A079008
a(n) is smallest number k such that the n successive values of phi(k+j) (j=0,..,n-1) are all distinct.
Original entry on oeis.org
1, 2, 5, 11, 11, 17, 17, 37, 46, 46, 112, 112, 123, 149, 149, 149, 257, 257, 257, 257, 257, 257, 257, 261, 658, 658, 685, 741, 741, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 1359, 4097, 4097, 4097, 4097, 4097
Offset: 1
a(8)=37, values of phi(k) for k=37,..,44 are {36, 18, 24, 16, 40, 12, 42, 20}.
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kul[x_] := Length[x]-Length[Union[x]]; frt[x_] := Table[EulerPhi[x+j], {j, 0, h-1}]; Table[fa=1; k=0; Do[s=frt[n]; s1=kul[s]; If[Equal[s1, 0]&&Equal[fa, 1], k=k+1; Print[{h, n, s}]; fa=0], {n, 1, 10000}], {h, 1, 50}]
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a(n) = if(n==1, 1, my(v=vector(n, i, eulerphi(i))); for(k=n, oo, if(#Set(v)==n, return(k-n)); v[k%n+1]=eulerphi(k))); \\ Jinyuan Wang, Feb 10 2021
A079009
Least k such that the 2^n successive values of phi(k+j) (j=0..2^n-1) are all distinct.
Original entry on oeis.org
1, 2, 11, 37, 149, 1359, 14130, 175327, 1218073, 108387730, 14305141265
Offset: 0
For n = 7: a(7) = 175327 because phi(175327+j), for j=0..127 are all distinct: {175326, 87648, ..., 175452, 85320}.
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isdist(v) = forstep(i = #v, 1, -1, forstep(j = i - 1, 1, -1, if(v[i] == v[j], return(j)))); 0;
a(n) = {my(m = 2^n, phis = vector(m, i, eulerphi(i)), k = m, ind = isdist(phis)); while(ind != 0, phis = concat(vecextract(phis, Str("^1.."ind)), vector(ind, i, eulerphi(k+i))); k += ind; ind = isdist(phis)); k - m + 1;} \\ Amiram Eldar, Aug 27 2024
A323743
Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.
Original entry on oeis.org
1, 3, 4, 5, 5, 7, 8, 9, 8, 9, 11, 12, 13, 14, 15, 10, 13, 15, 17, 18, 19, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 20, 22, 24, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39
Offset: 1
There is only one number with exactly 1 divisor (namely, k=1), but there are infinitely many numbers with j divisors for every j >= 2, so row 1 consists only of the single term 1.
The sequence of values tau(k) for k >= 1 is A000005, which begins 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ..., from which the sums of two consecutive terms are 1+2=3, 2+2=4, 2+3=5, 3+2=5, 2+4=6, 4+2=6, 2+4=6, 4+3=7, 3+4=7, ...; no number j < 3 appears as such a sum, every j >= 6 appears infinitely many times as such a sum, and each j in {3,4,5} appears as such a sum only finitely many times, so row 2 is {3, 4, 5}.
Row 3 does not contain 6 as a term because there exists no run of 3 consecutive numbers whose sum of tau values is exactly 6.
The first six rows of the table are as follows:
row 1: {1};
row 2: {3, 4, 5};
row 3: {5, 7, 8, 9};
row 4: {8, 9, 11, 12, 13, 14, 15};
row 5: {10, 13, 15, 17, 18, 19};
row 6: {14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27}.
Cf.
A000005,
A005237,
A006558,
A048892,
A072507,
A100366,
A119479,
A141621,
A284596,
A284597,
A292580,
A319037,
A319045,
A319046.
Showing 1-3 of 3 results.
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