A214912 -id:A214912 - cp's OEIS frontend

A215029 To get a(n), start with m=n, let p = largest prime < m, set m = m-p if m>0, or m=m+p if m <= 0; repeat until p=2 has been processed; set a(n) = m.

0, 1, 2, 1, -1, 0, 0, 1, 1, 2, -1, 0, 0, 1, 1, 2, -1, 0, 0, 1, 1, 2, -1, 0, 0, 1, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, 0, 1, 0, 1, -1, 0, 1, 2, -1, 0, 0, 1, 1, 2, -1, 0, 1, 2, 0, 1, 0, 1, -1, 0, 0, 1, 0, 1, 0, 1, 1, 2, -1, 0, 0, 1, 0, 1, 0, 1, 1, 2, -1, 0, 1, 2, -1, 0, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, -1, 0, 0, 1, 0, 1, 1

Offset: 0

N. J. A. Sloane, Aug 05 2012

sign

Antti Karttunen, Table of n, a(n) for n = 0..20000
Antti Karttunen, Data supplement: n, a(n) computed for n = 0..100003
StackExchange, A prime number pattern, Jul 29 2012.

Cf. A215030-A215035, A214912.

f:=proc(n) local m,a,i,p;
if n <= 2 then RETURN(n); fi;
m:=n; a:=n;
for i from 1 to n do
p:=prevprime(m);
if a>0 then a:=a-p else a:=a+p; fi;
m:=p;
if m <= 2 then RETURN(a); fi;
od;
# should never reach here
print("ERROR");
end;
[seq(f(i),i=0..120)];

A215029(n) = if(n<=2,n,my(mp=precprime(n-1),d=n); while(mp>0, if(d>0, d -= mp, d += mp); mp = precprime(mp-1)); (d)); \\ Antti Karttunen, Nov 28 2018

A215030 a(n) = A215029(A000040(n)), where A000040(n) is the n-th prime.

Original entry on oeis.org

2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1, 0, 1, 2

Offset: 1

N. J. A. Sloane, Aug 05 2012

nonn

For primes whose value is 0, see A214912; for value 1 see A031215; and for value 2 see A215031.

Antti Karttunen, Table of n, a(n) for n = 1..10000
StackExchange, A prime number pattern, Jul 29 2012.

Cf. A000040, A215029, A214912, A031215, A215031, A215032, A215033, A215034, A215035.

A215029(n) = if(n<=2,n,my(mp=precprime(n-1),d=n); while(mp>0, if(d>0, d -= mp, d += mp); mp = precprime(mp-1)); (d));
A215030(n) = A215029(prime(n)); \\ Antti Karttunen, Nov 28 2018

A215031 Primes p such that A215029(p) = 2.

Original entry on oeis.org

2, 31, 41, 47, 67, 103, 127, 149, 179, 241, 269, 283, 313, 347, 353, 367, 389, 401, 431, 449, 461, 467, 487, 499, 523, 563, 587, 607, 617, 631, 653, 739, 751, 761, 773, 811, 829, 859, 883, 907, 919, 937, 967, 1019, 1039, 1063, 1093, 1129, 1187, 1201, 1217, 1237, 1279, 1297, 1327, 1409, 1427, 1433, 1447, 1483, 1499, 1523, 1543, 1553, 1567, 1579, 1607, 1613, 1637

Offset: 1

N. J. A. Sloane, Aug 05 2012

nonn

StackExchange, A prime number pattern, Jul 29 2012.

Cf. A215029-A215035, A214912.

A215036 2 followed by "1,0" repeated.

Original entry on oeis.org

2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

N. J. A. Sloane, Aug 06 2012

Take the first n primes and combine them with coefficients +1 and -1; then a(n) is the smallest number (in absolute value) that can be obtained.
For example, a(1) = 2, a(2) = 1 from 3-2 = 1; a(3) = 0 from -2-3+5 = 0; a(11) = 0 from 2-3-5-7+11-13+17+19-23-29+31 = 0.
Comment from Franklin T. Adams-Watters, Aug 05 2012: Sketch of proof that the above sum of primes results in this sequence. If S_n is the set of possible values of the signed sums for the first n primes, then S_{n+1} = S_n U (S_n + prime(n+1)) U (S_n - prime(n+1)). Beyond about n=4, this will be everything even or everything odd in an interval around zero, and then a fringe on either side; the size of the interval will be 2 * A007504(n) - k for some small k. Recursively, since prime(n) << A007504(n), this will continue to hold. Hence the sequence continues to alternate 0's and 1's.  A quite modest estimate on the distribution of primes suffices to complete the proof.
For number of solutions see A022894, A113040; also A083309.

StackExchange, A prime number pattern, Jul 29 2012.
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A007504, A214912, A215029, A215030; A022894, A083309, A113040.

Essentially the same as A135528, A059841, A000035.

PadRight[{2},120,{0,1}] (* Harvey P. Dale, Nov 29 2018 *)

if(n%2,2*(n<2),1) \\ Charles R Greathouse IV, Aug 06 2012

A215033 Numbers n such that A215029(n) = 0.

Original entry on oeis.org

0, 5, 6, 11, 12, 17, 18, 23, 24, 26, 28, 32, 34, 36, 39, 42, 45, 48, 50, 52, 55, 59, 60, 65, 68, 70, 73, 74, 76, 78, 83, 84, 86, 88, 93, 97, 98, 100, 104, 106, 109, 110, 112, 117, 121, 125, 128, 130, 133, 137, 138, 143, 147, 150, 153, 157, 158, 160, 162, 167, 168, 170, 172, 177, 180, 183, 187, 191, 192, 197, 198, 203, 207, 211, 212, 214, 216, 218, 220, 222, 227

Offset: 1

N. J. A. Sloane, Aug 05 2012

nonn

StackExchange, A prime number pattern, Jul 29 2012.

Cf. A215029-A215035. A214912 is a subsequence.

cp's OEIS Frontend

A215029 To get a(n), start with m=n, let p = largest prime < m, set m = m-p if m>0, or m=m+p if m <= 0; repeat until p=2 has been processed; set a(n) = m.

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A215030 a(n) = A215029(A000040(n)), where A000040(n) is the n-th prime.

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A215031 Primes p such that A215029(p) = 2.

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A215036 2 followed by "1,0" repeated.

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A215033 Numbers n such that A215029(n) = 0.

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