

A217278


Sequences A124174 and A006454 interlaced, where A124174 are the Sophie Germain triangular numbers.


5



0, 0, 1, 3, 10, 15, 45, 120, 351, 528, 1540, 4095, 11935, 17955, 52326, 139128, 405450, 609960, 1777555, 4726275, 13773376, 20720703, 60384555, 160554240, 467889345, 703893960, 2051297326, 5454117903, 15894464365, 23911673955, 69683724540, 185279454480
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OFFSET

0,4


COMMENTS

35 =(a(n)  a(n12))/(a(n4)  a(n8)). It follows that this sequence is monotonic.
lim n > infinity a(2n)/a(2n  1) appears to approach (2cos^2(Pi/8))^2 = (3 + sqrt(8))/2 = (1+sqrt(2))^2/2 = (1/2)*(silver ratio)^2, where the silver ratio is associated with the limiting ratio at infinity between successive Pell numbers. For example, 15894464365/5454117903 = 2.914213562610..., accurate to 9 decimal places.
Alternative name: "Sophie Germain triangular numbers of the first kind" alternating with "Sophie Germain triangular numbers of the second kind"
a(2n) and 2*a(2n) + 1 are triangular [Sophie Germain triangular numbers of the first kind].
a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square) [Sophie Germain triangular numbers of the second kind].
Observation: In position n = 0, 1, 2, 3, 5, 11 of this sequence  where n + 1 maps 1:1 to the divisors (aka "subgroups") of 12  then a(n) is equal to the complete set of RamanujanNagell triangular numbers (A076046) with 0 repeated = {0, 0, 1, 3, 15, 4095}. Additionally, where pi(x) is the prime counting function and x = 35 = D(12), then pi(35) = 11, pi(pi(35)) = 5, pi(pi(pi(35))) = 3, pi(pi(pi(pi(35)))) = 2, pi(pi(pi(pi(pi(35))))) = 1, and pi(pi(pi(pi(pi(pi(35)))))) = 0.


LINKS

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,34,0,1,0,1).


FORMULA

a(n) = 35*(a(n4)  a(n8)) + a(n12).
From Raphie Frank, Dec 21 2015: (Start)
a(2*n + 1) = 1/64*(((4 + sqrt(2)) * (1  (1)^(n+1) * sqrt(2))^(2*floor((n+1)/2)) + (4  sqrt(2)) * (1+(1)^(n+1) * sqrt(2))^(2*floor((n+1)/2))))^2  1.
a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).
(End)


EXAMPLE

a(18) = 35*(52326  1540) + 45 = 1777555,
a(19) = 35*(139128  4095) + 120 = 4726275.


PROG

(PARI) concat([0, 0], Vec(x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x1)*(x+1)*(x^22*x1)*(x^2+2*x1)*(x^4+6*x^2+1)) + O(x^100))) \\ Colin Barker, Jun 23 2015


CROSSREFS

Cf. (sqrt(8a(2n) + 1)  1)/2 = A216134(n) = A216162(2n + 1).
Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).
Cf. (sqrt(8a(2n+1) + 1)  1)/2 = A006451(n).
Sequence in context: A020330 A023861 A037345 * A175336 A259877 A182334
Adjacent sequences: A217275 A217276 A217277 * A217279 A217280 A217281


KEYWORD

nonn,easy


AUTHOR

Raphie Frank, Sep 29 2012


STATUS

approved



