A249723 -id:A249723 - cp's OEIS frontend

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

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.

A249441 a(n) is the smallest prime whose square divides at least one entry in the n-th row of Pascal's triangle, or 0 if there is no such prime.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 2, 0, 2, 2, 2, 0, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Vladimir Shevelev, Oct 28 2014

a(n) = 3 for 15, 31, 47, 63, 95, 127, 191, 255, 383, 511, 767, 1023, 1535, 2047, 3071, etc.
The above values all occur in A249723 and from 31 onward seem to be given by A052955(n>=8). (Cf. also A249714 & A249715). - Antti Karttunen, Nov 04 2014
Using the Kummer theorem on carries, one can prove that, if a(n)>3 or 0, then n>23 takes the form of either 1...1 or 101...1 in base 2 and simultaneously 212...2 in base 3. However, it is easy to see that this leads to a contradiction. Thus there are no terms greater than 3 and only 8 zeros, i.e., there are only 8 rows in Pascal's triangle that contain all squarefree numbers. It turns out that the latter result has been known for a long time (see A048278).

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
E. E. Kummer, Über die Ergänzungssätze zu den allgemeinen Reciprocitätsgesetzen, J. Reine Angew. Math. 44 (1852), 93-146.
Mihai Prunescu, Sign-reductions, p-adic valuations, binomial coefficients modulo p^k and triangular symmetries. Preprint 2013.

Cf. A005117, A007913, A048278, A052955, A249695, A249714, A249715, A249723.

a_list := proc(len) local s; s := proc(L,p) local n; seq(max(op(map(b-> padic[ordp](b,p),{seq(binomial(n,k),k=0..n)}))),n=0..L); map(k-> `if`(k<2,0,p),[%]) end: zip((x,y)-> `if`(x=0,y,x),s(len,2),s(len,3)) end: a_list(86); # Peter Luschny, Nov 01 2014
# alternative
A249441 := proc(n)
    local p,wrks,bi,k;
    if n in [0,1,2,3,5,7,11,23] then
        return 0 ;
    end if;
    p :=2 ;
    while true do
        wrks := false;
        bi := 1 ;
        for k from 0 to n do
            if modp(bi,p^2) = 0 then
                wrks := true;
                break;
            end if;
            bi := bi*(n-k)/(1+k) ;
        end do:
        if wrks then
            return p;
        end if;
        p := nextprime(p) ;
    end do:
end proc: # R. J. Mathar, Nov 04 2014

row[n_] := Table[Binomial[n, k], {k, 1, (n-Mod[n, 2])/2}];
a[n_] := If[MemberQ[{0, 1, 2, 3, 5, 7, 11, 23}, n], 0, For[p = 2, True, p = NextPrime[p], If[AnyTrue[row[n], Divisible[#, p^2]&], Return[p]]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jul 30 2018 *)

a(n) = my(o=0); for(k=1,n\2, o+=valuation((n-k+1)/k, 2); if(o>1, return(2))); if(n<24 && n!=15, 0, 3) \\ Charles R Greathouse IV, Nov 03 2014

A249441(n) = { forprime(p=2,3,for(k=0,n\2,if((0==(binomial(n,k)%(p*p))),return(p)))); return(0); } \\ This is more straightforward, but a slower implementation - Antti Karttunen, Nov 03 2014

a(n)=if((n+1)>>valuation(n+1,2)<5, if(n<24 && setsearch([1,2,3,5,7,11,23],n), 0, 3), 2) \\ Charles R Greathouse IV, Nov 06 2014

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

A249695 a(n)=0, if A249441(n)=0; otherwise, a(n) is the smallest i such that A249441(n)^2 divides binomial(n,i).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 3, 0, 1, 2, 3, 0, 1, 6, 3, 7, 1, 2, 3, 4, 1, 6, 3, 0, 1, 2, 3, 12, 1, 6, 3, 5, 1, 2, 3, 4, 1, 6, 3, 8, 1, 2, 3, 12, 1, 6, 3, 21, 1, 2, 3, 4, 1, 6, 3, 24, 1, 2, 3, 12, 1, 6, 3, 1, 1, 2, 3, 4, 1, 6, 3, 8, 1, 2, 3, 12, 1, 6, 3, 16, 1, 2, 3, 4, 1, 6, 3

Offset: 0

Vladimir Shevelev, Nov 04 2014

nonn

After a(0) = 0, A048278 gives the positions of seven other zeros in the sequence. - Antti Karttunen, Nov 04 2014

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

A249714 and A249715 give the record values and their positions.
Cf. A048278, A249441, A249722, A249723, A249726.
Differs from A249442 for the first time at n=9.

