login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).
2

%I #23 Jul 14 2015 10:43:01

%S 1,7,13,23,33,47,61,79,97,119,141,167,193,223,253,287,321,359,397,439,

%T 481,527,573,623,673,727,781,839,897,959,1021,1087,1153,1223,1293,

%U 1367,1441,1519,1597,1679,1761,1847,1933,2023,2113,2207,2301,2399,2497,2599,2701

%N Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

%C Conjecture: from n>=2 onward, a(n) gives the positions of 2's in A227761.

%C a(29) = 897 = 3*13*23 is the first term which is neither prime nor semiprime, that is, has more than two prime divisors.

%H Antti Karttunen, <a href="/A227786/b227786.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F a(n) = A000290(n+1) - 2 - (n mod 2).

%F a(1)=1, and for n>1, a(n) = a(n-1)+A168276(n+1).

%F a(n) = (1/2) * (2*n^2 + 4*n -3 + (-1)^n) = 2*A116940(n-1) + 1. a(n-1) = 2*ceiling(n^2/2) - 3 = 2*A000985(n) - 3. G.f.: x*(-x^3 - x^2 + 5*x + 1)/((1-x)^3 * (1+x)). - _Ralf Stephan_, Aug 10 2013

%o (Scheme, two variants)

%o (definec (A227786 n) (if (< n 2) n (+ (A227786 (- n 1)) (A168276 (+ n 1)))))

%o (define (A227786v2 n) (- (A000290 (+ n 1)) 2 (modulo n 2)))

%Y Bisections: A082109, A073577. Cf. also A227761.

%K nonn

%O 1,2

%A _Antti Karttunen_, Jul 31 2013