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Numbers n such that there is a multiple of 9 on row n of Pascal's triangle with property that all multiples of 4 on the same row (if they exist) are larger than it.
+10
8
9, 10, 13, 15, 18, 19, 21, 27, 29, 31, 37, 39, 43, 45, 46, 47, 54, 55, 59, 63, 75, 79, 81, 82, 83, 85, 87, 90, 91, 93, 95, 99, 103, 109, 111, 117, 118, 119, 123, 126, 127, 135, 139, 151, 153, 154, 157, 159, 162, 163, 165, 167, 171, 175, 181, 183, 187, 189, 190, 191, 198, 199, 207, 219, 223, 225, 226, 229, 231, 234, 235, 237, 239, 243, 245, 247, 251, 253, 255
COMMENTS
All n such that 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 shorter than the distance from the edge to the nearest zero on row n of A034931 (Pascal's triangle reduced modulo 4), the latter distance taken to be infinite if there are no zeros on that row in the latter triangle. A052955 from its eight term onward, 31, 47, 63, 95, 127, ... seems to be a subsequence. See also the comments at A249441.
EXAMPLE
Row 13 of Pascal's triangle ( A007318) is: {1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1} and the term binomial(13, 5) = 1287 = 9*11*13 occurs before any term which is a multiple of 4. Note that one such term occurs right next to it, as binomial(13, 6) = 1716 = 4*3*11*13, but 1287 < 1716, thus 13 is included.
PROG
(PARI)
A249723list(upto_n) = { my(i=0, n=0); while(i<upto_n, for(k=0, n\2, if(!(binomial(n, k)%4), break, if(!(binomial(n, k)%9), i++; write("b249723.txt", i, " ", n); break))); n++); }
a(n)=0, if A249441(n)=0; otherwise, a(n) is the smallest i such that A249441(n)^2 divides binomial(n,i).
+10
7
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
COMMENTS
After a(0) = 0, A048278 gives the positions of seven other zeros in the sequence. - Antti Karttunen, Nov 04 2014
MAPLE
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:
MATHEMATICA
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]]]];
PROG
(PARI)
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.
for(n=0, 10000, write("b249695.txt", n, " ", A249695(n)));
CROSSREFS
Differs from A249442 for the first time at n=9.
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.
+10
5
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
COMMENTS
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.
PROG
(PARI)
A249724list(upto_n) = { my(i=0, n=0, dont_print=0); while(i<upto_n, for(k=0, n\2, if(!(binomial(n, k)%4), i++; write("b249724.txt", i, " ", n); dont_print=1; break, if(!(binomial(n, k)%9), dont_print=1; break))); if(!dont_print, i++; write("b249724.txt", i, " ", n), dont_print=0); n++); } \\ Ugly code.
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.
+10
5
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
COMMENTS
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.
PROG
(PARI)
A249726list(upto_n) = { my(i=0, n=0); while(i<upto_n, for(k=0, n\2, if(!(binomial(n, k)%36), i++; write("b249726.txt", i, " ", n); break, if((!(binomial(n, k)%4) || !(binomial(n, k)%9)), break))); n++); }
CROSSREFS
A044102 is a subsequence (after zero).
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