A249695 := proc(n)
    a41n := A249441(n) ;
    if a41n = 0 then
        return 0;
    end if;
    bi := 1;
    for i from 0 do
        if modp(bi,a41n^2)= 0 then
            return i;
        end if;
        bi := bi*(n-i)/(1+i) ;
    end do:
end proc: # R. J. Mathar, Nov 04 2014

bb[n_] := Table[Binomial[n, k], {k, 1, (n - Mod[n, 2])/2}];
a41[n_] := If[MemberQ[{0, 1, 2, 3, 5, 7, 11, 23}, n], 0, For[p = 2, True, p = NextPrime[p], If[AnyTrue[bb[n], Divisible[#, p^2]&], Return[p]]]];
a[n_] := If[(a41n = a41[n]) == 0, 0, For[i = 1, True, i++, If[Divisible[ Binomial[n, i], a41n^2], Return[i]]]];
a /@ Range[0, 100] (* Jean-François Alcover, Mar 27 2020 *)

A249695(n) = { forprime(p=2,3,for(k=0,floor(n/2),if((0==(binomial(n,k)%(p*p))),return(k)))); return(0); } \\ Straightforward and unoptimized version. But fast enough for 10000 terms.
A249695(n) = { for(p=2,3, my(o=0); for(k=1, n\2, o+=valuation((n-k+1)/k, p); if(o>1, return(k)))); return(0); } \\ This version is based on Charles R Greathouse IV's code for A249441.
for(n=0, 10000, write("b249695.txt", n, " ", A249695(n)));
\\ Antti Karttunen, Nov 04 2014

A249722 Numbers n such that there is a multiple of 4 on row n of Pascal's triangle with property that all multiples of 9 on the same row (if they exist) are larger than it.

Original entry on oeis.org

4, 6, 8, 12, 14, 16, 17, 20, 22, 24, 25, 26, 28, 30, 32, 33, 34, 35, 38, 40, 41, 42, 44, 48, 49, 50, 51, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 74, 76, 77, 78, 80, 84, 86, 88, 89, 92, 94, 96, 97, 98, 100, 101, 102, 104, 105, 106, 107, 112, 113, 114, 115, 116, 120, 121, 122, 124, 125

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

All n such that on row n of A034931 (Pascal's triangle reduced modulo 4) there is at least one zero and the distance from the edge to the nearest zero is shorter than the distance from the edge to the nearest zero on row n of A095143 (Pascal's triangle reduced modulo 9), the latter distance taken to be infinite if there are no zeros on that row in the latter triangle.

			Row 4 of Pascal's triangle (A007318) is {1,4,6,4,1}. The least multiple of 4 occurs as C(4,1) = 4, and there are no multiples of 9 present, thus 4 is included among the terms.
Row 12 of Pascal's triangle is {1,12,66,220,495,792,924,792,495,220,66,12,1}. The least multiple of 4 occurs as C(12,1) = 12, which is less than the least multiple of 9 present at C(12,4) = 495 = 9*55, thus 12 is included among the terms.

Antti Karttunen, Table of n, a(n) for n = 1..10000

A subsequence of A249724.
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A095143, A048645, A051382.

A249722list(upto_n) = { my(i=0, n=0); while(i

A249724 Numbers k such that on row k of Pascal's triangle there is no multiple of 9 which would be less than any (potential) multiple of 4 on the same row.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 14, 16, 17, 20, 22, 23, 24, 25, 26, 28, 30, 32, 33, 34, 35, 36, 38, 40, 41, 42, 44, 48, 49, 50, 51, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 80, 84, 86, 88, 89, 92, 94, 96, 97, 98, 100, 101, 102, 104, 105, 106, 107, 108, 110, 112, 113, 114, 115, 116, 120, 121

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

Disjoint union of {0} and the following sequences: A048278 (gives 7 other cases where there are neither multiples of 4 nor 9 on row k), A249722 (rows where a multiple of 4 is found before a multiple of 9), A249726 (cases where the least term on row k which is a multiple of 4 is also a multiple of 9, and vice versa, i.e., such a term a multiple of 36).

