A057539 Birthday set of order 7, i.e., numbers congruent to +- 1 modulo 2, 3, 4, 5, 6 and 7.

1, 29, 41, 71, 139, 169, 181, 209, 211, 239, 251, 281, 349, 379, 391, 419, 421, 449, 461, 491, 559, 589, 601, 629, 631, 659, 671, 701, 769, 799, 811, 839, 841, 869, 881, 911, 979, 1009, 1021, 1049, 1051, 1079, 1091, 1121, 1189, 1219, 1231, 1259, 1261, 1289

Offset: 1

Author

Andrew R. Feist (andrewf(AT)math.duke.edu), Sep 06 2000

Keywords

• easy
• nonn

Comments

Integers of the form sqrt(840*k+1) for k >= 0. - Boyd Blundell, Jul 10 2021

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• A. Feist, On the Density of Birthday Sets, The Pentagon, 60 (No. 1, Fall 2000), 31-35.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,1,-1).

Crossrefs

Cf. A057538, A057540, A057541.

Programs

Mathematica

LinearRecurrence[{1,0,0,0,0,0,0,1,-1},{1,29,41,71,139,169,181,209,211},50] (* Harvey P. Dale, Sep 24 2014 *)

PARI

is_A057539(n,m=[2,3,4,5,6,7])=!for(i=1,#m,abs((n+1)%m[i]-1)==1||return)

PARI

is(n)=for(i=4,7,if(abs(centerlift(Mod(n,i)))!=1, return(0))); 1 \\ Charles R Greathouse IV, Oct 20 2014

Python

def ok(n): return all(n%d in [1, d-1] for d in range(2, 8))
def aupto(nn): return [m for m in range(1, nn+1) if ok(m)]
print(aupto(1300)) # Michael S. Branicky, Jan 29 2021

Formula

G.f.: x*(1 + 28*x + 12*x^2 + 30*x^3 + 68*x^4 + 30*x^5 + 12*x^6 + 28*x^7 + x^8) / ((1+x)*(x^2+1)*(x^4+1)*(x-1)^2). - R. J. Mathar, Oct 08 2011

a(n) = a(n-8) + 210 = a(n-1) + a(n-8) - a(n-9). - Charles R Greathouse IV, Oct 20 2014

a(n) = 105n/4 + O(1). - Charles R Greathouse IV, Oct 20 2014

Extensions

Offset corrected to 1 by Ray Chandler, Jul 29 2019