[ m ], {n, 0, 6} ]

a(n, v) = {for (i=1, #v, if (v[i] == n, return (i-1));); return (-1);} \\ where v is vector A048277; Michel Marcus, Mar 05 2014

Corrected and extended by Michel Marcus, Mar 05 2014

cp's OEIS Frontend

A048278 Positive numbers n such that the numbers binomial(n,k) are squarefree for all k = 0..n.

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A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

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A249442 a(n) is the smallest m such that binomial(n,m) is not squarefree, or a(n)=0, if there is no such m.

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A048276 a(n) = number of squarefree numbers among C(n,k), k=0..n.

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A249732 Number of (not necessarily distinct) multiples of 4 on row n of Pascal's triangle.

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A249716 The least nonsquarefree number on row n of Pascal's triangle, or 1 if all the terms on that row are squarefree.

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A064460 Number of distinct nonsquarefree entries in n-th row of Pascal's triangle.

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