If A249717(k) < 3 then k is included in this sequence. This is a sufficient but not necessary condition, e.g., A249717(25) = 5, but 25 is also included in this sequence.

Antti Karttunen, Table of n, a(n) for n = 1..23590

Complement: A249723.

Cf. A007318, A052955, A249441, A249695, A249717, A048278, A048645, A051382, A249722, A249726.

A249724list(upto_n) = { my(i=0, n=0, dont_print=0); while(i

A249726 Numbers n such that there is a multiple of 36 on row n of Pascal's triangle with property that it is also the least multiple of 4 and the least multiple of 9 on the same row.

Original entry on oeis.org

36, 72, 73, 108, 110, 144, 145, 147, 180, 216, 217, 218, 221, 252, 288, 289, 291, 295, 324, 326, 360, 361, 396, 432, 433, 434, 435, 437, 443, 468, 504, 505, 540, 542, 576, 577, 579, 583, 612, 648, 649, 650, 653, 684, 720, 721, 723, 756, 758, 792, 793, 828, 864, 865, 866, 867, 869, 871, 875, 887, 900, 936, 937, 972, 974, 1008, 1009, 1011, 1044, 1080

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

All n such that both on row n of A034931 (Pascal's triangle reduced modulo 4) and on row n of A095143 (Pascal's triangle reduced modulo 9) there is at least one zero and the distance from the edge to the nearest zero is same on both rows.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Subsequence of A249724.
A044102 is a subsequence (after zero).
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A095143, A048645, A051382.

A249726list(upto_n) = { my(i=0, n=0); while(i

A249731 Number of multiples of 4 on row n of Pascal's triangle minus the number of multiples of 9 on the same row: a(n) = A249732(n) - A249733(n).

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, -2, 0, 0, 3, 0, 3, -2, 13, 12, -1, 2, 13, -2, 3, 0, 15, 12, 11, -20, -4, -12, -12, -14, 21, 14, 20, 24, 1, 2, 11, -4, 20, 20, 11, 6, 29, -18, -4, -6, 22, 26, 32, 18, 32, 22, -25, -34, 9, -4, -1, -6, 9, 0, 15, -50, 25, 36, 23, 32, 49, 32, 44, 48, 13, 26, 43, 10, 41, 40, 31, 24, 73, -12

Offset: 0

Antti Karttunen, Nov 05 2014

sign

Antti Karttunen, Table of n, a(n) for n = 0..6561

Cf. A007318, A034931, A095143, A249723, A249732, A249733.

import re
from gmpy2 import digits
def A249731(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 (((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<>1)&~n).bit_count()<>1) # Chai Wah Wu, Jul 24 2025

(define (A249731 n) (- (A249732 n) (A249733 n)))

a(n) = A249732(n) - A249733(n).

cp's OEIS Frontend

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

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A249441 a(n) is the smallest prime whose square divides at least one entry in the n-th row of Pascal's triangle, or 0 if there is no such prime.

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A249695 a(n)=0, if A249441(n)=0; otherwise, a(n) is the smallest i such that A249441(n)^2 divides binomial(n,i).

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A249722 Numbers n such that there is a multiple of 4 on row n of Pascal's triangle with property that all multiples of 9 on the same row (if they exist) are larger than it.

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A249724 Numbers k such that on row k of Pascal's triangle there is no multiple of 9 which would be less than any (potential) multiple of 4 on the same row.

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A249726 Numbers n such that there is a multiple of 36 on row n of Pascal's triangle with property that it is also the least multiple of 4 and the least multiple of 9 on the same row.

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A249731 Number of multiples of 4 on row n of Pascal's triangle minus the number of multiples of 9 on the same row: a(n) = A249732(n) - A249733(n).

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