A048058 -id:A048058 - cp's OEIS frontend

A318126 a(n) is the number of pieces of the simplest continuous piecewise linear function that agrees with n mod k for all positive integer k.

1, 2, 3, 4, 5, 4, 5, 6, 7, 7, 7, 6, 7, 8, 9, 10, 10, 9, 10, 11, 11, 12, 11, 10, 11, 12, 13, 13, 14, 13, 13, 14, 15, 15, 14, 13, 14, 15, 16, 16, 17, 16, 16, 17, 17, 18, 17, 16, 16, 17, 19, 19, 18, 17, 17, 20, 19, 18, 17, 16, 17, 18, 19, 20, 21, 19, 20, 21, 22

Offset: 0

Luc Rousseau, Aug 18 2018

nonn

For a fixed n, the list of values (n mod k) can be modeled by a continuous piecewise linear function. Its simplest form consists of choosing the least possible number of intervals with integer endpoints. By definition a(n) is this number of intervals.

It appears that a(n) is asymptotically sqrt(8n) and that a(n) <= sqrt(8n) for all n >= 1.

			With n=5, the list of values of (n mod k), i.e., {0, 1, 2, 1, 0, 5, 5, 5, ...} is modeled by:
- {0, 1, 2} = k - 1 between k=1 and k=3,
- {2, 1, 0} = 5 - k between k=3 and k=5,
- {0, 5} = 5*k - 25 between k=5 and k=6,
- {5, 5, 5, ...} = 5 between k=6 and positive infinity.
Four intervals are involved, so a(5) = 4.

Luc Rousseau, Diagram illustrating a(11)=6 and a(24)=11.
Luc Rousseau, Plot of a(n) and sqrt(8*n) for n in 0..163
Wikipedia, Piecewise linear function

Cf. A048058 (the table of n mod k).

a[n_] := Module[{d = Differences[(Mod[n, #] &) /@ Range[n + 2]],
   r = 1, k},
  For[k = 2, k <= Length[d], k++, If[d[[k]] != d[[k - 1]], r++]];
  r]; a /@ Range[0, 68]

nxt(n,x)=my(y=floor(n/floor(n/x)));if(y==x,x+1,y)
a(n)=my(r=1,x=1,t=n,s=-1,xx,tt,ss);while(t,xx=nxt(n,x);tt=floor(n/xx);ss=(t*x-tt*xx)/(xx-x);if(ss!=s,r++);x=xx;t=tt;s=ss);r
for(n=0,68,print1(a(n), ", "))

A173757 Numbers k such that exactly one of k^2 + k + 1 and k^2 + k + 11 is prime.

Original entry on oeis.org

0, 4, 7, 9, 13, 14, 16, 18, 21, 23, 24, 25, 27, 28, 29, 30, 33, 34, 39, 45, 47, 50, 51, 52, 54, 56, 57, 58, 59, 60, 61, 66, 67, 68, 69, 71, 73, 77, 81, 83, 84, 85, 89, 90, 93, 94, 96, 99, 100, 103, 105, 106, 108, 110, 111, 113, 114, 117, 119, 122, 123, 124, 125, 127, 130

Offset: 1

Juri-Stepan Gerasimov, Feb 23 2010

nonn

Numbers k such that either k^2+k+1 or k^2+k+11 is prime, but not both. - R. J. Mathar, Mar 01 2010

			0 is in the sequence because 0^2+0+1 = 1 is nonprime and 0^2+0+11 = 11 is prime; 1 is not in the sequence because 1^2+1+1 = 3 is prime and 1^2+1+11 = 13 is also prime, 10 is not in the sequence because 10^2+10+1 = 111 is nonprime and 10^2+10+11 = 121 is also nonprime; 14 is in the sequence because 14^2+14+1 = 211 is prime and 14^2+14+11 = 221 is nonprime.

Cf. A048058.

