Original entry on oeis.org
3, 7, 11, 20, 28, 63, 72, 79, 688, 795, 999, 2716, 13220, 15940, 17903, 26832, 28416, 33448, 117923
Offset: 1
Cf.
A001590,
A100683,
A231574,
A231575,
A232542,
A214899,
A230607,
A020992,
A232498,
A214727,
A081172,
A214752,
A141523,
A214825,
A235862,
A214827,
A214828,
A214829,
A243622,
A243623.
-
a={1,8,8}; For[n=3, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[3]]=sum]
Original entry on oeis.org
0, 5, 14, 26, 33, 222, 234, 482, 937, 1170, 1290, 1877, 1897, 3413, 6017, 9365, 47470, 48254, 97421, 102057, 119689, 132418, 192517, 194442
Offset: 1
Cf.
A001590,
A100683,
A231574,
A231575,
A232542,
A214899,
A230607,
A020992,
A232498,
A214727,
A081172,
A214752,
A141523,
A214825,
A235862,
A214827-
A214829,
A243622,
A243623,
A214831,
A141036.
-
a246517 n = a246517_list !! (n-1)
a246517_list = filter ((== 1) . a010051'' . a141036) [0..]
-- Reinhard Zumkeller, Sep 15 2014
-
a={2,1,1}; Print[0]; For[n=3, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[3]]=sum]
A381804
Number of residues r mod n congruent to k such that rad(k) | n but rad(r) does not divide n, with rad = A007947.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 2, 0, 1, 4, 6, 0, 0, 0, 8, 0, 1, 0, 1, 0, 0, 3, 3, 7, 2, 0, 13, 0, 1, 0, 5, 0, 7, 7, 6, 0, 1, 0, 16, 14, 8, 0, 15, 3, 1, 15, 23, 0, 2, 0, 0, 5, 0, 5, 10, 0, 3, 9, 15, 0, 2, 0, 30, 20, 14, 10, 10, 0, 3, 0, 14, 0
Offset: 1
Let S(n) = row n of A381801 and R(n) = row n of A162306, with n in R(n) instead written as n mod n = 0.
Define quality Q between natural numbers k and n to be rad(k) does not divide n.
a(10) = 1 since S(10) = {0,1,2,4,5,6,8} only contains r = 6 with quality Q.
a(15) = 3 since S(15) = {0,1,3,5,6,9,10,12} and R(15) = {0,1,3,5,9} = {6,10,12}.
a(18) = 2 since S(18) = {0,1,2,3,4,6,8,9,10,12,14,16} and R(18) = {1,2,3,4,6,8,9,12,16,18} = {10,14}.
a(20) = 1 since S(20) = {0,1,2,4,5,8,10,12,16} and R(20) = {0,1,2,4,5,8,10,16} = {12}, etc.
-
f[x_] := Block[{c, ff, m, r, p, s, w},
c[_] := True; ff = FactorInteger[x][[All, 1]]; w = Length[ff];
s = {1};
Do[Set[p[i], ff[[i]]], {i, w}];
Do[Set[s, Union@ Flatten@ Join[s, #[[-1, 1]]]] &@ Reap@
Do[m = s[[j]];
While[Sow@ Set[r, Mod[m*p[i], x]];
c[r],
c[r] = False;
m *= p[i]],
{j, Length[s]}],
{i, w}]; s ];
rad[x_] := Times @@ FactorInteger[x][[All, 1]];
{0}~Join~Table[Length@ Complement[f[n], {0}~Join~Select[Range[n - 1], Divisible[#, rad[#]] &]], {n, 2, 120}]
Original entry on oeis.org
3, 4, 7, 8, 16, 26, 34, 42, 78, 94, 101, 107, 216, 255, 543, 562, 851, 981, 1099, 1528, 1824, 1955, 2122, 2488, 2500, 15331, 15961, 24107, 24938, 26051, 58504, 61617, 81034, 85119, 89768, 90597, 97191, 116899, 195346
Offset: 1
Cf.
A001590,
A100683,
A231574,
A231575,
A232542,
A214899,
A230607,
A020992,
A232498,
A214727,
A081172,
A214752,
A141523,
A214825,
A235862,
A214827,
A214828,
A214829,
A243622,
A243623,
A214831.
-
a={1,9,9}; For[n=3, n<=1000, n++, sum=Plus@@a; If[PrimeQ[sum], Print[n]]; a=RotateLeft[a]; a[[3]]=sum]
Flatten[Position[LinearRecurrence[{1,1,1},{1,9,9},200000],?PrimeQ]]-1 (* The prime number at index 195346 (term a(39) of this sequence) has 51699 digits, so this program takes a long time to run. *) (* _Harvey P. Dale, Dec 29 2019 *)
Showing 1-4 of 4 results.
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