[ n: n in [0..130] | IsPrime(k+1) ne IsPrime(k+11) where k is n^2+n ]; // Klaus Brockhaus, Feb 26 2010

Edited and extended by Klaus Brockhaus, Feb 26 2010

More terms from R. J. Mathar, Mar 01 2010

A275591 a(n) = n^2 + 9*n + 1.

Original entry on oeis.org

1, 11, 23, 37, 53, 71, 91, 113, 137, 163, 191, 221, 253, 287, 323, 361, 401, 443, 487, 533, 581, 631, 683, 737, 793, 851, 911, 973, 1037, 1103, 1171, 1241, 1313, 1387, 1463, 1541, 1621, 1703, 1787, 1873, 1961, 2051, 2143, 2237, 2333, 2431, 2531, 2633, 2737

Offset: 0

Miquel Cerda, Aug 02 2016

Also, nonnegative integers m such that 4*m + 77 is a square. The negative values of m are -7, -13, -17, -19.

The product of two consecutive terms belongs to the sequence. In fact: a(k)*a(k+1) = a(k*(k+1)+9*k+1).

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A028569.

Subsequence of A007775.

Table[n^2 + 9 n + 1, {n, 0, 50}] (* Bruno Berselli, Aug 05 2016 *)

a(n) = n^2 + 9*n + 1 \\ Charles R Greathouse IV, Aug 03 2016

Vec((1+8*x-7*x^2)/(1-x)^3 + O(x^100)) \\ Colin Barker, Aug 04 2016

O.g.f.: (1 + 8*x - 7*x^2)/(1 - x)^3. - Colin Barker, Aug 03 2016
E.g.f.: (1 + 10*x + x^2)*exp(x).
a(n) = a(-n-9) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Colin Barker, Aug 03 2016
a(n) = A048058(n-1) + A008592(n-1) for n>0.
a(n) = 1 + A028569(n). - Omar E. Pol, Aug 02 2016
a(n) + a(-n) = (n-1)^2 + (n+1)^2.
Sum_{i>=0} 1/a(i) = 9736/29393 + tan(sqrt(77)*Pi/2)*Pi/sqrt(77) = 1.301517...

Edited and extended by Bruno Berselli, Aug 05 2016

A382210 Irregular triangle read by rows: T(n,k) = k^2 - k + (A003173(n) + 1)/4 with 1 <= k < (A003173(n) + 1)/4.

Original entry on oeis.org

2, 3, 5, 5, 7, 11, 17, 11, 13, 17, 23, 31, 41, 53, 67, 83, 101, 17, 19, 23, 29, 37, 47, 59, 73, 89, 107, 127, 149, 173, 199, 227, 257, 41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 173, 197, 223, 251, 281, 313, 347, 383, 421, 461, 503, 547, 593, 641, 691, 743, 797, 853, 911, 971, 1033, 1097, 1163, 1231, 1301, 1373, 1447, 1523, 1601

Offset: 4

Stefano Spezia, Mar 18 2025

			The irregular triangle begins as:
   2;
   3,  5;
   5,  7, 11, 17;
  11, 13, 17, 23, 31, 41, 53, 67, 83, 101;
  17, 19, 23, 29, 37, 47, 59, 73, 89, 107, 127, 149, 173, 199, 227, 257;
  ...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 225.

Cf. A003173, A048058, A302445.

Heegner:={1, 2, 3, 7, 11, 19, 43, 67, 163};A003173[n_]:=Part[Heegner,n]; T[n_,k_]:=k^2-k+(A003173[n]+1)/4;Table[T[n,k],{n,4,9},{k,(A003173[n]+1)/4-1}]//Flatten

cp's OEIS Frontend

A318126 a(n) is the number of pieces of the simplest continuous piecewise linear function that agrees with n mod k for all positive integer k.

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A173757 Numbers k such that exactly one of k^2 + k + 1 and k^2 + k + 11 is prime.

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A275591 a(n) = n^2 + 9*n + 1.

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A382210 Irregular triangle read by rows: T(n,k) = k^2 - k + (A003173(n) + 1)/4 with 1 <= k < (A003173(n) + 1)/4.

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